[R] t.test in a loop
Pushpike J Thilakarathne
pushpike at med.kuleuven.be
Thu Jan 29 14:09:07 CET 2009
Hi, can u send a sample data set then I can setup it out.
Pushpike.
________________________________________________________________
O. Pushpike J. Thilakarathne,
L-BioStat, Catholic University of Leuven.
U.Z. Sint-Rafaël (2nd Floor)
Kapucijnenvoer 35,
B-3000 Leuven,
BELGIUM.
Tel : + 32 (0) 16 33 68 87
Fax: + 32 (0) 16 33 70 15
URL [L-BioStat] : http://med.kuleuven.be/biostat/
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Today's Topics:
1. t.test in a loop (Michael Pearmain)
2. Re: evaluation revisited (Wacek Kusnierczyk)
3. Re: evaluation revisited (Wacek Kusnierczyk)
4. Re: t.test in a loop (Thomas Lumley)
5. Re: for/if loop (Zhou Fang)
6. putting match.call to good use (Harald Eikrem)
7. Re: for/if loop (jim holtman)
8. Re: evaluation revisited (Gabor Grothendieck)
9. Re: evaluation revisited (Wacek Kusnierczyk)
10. Character SNP data to binary MAF data (Hadassa Brunschwig)
11. initial value in 'vmmin' is not finite (June Wong)
12. putting match.call to good use (Harald Eikrem)
13. Re: Mystery Error in midnightStandard (Ted Byers)
14. Re: initial value in 'vmmin' is not finite (Prof Brian Ripley)
15. plot slideshow (diego Diego)
16. Re: Re : Need help on running Heckman Correction Estimation
using R (Arne Henningsen)
17. StepAIC with coxph (Michele Santacatterina)
18. Merge two vectors into one. (patricia garc?a gonz?lez)
19. Re: putting match.call to good use (Prof Brian Ripley)
20. Re: Merge two vectors into one. (G?bor Cs?rdi)
21. Re: Merge two vectors into one. (Dimitris Rizopoulos)
22. Re: Merge two vectors into one. (patricia garc?a gonz?lez)
23. Re: plot slideshow (stephen sefick)
24. Re: Mystery Error in midnightStandard (Yohan Chalabi)
25. Re: putting match.call to good use (Peter Dalgaard)
26. Re: plot slideshow (David Winsemius)
27. Re: putting match.call to good use (Dieter Menne)
28. Re: OT: Adding verbatim R code text into LaTeX documents:
texttt; verb or url? (JLucke at ria.buffalo.edu)
29. Re: putting match.call to good use (Prof Brian Ripley)
30. Grouping problem (venkata kirankumar)
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32. Newbie question about "grouping" (Rixon, John C.)
33. Logical subset of the columns in a dataframe (Mark Na)
34. Re: Grouping problem (David Winsemius)
35. Re: Grouping problem (hadley wickham)
36. Re: Newbie question about "grouping" (David Winsemius)
37. Re: Newbie question about "grouping" (Thomas Lumley)
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39. Re: Logical subset of the columns in a dataframe
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40. Re: Mystery Error in midnightStandard (Ted Byers)
41. Re: Logical subset of the columns in a dataframe (David Winsemius)
42. Re: Mystery Error in midnightStandard (Yohan Chalabi)
43. constrainOptim (June Wong)
44. Re: constrainOptim (Ravi Varadhan)
45. repeated measures design for GAM? (Strubbe Diederik)
46. Repeated measures design for GAM? - corrected question...
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55. Re: constrainOptim (Ben Bolker)
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57. Get median of each column (Frank Zhang)
58. R compilation (Attiglah, Mama)
59. Re: Get median of each column (jim holtman)
60. Re: Saving plot into file without showing it (jim holtman)
61. Re: Get median of each column (Stephan Kolassa)
62. Re: Saving plot into file without showing it (Stephan Kolassa)
63. Re: Repeated measures design for GAM? - corrected question...
(Strubbe Diederik)
64. Re: Get median of each column (Rolf Turner)
65. Re: Power analysis for MANOVA? (Stephan Kolassa)
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74. Re: Newbie Question About Histograms (Peter Alspach)
75. Changing histogram stack in qplot (Jason Rupert)
76. Re: Neighborhood distance calculator (Antonio, Fabio Di Narzo)
77. Re: Text data (jim holtman)
78. Re: Help with normal distribution in random samples...
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79. Re: Text data (Nutter, Benjamin)
80. Re: Help with normal distribution in random samples...
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81. Re: Changing histogram stack in qplot (hadley wickham)
82. Re: Memory issue? (Ubuntu Diego)
83. Re: Newbie Question About Histograms (pfc_ivan)
84. Re: rproxy.dll (Cl?ment D)
85. Re: for/if loop (SnowManPaddington)
86. Re: Newbie Question About Histograms (Eik Vettorazzi)
87. Re: Using R in a web application (Gad Abraham)
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90. Re: [SPAM] - Re: for/if loop - Bayesian Filter detected spam
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91. questions about histogram (Wenxia Li)
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98. Re: using Sweave with a master file that has several iputted
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99. standard error of logit parameters (Bomee Park)
100. Ignore text when reading data (beyar)
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102. Re: Ignore text when reading data (Remko Duursma)
103. Re: Ignore text when reading data (jim holtman)
104. Re: Ignore text when reading data (beyar)
105. Question about collapse/aggregate and avoidance of loops
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106. Re : standard error of logit parameters (justin bem)
107. Re: t.test in a loop (Petr PIKAL)
108. Re: Re : standard error of logit parameters
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109. Odp: Question about collapse/aggregate and avoidance of
loops (Petr PIKAL)
110. Odp: stack data sets (Petr PIKAL)
111. Re: Character SNP data to binary MAF data (Hadassa Brunschwig)
112. Re: Odp: Question about collapse/aggregate and avoidance of
loops (Bernd Weiss)
113. Re: Character SNP data to binary MAF data (Barry Rowlingson)
114. Re: Character SNP data to binary MAF data (Thomas Lumley)
115. Re: Character SNP data to binary MAF data (Patrick Aboyoun)
116. Re: Question about collapse/aggregate and avoidance of loops
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117. Re: Character SNP data to binary MAF data (Barry Rowlingson)
118. Multiple tables (Gerit Offermann)
119. Text in a character vector to indicate "ifelse" argument
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120. Re: help with plot layout (Jim Lemon)
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----------------------------------------------------------------------
Message: 1
Date: Wed, 28 Jan 2009 11:25:22 +0000
From: Michael Pearmain <mpearmain at google.com>
Subject: [R] t.test in a loop
To: r-help at r-project.org
Message-ID:
<2763e000901280325y4c8801dctc02a0aa48a9d45b6 at mail.gmail.com>
Content-Type: text/plain
Hi All,
I've been having a little trouble with creating a loop that will run a a
series of t.tests for inspection,
Below is the code i've tried, and some checks i've looked at.
I've used the get(paste()) idea as i was told previously that the use of the
eval should try and be avoided.
I've run a single syntax to check that my systax is correct and works
without any problems
> t.test(channel.data.train$News~channel.data.train$power)
Can anyone offer any advice?
Many thanks
Mike
> str(channel.data.train$power)
num [1:9913] 0 0 0 0 0 0 0 0 0 0 ...
> summary(channel.data.train$power)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0000 0.0000 0.0000 0.2368 0.0000 1.0000
> names(channel.data.train)
[1] "News" "Entertainment" "Communicate"
[4] "Lifestyle" "Games" "Music"
[7] "Money" "Celebrity" "Shopping"
[10] "Sport" "Film" "Travel"
[13] "Cars" "Property" "Chat"
[16] "Bet.Play.Win" "config" "exposed"
[19] "site" "referrer" "started"
[22] "last_viewed" "num_views" "secs_since_viewed"
[25] "register" "secs.na" "power"
[28] "tt"
> for(i in names(channel.data.train[,c(1:16)])){
+
t.test(get(paste("channel.data.train$",i,"~channel.data.train$power",sep="")))
+ }
Error in get(paste("channel.data.train$", i, "~channel.data.train$power",
:
variable "channel.data.train$News~channel.data.train$power" was not found
--
Michael Pearmain
Senior Analytics Research Specialist
Google UK Ltd
Belgrave House
76 Buckingham Palace Road
London SW1W 9TQ
United Kingdom
t +44 (0) 2032191684
mpearmain at google.com
If you received this communication by mistake, please don't forward it to
anyone else (it may contain confidential or privileged information), please
erase all copies of it, including all attachments, and please let the sender
know it went to the wrong person. Thanks.
[[alternative HTML version deleted]]
------------------------------
Message: 2
Date: Wed, 28 Jan 2009 12:26:53 +0100
From: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no>
Subject: Re: [R] evaluation revisited
To: Gabor Grothendieck <ggrothendieck at gmail.com>
Cc: r-help at r-project.org
Message-ID: <498040FD.1010506 at idi.ntnu.no>
Content-Type: text/plain; charset=ISO-8859-1
Gabor Grothendieck wrote:
> The argument to eval.parent is evaluated before eval.parent
> ever sees it.
really? eval.parent is just a regular r function, a wrapper for eval
with envir=parent.frame(). the arguments to eval.parent are passed to
eval *unevaluated* (as promises), and are only evaluated when eval needs
them. here's a modified eval.parent:
my.eval.parent = function(expr, n=1) {
print('foo')
p = parent.frame(n+1)
eval(expr, p) }
my.eval.parent({print(1); 2})
# prints 'foo' before printing 1 and returning 2
> Try issuing this command before you run your
> code:
>
> debug(eval.parent)
>
> and look at the value of the arguments as passed to eval.parent
> in the debugger.
>
well, when you are in the debugger and look at the value of the
arguments you actually force the promises, so no wonder you see them
evaluated. if you don't look at them, they're not evaluated:
trace(eval)
trace(parent.frame)
eval.parent({print(1);2})
# calling parent.frame
# calling eval
# printing 1 (after parent.frame and eval have been called)
# returning 2
vQ
------------------------------
Message: 3
Date: Wed, 28 Jan 2009 12:35:03 +0100
From: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no>
Subject: Re: [R] evaluation revisited
To: Gabor Grothendieck <ggrothendieck at gmail.com>
Cc: R help <R-help at stat.math.ethz.ch>
Message-ID: <498042E7.9050600 at idi.ntnu.no>
Content-Type: text/plain; charset=ISO-8859-1
Wacek Kusnierczyk wrote:
> Gabor Grothendieck wrote:
>
>> The argument to eval.parent is evaluated before eval.parent
>> ever sees it.
>>
>
> really? eval.parent is just a regular r function, a wrapper for eval
> with envir=parent.frame(). the arguments to eval.parent are passed to
> eval *unevaluated* (as promises), and are only evaluated when eval needs
> them.
to be strict, the argument n to eval.parent is not further passed to
eval, and is evaluated before eval is called. the above referred to the
'expr' argument to eval.parent. one more example:
my.eval.parent = function(expr, n=1) {
print('foo')
p = parent.frame(n+1)
eval(expr, p) }
trace(eval)
my.eval.parent({print('expr'); 1}, {print('n'); 1})
# "foo"
# "n"
# trace eval(expr, p)
# "expr"
# 1
vQ
------------------------------
Message: 4
Date: Wed, 28 Jan 2009 03:57:55 -0800 (PST)
From: Thomas Lumley <tlumley at u.washington.edu>
Subject: Re: [R] t.test in a loop
To: Michael Pearmain <mpearmain at google.com>
Cc: r-help at r-project.org
Message-ID:
<Pine.LNX.4.43.0901280357550.12037 at hymn14.u.washington.edu>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 28 Jan 2009, Michael Pearmain wrote:
> Hi All,
> I've been having a little trouble with creating a loop that will run a a
> series of t.tests for inspection,
> Below is the code i've tried, and some checks i've looked at.
>
> I've used the get(paste()) idea as i was told previously that the use of the
> eval should try and be avoided.
>
> I've run a single syntax to check that my systax is correct and works
> without any problems
>> t.test(channel.data.train$News~channel.data.train$power)
>
> Can anyone offer any advice?
There's the additional problem that if your code worked it would do 16 t-tests but only report the last one.
Assuming you want them printed
for(v in names(channel.data.train)[1:16]) {
print(v)
print(t.test(channel.data.train[[v]]~channel.data.train$power)
}
or
for(v in names(channel.data.train)[1:16]){
test <- eval(bquote(.(v)~power, data=channel.data.train)
print(eval(test))
}
This sort of use of eval is fairly harmless.
-thomas
> Many thanks
>
> Mike
>
>> str(channel.data.train$power)
> num [1:9913] 0 0 0 0 0 0 0 0 0 0 ...
>> summary(channel.data.train$power)
> Min. 1st Qu. Median Mean 3rd Qu. Max.
> 0.0000 0.0000 0.0000 0.2368 0.0000 1.0000
>> names(channel.data.train)
> [1] "News" "Entertainment" "Communicate"
> [4] "Lifestyle" "Games" "Music"
> [7] "Money" "Celebrity" "Shopping"
> [10] "Sport" "Film" "Travel"
> [13] "Cars" "Property" "Chat"
> [16] "Bet.Play.Win" "config" "exposed"
> [19] "site" "referrer" "started"
> [22] "last_viewed" "num_views" "secs_since_viewed"
> [25] "register" "secs.na" "power"
> [28] "tt"
>> for(i in names(channel.data.train[,c(1:16)])){
> +
> t.test(get(paste("channel.data.train$",i,"~channel.data.train$power",sep="")))
> + }
> Error in get(paste("channel.data.train$", i, "~channel.data.train$power",
> :
> variable "channel.data.train$News~channel.data.train$power" was not found
>
>
>
> --
> Michael Pearmain
> Senior Analytics Research Specialist
>
>
> Google UK Ltd
> Belgrave House
> 76 Buckingham Palace Road
> London SW1W 9TQ
> United Kingdom
> t +44 (0) 2032191684
> mpearmain at google.com
>
> If you received this communication by mistake, please don't forward it to
> anyone else (it may contain confidential or privileged information), please
> erase all copies of it, including all attachments, and please let the sender
> know it went to the wrong person. Thanks.
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Thomas Lumley Assoc. Professor, Biostatistics
tlumley at u.washington.edu University of Washington, Seattle
------------------------------
Message: 5
Date: Wed, 28 Jan 2009 12:07:40 +0000
From: Zhou Fang <zhou.zfang at gmail.com>
Subject: Re: [R] for/if loop
To: r-help at r-project.org
Message-ID: <49804A8C.1030807 at gmail.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
What are you trying to do with
> for (pp in 1:pp+1){
?
Also, note that 1:rr+1 and 1:(rr+1) mean different things.
Zhou
------------------------------
Message: 6
Date: Wed, 28 Jan 2009 13:34:55 +0100
From: "Harald Eikrem" <heikrem at c2i.net>
Subject: [R] putting match.call to good use
To: r-help at r-project.org
Message-ID: <web-124267531 at mailbe01.swip.net>
Content-Type: text/plain; charset="iso-8859-15"
------------------------------
Message: 7
Date: Wed, 28 Jan 2009 07:36:52 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] for/if loop
To: SnowManPaddington <wiwiana at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<644e1f320901280436w24c9f77chc9aeae47aeff5f4a at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Within the loops you are changing the loop variables (pp & rr). Why
are you doing this? THis might be causing your problem of what sounds
like an infinite loop. You probably want to rethink what you are
trying to do in the loop.
On Wed, Jan 28, 2009 at 3:21 AM, SnowManPaddington <wiwiana at gmail.com> wrote:
>
> Hi, it's my first time to write a loop with R for my homework. This loop is
> part of the function. I wanna assign values for hll according to panel
> [ii,1]=pp. I didn't get any error message in this part. but then when I
> further calculate another stuff with hll, the function can't return. I think
> it must be some problem in my loop. Probably something stupid or easy. But I
> tried to look for previous posts in forum and read R language help. But none
> can help.. Thanks!
>
>
>
> for (ii in 1:100){
> for (pp in 1:pp+1){
> for (rr in 1:rr+1){
> if (panel[ii,1]!=pp)
> {
> hll(pp,1)=ColSums(lselb1(rr:ii-1,1))
> hll(pp,2)=ColSums(lselb2(rr:ii-1,1))
> rr=ii
> pp=pp+1
> }
> else
> {
> hll(pp,1)=ColSums(lselb1(rr:ii,1))
> hll(pp,2)=ColSums(lselb2(rr:ii,1))
> rr=ii
> pp=pp+1}
> }
> }}}
>
>
> in fact I have the corresponding Gauss code here. But I really don't know
> how to write such loop in R.
>
> rr=1;
> ii=1;
> pp=1;
> do until ii==n+1;
> if pan[ii,1] ne pp;
> hll[pp,1]=sumc(lselb1[rr:ii-1,1]);
> hll[pp,2]=sumc(lselb2[rr:ii-1,1]);
> rr=ii;
> pp=pp+1;
> endif;
> if ii==n;
> hll[pp,1]=sumc(lselb1[rr:ii,1]);
> hll[pp,2]=sumc(lselb2[rr:ii,1]);
> rr=ii;
> pp=pp+1;
> endif;
> ii=ii+1;
> endo;
>
> --
> View this message in context: http://www.nabble.com/for-if-loop-tp21701496p21701496.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 8
Date: Wed, 28 Jan 2009 08:04:03 -0500
From: Gabor Grothendieck <ggrothendieck at gmail.com>
Subject: Re: [R] evaluation revisited
To: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no>
Cc: r-help at r-project.org
Message-ID:
<971536df0901280504pf93398fjbbc2153eed8f3264 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
On Wed, Jan 28, 2009 at 6:26 AM, Wacek Kusnierczyk
<Waclaw.Marcin.Kusnierczyk at idi.ntnu.no> wrote:
> Gabor Grothendieck wrote:
>> The argument to eval.parent is evaluated before eval.parent
>> ever sees it.
>
> really? eval.parent is just a regular r function, a wrapper for eval
> with envir=parent.frame(). the arguments to eval.parent are passed to
> eval *unevaluated* (as promises), and are only evaluated when eval needs
> them. here's a modified eval.parent:
Yes, you're right about the mechanism although quoting the
help page its nevertheless true that it:
"evaluates its first argument in the current scope before
passing it to the evaluator"
------------------------------
Message: 9
Date: Wed, 28 Jan 2009 14:29:02 +0100
From: Wacek Kusnierczyk <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no>
Subject: Re: [R] evaluation revisited
To: Gabor Grothendieck <ggrothendieck at gmail.com>
Cc: r-help at r-project.org
Message-ID: <49805D9E.9000901 at idi.ntnu.no>
Content-Type: text/plain; charset=ISO-8859-1
Gabor Grothendieck wrote:
> On Wed, Jan 28, 2009 at 6:26 AM, Wacek Kusnierczyk
> <Waclaw.Marcin.Kusnierczyk at idi.ntnu.no> wrote:
>
>> Gabor Grothendieck wrote:
>>
>>> The argument to eval.parent is evaluated before eval.parent
>>> ever sees it.
>>>
>> really? eval.parent is just a regular r function, a wrapper for eval
>> with envir=parent.frame(). the arguments to eval.parent are passed to
>> eval *unevaluated* (as promises), and are only evaluated when eval needs
>> them. here's a modified eval.parent:
>>
>
> Yes, you're right about the mechanism although quoting the
> help page its nevertheless true that it:
> "evaluates its first argument in the current scope before
> passing it to the evaluator"
>
... where 'current scope' is as clear as the sky over trondheim right
now [1], the issue being:
- is 'current scope' the scope in which eval (the above quote refers to
eval) is called (as it seems to be meant), or
- the scope *within* the call to eval (which would be intuitively
obvious, since when eval 'evaluates' it must have already been entered
and not yet left, so we're inside the eval-call scope).
another example of how quoting an r help page helps provided you already
know the answer.
must admit that 'eval evaluates its argument before passing it to the
evaluator' is quite funny a quote; so eval is able to evaluate without
an evaluator? magic!
*what* is it that is true, quoting the help page?
vQ
[1]
http://www.yr.no/place/Norway/S%C3%B8r-Tr%C3%B8ndelag/Trondheim/Trondheim/
------------------------------
Message: 10
Date: Wed, 28 Jan 2009 15:36:03 +0200
From: Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>
Subject: [R] Character SNP data to binary MAF data
To: r-help at r-project.org
Message-ID:
<db80b30d0901280536v610e265w105cd60a133b03ef at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Hi
I am sure there is a function out there already but I couldn't find it.
I have SNP data, that is, a matrix which contains in each row two
characters (they are different in each row) and I would like to
convert this matrix to a binary one according to the minor allele
frequency. For non-geneticists: I want to have a binary matrix
for which in each row the 0 stands for the less frequent character
and 1 for the more frequent character.
Thanks for any suggestions.
Hadassa
--
Hadassa Brunschwig
PhD Student
Department of Statistics
The Hebrew University of Jerusalem
http://www.stat.huji.ac.il
------------------------------
Message: 11
Date: Wed, 28 Jan 2009 14:15:20 +0000
From: June Wong <neptune545 at hotmail.com>
Subject: [R] initial value in 'vmmin' is not finite
To: <r-help at r-project.org>
Message-ID: <BAY136-W36F4FAD45683269D9C3DBC8AC80 at phx.gbl>
Content-Type: text/plain
Dear r helpers
I run the following code for nested logit and got a message that
Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
What does this mean? and how can I correct it?
Thank you
June
> yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)> dim(yogurt)[1] 12784 25> choice = yogurt[,2:5]> price=yogurt[,14:17]> feature=yogurt[,6:9]> n = nrow(yogurt)> constant = rep(1,each=n)> yop=cbind(constant,feature[,1],price[,1])> dan=cbind(constant,feature[,2],price[,2])> hil=cbind(constant,feature[,3],price[,3])> wt=cbind(feature[,4],price[,4])> > fr <- function(x) { + x1 = x[1]+ x2 = x[2]+ x3 = x[3]+ x4 = x[4]+ x5 = x[5]+ x6 = x[6]+ x7 = x[7]+ con1 = rbind(x[1],x[5],x[6])+ con2 = rbind(x[2],x[5],x[6])+ con3 = rbind(x[3],x[5],x[6])+ con4 = rbind(x[5],x[6])+ rho=exp(x[7])/(1+exp(x[7]))+ ey = exp((yop%*%con1)/rho)+ ed = exp((dan%*%con2)/rho)+ eh = exp((hil%*%con3)/rho)+ ew = exp((wt%*%con4)/rho)+ ev = ey+ed+eh+ew+ den=(ey+ed+eh+ew)+ iv = rho*log(den)+ pp=exp(x[4]+iv)/(1+exp(x[4]+iv))+ pr1 =pp* ey/den+ pr2 =pp* ed/den+ pr3 =pp* eh/den+ pr4 =pp* ew/den+ pnp=1/(1+exp(x[4]+iv))+ likelihood = (pnp*yogurt[,1])+(pr1*yogurt[,2])+(pr2*yogurt[,3])+(pr3*yogurt[,!
4])+(pr4*yogurt[,4])+ lsum = log(likelihood)+ return(-colSums(lsum))+ }> p = optim(c(0,0,0,0,0.1,-2,-0.2),fr, hessian = TRUE, method = "BFGS")Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
_________________________________________________________________
[[alternative HTML version deleted]]
------------------------------
Message: 12
Date: Wed, 28 Jan 2009 15:29:44 +0100
From: Harald Eikrem <heikrem at c2i.net>
Subject: [R] putting match.call to good use
To: r-help at r-project.org
Message-ID: <49806BD8.1090804 at c2i.net>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
( I just became aware the mailer enforces html bodies, as such removed
by the list handler. Sorry about that. My message was )
I have this function
slm <- function(fun=lm, ...) {
#ilm <- eval(match.call()[-1]); # no way
ilm <- eval(parse(text=sub("^list", deparse(substitute(fun)),
deparse(substitute(...())))));
...
The latter actually does the trick, but recognising how some gurus hate
parse, I would like to know if this can anyhow be done with match.call,
or any other reasonable solution.
The issue here is that lm (and likewise glm, bayesglm, etc.) returns the
function call, which needs to show up as the original args to slm of course.
~~harald e
------------------------------
Message: 13
Date: Wed, 28 Jan 2009 09:30:58 -0500
From: Ted Byers <r.ted.byers at gmail.com>
Subject: Re: [R] Mystery Error in midnightStandard
To: Yohan Chalabi <chalabi at phys.ethz.ch>
Cc: R-help Forum <r-help at r-project.org>
Message-ID:
<4f1819890901280630l24b48a9ah66273c6425620fad at mail.gmail.com>
Content-Type: text/plain
Hi Yohan, Thanks.
On Wed, Jan 28, 2009 at 4:57 AM, Yohan Chalabi <chalabi at phys.ethz.ch> wrote:
> >>>> "TB" == Ted Byers <r.ted.byers at gmail.com>
> >>>> on Tue, 27 Jan 2009 16:00:27 -0500
>
> TB> I wasn't even aware I was using midnightStandard. You won't
> TB> find it in my
> TB> script.
> TB>
> TB> Here is the relevant loop:
> TB>
> TB> date1 = timeDate(charvec = Sys.Date(), format = %Y-%m-%d)
> TB> date1
> TB> dow = 3;
> TB> for (i in 1:length(V4) ) {
> TB> x = read.csv(as.character(V4[[i]]), header = FALSE,
> TB> na.strings=);
> TB> y = x[,1];
> TB> year = V2[[i]];
> TB> week = V3[[i]];
> TB> dtstr = sprintf(%i-%i-%i,year,week,dow);
> TB> date2 = timeDate(dtstr, format = %Y-%U-%w);
> TB> resultsdataframe[[i]] <- difftimeDate(date1,date2,units =
> TB> weeks);
> TB> fp = fitdistr(y,exponential);
> TB> print(c(V1[[i]],V2[[i]],V3[[i]],fp,fp));
> TB> print(c(year,week,date2,resultsdataframe[[i]]));
> TB> resultsdataframe[[i]] <- fp;
> TB> resultsdataframe[[i]] <- fp;
> TB> }
> TB>
> TB> It fails with a little more than 100 records left in V4.
> TB>
> TB> The full error message is:
> TB>
> TB> Error in midnightStandard(charvec, format) :
> TB> 'charvec' has non-NA entries of different number of characters
>
> timeDate() uses the midnight standard. The function 'midnightStandard'
> assumes that all entries in 'charvec' have the same 'format'. Can you
> please check if this is the case?
>
It is certain that all entries have the same format, but I'm starting to
think that the error message is something of a red herring. Consider this:
> year = 2009
> week = 0
> day = 3
> datestr = sprintf("%i-%i-%i",year,week,day);datestr
[1] "2009-0-3"
> date1 = timeDate(datestr, format = "%Y-%U-%w");
> date1
GMT
[1] [NA]
> day = 4
> datestr = sprintf("%i-%i-%i",year,week,day);datestr
[1] "2009-0-4"
> date1 = timeDate(datestr, format = "%Y-%U-%w");
> date1
GMT
[1] [2009-01-01]
>
> datestr = sprintf("%i-%i-%i",year,week,3);datestr
[1] "2009-0-3"
> date2 = timeDate(datestr, format = "%Y-%U-%w");date2
GMT
[1] [NA]
> difftimeDate(date2,date1, units = "weeks")
Error in midnightStandard(charvec, format) :
'charvec' has non-NA entries of different number of characters
In addition: Warning messages:
1: In min(x) : no non-missing arguments to min; returning Inf
2: In max(x) : no non-missing arguments to max; returning -Inf
The first values for year, week and day are the values on which my loop
dies. It returns 'NA' here. It seems clear that it is returning NA because
the date that data corresponds to is 2008-12-31.
The error is being produced by difftimeDate rather than timeDate (as shown
by the above session). But that represents a flaw in the function design.
It should fail when taking the elapsed time between a null and the present,
but if I wrote such a function, I'd have it return null (perhaps with a
warning) rather than just die.
A bigger issue is that timeDate ought never give null here (which is what I
assume 'NA' means), since all the data comes from transaction data with real
dates, so the elapsed time, measured in weeks, ought to always be a valid
real number that is positive semidefinite. I have not yet come to any
conclusions as to how it ought to behave (whether to return new years day,
along with a warning, or to return the date requested by reinvoking itself
with the year and week adjusted so a valid date is returned).
On a practical side, how would I test date2 to see if it is null, so I can
give it a sensible default value?
A more troubling thought is that with this handling of dates in this
combination of SQL (my group by clause uses
YEAR(transaction_date),WEEK(transaction_date)) to get the data and R to
process it, the week containing new years day will ALWAYS be split in two at
the first second of the new year. I'm going to have to either figure out a
way to correct this, or ignore it (as it doesn't actually make things wrong,
but rather it splits a sample into two unequal parts).
Thoughts?
Thanks
Ted
[[alternative HTML version deleted]]
------------------------------
Message: 14
Date: Wed, 28 Jan 2009 14:35:11 +0000 (GMT)
From: Prof Brian Ripley <ripley at stats.ox.ac.uk>
Subject: Re: [R] initial value in 'vmmin' is not finite
To: June Wong <neptune545 at hotmail.com>
Cc: r-help at r-project.org
Message-ID: <alpine.OSX.1.00.0901281432210.97827 at tystie.local>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 28 Jan 2009, June Wong wrote:
>
> Dear r helpers
>
> I run the following code for nested logit and got a message that
>
> Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
> What does this mean? and how can I correct it?
It means that your function at your starting values is evaluating to a
non-finite value (+/-Inf, NA, NaN).
Your example is unreadable, and we don't have the file so cannot help
you debug this.
>
> Thank you
> June
>
>> yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)> dim(yogurt)[1] 12784 25> choice = yogurt[,2:5]> price=yogurt[,14:17]> feature=yogurt[,6:9]> n = nrow(yogurt)> constant = rep(1,each=n)> yop=cbind(constant,feature[,1],price[,1])> dan=cbind(constant,feature[,2],price[,2])> hil=cbind(constant,feature[,3],price[,3])> wt=cbind(feature[,4],price[,4])> > fr <- function(x) { + x1 = x[1]+ x2 = x[2]+ x3 = x[3]+ x4 = x[4]+ x5 = x[5]+ x6 = x[6]+ x7 = x[7]+ con1 = rbind(x[1],x[5],x[6])+ con2 = rbind(x[2],x[5],x[6])+ con3 = rbind(x[3],x[5],x[6])+ con4 = rbind(x[5],x[6])+ rho=exp(x[7])/(1+exp(x[7]))+ ey = exp((yop%*%con1)/rho)+ ed = exp((dan%*%con2)/rho)+ eh = exp((hil%*%con3)/rho)+ ew = exp((wt%*%con4)/rho)+ ev = ey+ed+eh+ew+ den=(ey+ed+eh+ew)+ iv = rho*log(den)+ pp=exp(x[4]+iv)/(1+exp(x[4]+iv))+ pr1 =pp* ey/den+ pr2 =pp* ed/den+ pr3 =pp* eh/den+ pr4 =pp* ew/den+ pnp=1/(1+exp(x[4]+iv))+ likelihood = (pnp*yogurt[,1])+(pr1*yogurt[,2])+(pr2*yogurt[,3])+(pr3*yogurt[!
,!
> 4])+(pr4*yogurt[,4])+ lsum = log(likelihood)+ return(-colSums(lsum))+ }> p = optim(c(0,0,0,0,0.1,-2,-0.2),fr, hessian = TRUE, method = "BFGS")Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
>
> _________________________________________________________________
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
------------------------------
Message: 15
Date: Wed, 28 Jan 2009 11:44:31 -0300
From: diego Diego <dhabbyc at gmail.com>
Subject: [R] plot slideshow
To: r-help at r-project.org
Message-ID:
<e76ec2580901280644q24a8fe9cm2fdd3b919996f413 at mail.gmail.com>
Content-Type: text/plain
Dear R experts:
I've seen that it's possible to make a sort of "slideshow" with several
R-plots (each slide is activated by a click on the mouse). How can I put
this on a R-script???
Regards.
D.
[[alternative HTML version deleted]]
------------------------------
Message: 16
Date: Wed, 28 Jan 2009 09:48:06 -0500
From: Arne Henningsen <arne.henningsen at googlemail.com>
Subject: Re: [R] Re : Need help on running Heckman Correction
Estimation using R
To: r-help at r-project.org, justin bem <justin_bem at yahoo.fr>, Kishore
<gladikishore at gmail.com>
Cc: Ott Toomet <ott.toomet at ut.ee>
Message-ID: <200901280948.06591.arne.henningsen at googlemail.com>
Content-Type: text/plain; charset="iso-8859-15"
Hi Kishore and Justin,
The sample selection stuff has been separated from the micEcon package about
one year ago. It is available in the sampleSelection package [1,2,3] now. The
sample selection package is thoroughly described in a (freely available) paper
published in the Journal of Statistical Software [4]. We recommend using the
"selection" command rather than the "heckit" command, because the former can
be used to estimate the model not only by the two-step method but also by ML.
[1] http://www.sampleselection.org/
[2] http://r-forge.r-project.org/projects/sampleselection/
[3] http://cran.r-project.org/web/packages/sampleSelection/index.html
[4] http://www.jstatsoft.org/v27/i07
Best wishes,
Arne
On Tuesday 27 January 2009 06:02:46, justin bem wrote:
> See the micEcon package. there is and heckit function
> ??Justin BEM
> BP 1917 Yaound??
> T??l (237) 99597295
> (237) 22040246
>
>
>
>
> ________________________________
> De : Kishore <gladikishore at gmail.com>
> ?? : r-help at r-project.org; r-help at stat.math.ethz.ch
> Envoy?? le : Mardi, 27 Janvier 2009, 11h54mn 00s
> Objet??: [R] Need help on running Heckman Correction Estimation using R
>
> Team,
>
> I am trying to resolve the self-selection bias of a sample in an experiment
> and would like to run the Heckman Correction Estimation using R.?? Can
> someone help me with the R-Code... I tried searching for the discussion,
> but not successful. Thanks in advance,
>
> Best,
>
> Kishore/..
> http://kaykayatisb.blogspot.com
>
> ?????? [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html and provide commented, minimal,
> self-contained, reproducible code.
>
>
>
>
> [[alternative HTML version deleted]]
--
Arne Henningsen
http://www.arne-henningsen.name/
------------------------------
Message: 17
Date: Wed, 28 Jan 2009 15:50:11 +0100
From: Michele Santacatterina <miksanta at gmail.com>
Subject: [R] StepAIC with coxph
To: R-help at r-project.org
Message-ID:
<1f0555cf0901280650q4dae0443mca01c02e6d4c0582 at mail.gmail.com>
Content-Type: text/plain
Hi,
i'm trying to apply StepAIC with coxph...but i have the same error:
stepAIC(fitBMT)
Start: AIC=327.77
Surv(TEMPO,morto==1) ˜ VOD + SESSO + ETA + ........
Error in dropterm.default(fit,scope$drop, scale=scale,trace=max(0, :
number of rows in use has changed: remove missing values?
anybody know this error??
Thanks.
Michele
[[alternative HTML version deleted]]
------------------------------
Message: 18
Date: Wed, 28 Jan 2009 14:54:19 +0000
From: patricia garc?a gonz?lez <kurtney_84 at hotmail.com>
Subject: [R] Merge two vectors into one.
To: <r-help at r-project.org>
Message-ID: <BAY106-W17030B9E71F42E0796554CF6C80 at phx.gbl>
Content-Type: text/plain
Hi all,
I have two vectors like this:
x <- c( "Y", "H", NA, NA )
y <- c( NA, "H", NA, "B" )
And would like to make one vector with the common elements, and the element available only in one of the vectors.
res <- c( "Y", "H", NA, "B" )
Thanks,
Patricia
> From: neptune545 at hotmail.com
> To: r-help at r-project.org
> Date: Wed, 28 Jan 2009 14:15:20 +0000
> Subject: [R] initial value in 'vmmin' is not finite
>
>
> Dear r helpers
>
> I run the following code for nested logit and got a message that
>
> Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
> What does this mean? and how can I correct it?
>
> Thank you
> June
>
> > yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)> dim(yogurt)[1] 12784 25> choice = yogurt[,2:5]> price=yogurt[,14:17]> feature=yogurt[,6:9]> n = nrow(yogurt)> constant = rep(1,each=n)> yop=cbind(constant,feature[,1],price[,1])> dan=cbind(constant,feature[,2],price[,2])> hil=cbind(constant,feature[,3],price[,3])> wt=cbind(feature[,4],price[,4])> > fr <- function(x) { + x1 = x[1]+ x2 = x[2]+ x3 = x[3]+ x4 = x[4]+ x5 = x[5]+ x6 = x[6]+ x7 = x[7]+ con1 = rbind(x[1],x[5],x[6])+ con2 = rbind(x[2],x[5],x[6])+ con3 = rbind(x[3],x[5],x[6])+ con4 = rbind(x[5],x[6])+ rho=exp(x[7])/(1+exp(x[7]))+ ey = exp((yop%*%con1)/rho)+ ed = exp((dan%*%con2)/rho)+ eh = exp((hil%*%con3)/rho)+ ew = exp((wt%*%con4)/rho)+ ev = ey+ed+eh+ew+ den=(ey+ed+eh+ew)+ iv = rho*log(den)+ pp=exp(x[4]+iv)/(1+exp(x[4]+iv))+ pr1 =pp* ey/den+ pr2 =pp* ed/den+ pr3 =pp* eh/den+ pr4 =pp* ew/den+ pnp=1/(1+exp(x[4]+iv))+ likelihood = (pnp*yogurt[,1])+(pr1*yogurt[,2])+(pr2*yogurt[,3])+(pr3*yogurt!
[,!
> 4])+(pr4*yogurt[,4])+ lsum = log(likelihood)+ return(-colSums(lsum))+ }> p = optim(c(0,0,0,0,0.1,-2,-0.2),fr, hessian = TRUE, method = "BFGS")Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
>
> _________________________________________________________________
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
_________________________________________________________________
[[alternative HTML version deleted]]
------------------------------
Message: 19
Date: Wed, 28 Jan 2009 15:00:42 +0000 (GMT)
From: Prof Brian Ripley <ripley at stats.ox.ac.uk>
Subject: Re: [R] putting match.call to good use
To: Harald Eikrem <heikrem at c2i.net>
Cc: r-help at r-project.org
Message-ID: <alpine.OSX.1.00.0901281452150.97868 at tystie.local>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 28 Jan 2009, Harald Eikrem wrote:
> ( I just became aware the mailer enforces html bodies, as such removed by the
> list handler. Sorry about that. My message was )
>
> I have this function
>
> slm <- function(fun=lm, ...) {
> #ilm <- eval(match.call()[-1]); # no way
> ilm <- eval(parse(text=sub("^list", deparse(substitute(fun)),
> deparse(substitute(...())))));
> ...
>
> The latter actually does the trick, but recognising how some gurus hate
> parse, I would like to know if this can anyhow be done with match.call, or
> any other reasonable solution.
>
> The issue here is that lm (and likewise glm, bayesglm, etc.) returns the
> function call, which needs to show up as the original args to slm of course.
The way to do this is eval(substitute()). E.g. from the new Rd2HTML
Rd <- eval(substitute(parse_Rd(f, encoding = enc),
list(f = Rd,enc = encoding)))
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
------------------------------
Message: 20
Date: Wed, 28 Jan 2009 16:01:11 +0100
From: G?bor Cs?rdi <csardi at rmki.kfki.hu>
Subject: Re: [R] Merge two vectors into one.
To: patricia garc?a gonz?lez <kurtney_84 at hotmail.com>
Cc: r-help at r-project.org
Message-ID:
<d70c15d40901280701t7ab41b20rf14de15ef41105e8 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Is position important? The vectors always have the same length? They
always have the same entry if both are not NA?
If yes, yes and yes, then
res <- ifelse( is.na(x), y, x)
does what you want. Otherwise please explain better what you want.
Gabor
On Wed, Jan 28, 2009 at 3:54 PM, patricia garc?a gonz?lez
<kurtney_84 at hotmail.com> wrote:
>
> Hi all,
>
> I have two vectors like this:
>
>
> x <- c( "Y", "H", NA, NA )
>
> y <- c( NA, "H", NA, "B" )
>
> And would like to make one vector with the common elements, and the element available only in one of the vectors.
>
>
> res <- c( "Y", "H", NA, "B" )
>
>
> Thanks,
>
> Patricia
>
--
Gabor Csardi <Gabor.Csardi at unil.ch> UNIL DGM
------------------------------
Message: 21
Date: Wed, 28 Jan 2009 16:01:08 +0100
From: Dimitris Rizopoulos <d.rizopoulos at erasmusmc.nl>
Subject: Re: [R] Merge two vectors into one.
To: patricia garc?a gonz?lez <kurtney_84 at hotmail.com>
Cc: r-help at r-project.org
Message-ID: <49807334.2070501 at erasmusmc.nl>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
you could start by something like the following:
x <- c("Y", "H", NA, NA)
y <- c(NA, "H", NA, "B")
ifelse(is.na(x), y, x)
I hope it helps.
Best,
Dimitris
patricia garc?a gonz?lez wrote:
> Hi all,
>
> I have two vectors like this:
>
>
> x <- c( "Y", "H", NA, NA )
>
> y <- c( NA, "H", NA, "B" )
>
> And would like to make one vector with the common elements, and the element available only in one of the vectors.
>
>
> res <- c( "Y", "H", NA, "B" )
>
>
> Thanks,
>
> Patricia
>
>
>
>> From: neptune545 at hotmail.com
>> To: r-help at r-project.org
>> Date: Wed, 28 Jan 2009 14:15:20 +0000
>> Subject: [R] initial value in 'vmmin' is not finite
>>
>>
>> Dear r helpers
>>
>> I run the following code for nested logit and got a message that
>>
>> Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
>> What does this mean? and how can I correct it?
>>
>> Thank you
>> June
>>
>>> yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)> dim(yogurt)[1] 12784 25> choice = yogurt[,2:5]> price=yogurt[,14:17]> feature=yogurt[,6:9]> n = nrow(yogurt)> constant = rep(1,each=n)> yop=cbind(constant,feature[,1],price[,1])> dan=cbind(constant,feature[,2],price[,2])> hil=cbind(constant,feature[,3],price[,3])> wt=cbind(feature[,4],price[,4])> > fr <- function(x) { + x1 = x[1]+ x2 = x[2]+ x3 = x[3]+ x4 = x[4]+ x5 = x[5]+ x6 = x[6]+ x7 = x[7]+ con1 = rbind(x[1],x[5],x[6])+ con2 = rbind(x[2],x[5],x[6])+ con3 = rbind(x[3],x[5],x[6])+ con4 = rbind(x[5],x[6])+ rho=exp(x[7])/(1+exp(x[7]))+ ey = exp((yop%*%con1)/rho)+ ed = exp((dan%*%con2)/rho)+ eh = exp((hil%*%con3)/rho)+ ew = exp((wt%*%con4)/rho)+ ev = ey+ed+eh+ew+ den=(ey+ed+eh+ew)+ iv = rho*log(den)+ pp=exp(x[4]+iv)/(1+exp(x[4]+iv))+ pr1 =pp* ey/den+ pr2 =pp* ed/den+ pr3 =pp* eh/den+ pr4 =pp* ew/den+ pnp=1/(1+exp(x[4]+iv))+ likelihood = (pnp*yogurt[,1])+(pr1*yogurt[,2])+(pr2*yogurt[,3])+(pr3*yogurt
!
> [,!
>> 4])+(pr4*yogurt[,4])+ lsum = log(likelihood)+ return(-colSums(lsum))+ }> p = optim(c(0,0,0,0,0.1,-2,-0.2),fr, hessian = TRUE, method = "BFGS")Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
>>
>> _________________________________________________________________
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> _________________________________________________________________
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
------------------------------
Message: 22
Date: Wed, 28 Jan 2009 15:04:14 +0000
From: patricia garc?a gonz?lez <kurtney_84 at hotmail.com>
Subject: Re: [R] Merge two vectors into one.
To: <csardi at rmki.kfki.hu>
Cc: r-help at r-project.org
Message-ID: <BAY106-W11D32EFEDD34EB887498B4F6C80 at phx.gbl>
Content-Type: text/plain
Hi,
Sorry, the answers are yes yes yes. And thank you for your idea it works perfectly.
Regards
Patricia
> Date: Wed, 28 Jan 2009 16:01:11 +0100
> Subject: Re: [R] Merge two vectors into one.
> From: csardi at rmki.kfki.hu
> To: kurtney_84 at hotmail.com
> CC: r-help at r-project.org
>
> Is position important? The vectors always have the same length? They
> always have the same entry if both are not NA?
>
> If yes, yes and yes, then
>
> res <- ifelse( is.na(x), y, x)
>
> does what you want. Otherwise please explain better what you want.
>
> Gabor
>
> On Wed, Jan 28, 2009 at 3:54 PM, patricia garcía gonzález
> <kurtney_84 at hotmail.com> wrote:
> >
> > Hi all,
> >
> > I have two vectors like this:
> >
> >
> > x <- c( "Y", "H", NA, NA )
> >
> > y <- c( NA, "H", NA, "B" )
> >
> > And would like to make one vector with the common elements, and the element available only in one of the vectors.
> >
> >
> > res <- c( "Y", "H", NA, "B" )
> >
> >
> > Thanks,
> >
> > Patricia
> >
>
> --
> Gabor Csardi <Gabor.Csardi at unil.ch> UNIL DGM
_________________________________________________________________
[[alternative HTML version deleted]]
------------------------------
Message: 23
Date: Wed, 28 Jan 2009 10:10:40 -0500
From: stephen sefick <ssefick at gmail.com>
Subject: Re: [R] plot slideshow
To: diego Diego <dhabbyc at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<c502a9e10901280710t40ddac82n137de355fe620613 at mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
I know you probably want to do this in R, but you could do this in
power point or the openoffice variant rather easily.
Stephen
On Wed, Jan 28, 2009 at 9:44 AM, diego Diego <dhabbyc at gmail.com> wrote:
> Dear R experts:
> I've seen that it's possible to make a sort of "slideshow" with several
> R-plots (each slide is activated by a click on the mouse). How can I put
> this on a R-script???
>
>
> Regards.
>
> D.
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Stephen Sefick
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
------------------------------
Message: 24
Date: Wed, 28 Jan 2009 16:28:28 +0100
From: Yohan Chalabi <chalabi at phys.ethz.ch>
Subject: Re: [R] Mystery Error in midnightStandard
To: Ted Byers <r.ted.byers at gmail.com>
Cc: R-help Forum <r-help at r-project.org>
Message-ID: <20090128162828.36d5648f at mimi>
Content-Type: text/plain; charset=US-ASCII
>>>> "TB" == Ted Byers <r.ted.byers at gmail.com>
>>>> on Wed, 28 Jan 2009 09:30:58 -0500
TB> It is certain that all entries have the same format, but I'm
TB> starting to
TB> think that the error message is something of a red herring.
TB> Consider this:
TB>
TB> > year = 2009
TB> > week = 0
TB> > day = 3
TB> > datestr = sprintf(%i-%i-%i,year,week,day);datestr
TB> [1] 2009-0-3
TB> > date1 = timeDate(datestr, format = %Y-%U-%w);
TB> > date1
TB> GMT
TB> [1] [NA]
TB> > day = 4
TB> > datestr = sprintf(%i-%i-%i,year,week,day);datestr
TB> [1] 2009-0-4
TB> > date1 = timeDate(datestr, format = %Y-%U-%w);
TB> > date1
TB> GMT
TB> [1] [2009-01-01]
TB> >
TB> > datestr = sprintf(%i-%i-%i,year,week,3);datestr
TB> [1] 2009-0-3
TB> > date2 = timeDate(datestr, format = %Y-%U-%w);date2
TB> GMT
TB> [1] [NA]
TB> > difftimeDate(date2,date1, units = weeks)
TB> Error in midnightStandard(charvec, format) :
TB> 'charvec' has non-NA entries of different number of characters
TB> In addition: Warning messages:
TB> 1: In min(x) : no non-missing arguments to min; returning Inf
TB> 2: In max(x) : no non-missing arguments to max; returning -Inf
TB>
TB>
TB>
TB> The first values for year, week and day are the values on
TB> which my loop
TB> dies. It returns 'NA' here. It seems clear that it is
TB> returning NA because
TB> the date that data corresponds to is 2008-12-31.
TB>
TB> The error is being produced by difftimeDate rather than timeDate
TB> (as shown
TB> by the above session). But that represents a flaw in the
TB> function design.
This is not a flaw in timeDate. it behaves the same way as
'as.POSIXct'
strptime(datestr, format = "%Y-%U-%w")
Instead of claiming that there is a flaw in the function you could have
suggested an 'is.na' method for 'timeDate'.
I will add an 'is.na' method in the dev version of 'timeDate'.
regards,
Yohan
TB> It should fail when taking the elapsed time between a null
TB> and the present,
TB> but if I wrote such a function, I'd have it return null
TB> (perhaps with a
TB> warning) rather than just die.
TB>
TB> A bigger issue is that timeDate ought never give null here
TB> (which is what I
TB> assume 'NA' means), since all the data comes from transaction
TB> data with real
TB> dates, so the elapsed time, measured in weeks, ought to always
TB> be a valid
TB> real number that is positive semidefinite. I have not yet
TB> come to any
TB> conclusions as to how it ought to behave (whether to return
TB> new years day,
TB> along with a warning, or to return the date requested by
TB> reinvoking itself
TB> with the year and week adjusted so a valid date is returned).
TB>
TB> On a practical side, how would I test date2 to see if it is
TB> null, so I can
TB> give it a sensible default value?
TB>
TB> A more troubling thought is that with this handling of dates
TB> in this
TB> combination of SQL (my group by clause uses
TB> YEAR(transaction_date),WEEK(transaction_date)) to get the data
TB> and R to
TB> process it, the week containing new years day will ALWAYS be
TB> split in two at
TB> the first second of the new year. I'm going to have to either
TB> figure out a
TB> way to correct this, or ignore it (as it doesn't actually make
TB> things wrong,
TB> but rather it splits a sample into two unequal parts).
--
PhD student
Swiss Federal Institute of Technology
Zurich
www.ethz.ch
------------------------------
Message: 25
Date: Wed, 28 Jan 2009 16:29:37 +0100
From: Peter Dalgaard <P.Dalgaard at biostat.ku.dk>
Subject: Re: [R] putting match.call to good use
To: Prof Brian Ripley <ripley at stats.ox.ac.uk>
Cc: r-help at r-project.org, Harald Eikrem <heikrem at c2i.net>
Message-ID: <498079E1.4020109 at biostat.ku.dk>
Content-Type: text/plain; charset=UTF-8
Prof Brian Ripley wrote:
> On Wed, 28 Jan 2009, Harald Eikrem wrote:
>
>> ( I just became aware the mailer enforces html bodies, as such removed
>> by the list handler. Sorry about that. My message was )
>>
>> I have this function
>>
>> slm <- function(fun=lm, ...) {
>> #ilm <- eval(match.call()[-1]); # no way
>> ilm <- eval(parse(text=sub("^list", deparse(substitute(fun)),
>> deparse(substitute(...())))));
>> ...
>>
>> The latter actually does the trick, but recognising how some gurus
>> hate parse, I would like to know if this can anyhow be done with
>> match.call, or any other reasonable solution.
>>
>> The issue here is that lm (and likewise glm, bayesglm, etc.) returns
>> the function call, which needs to show up as the original args to slm
>> of course.
>
> The way to do this is eval(substitute()). E.g. from the new Rd2HTML
>
> Rd <- eval(substitute(parse_Rd(f, encoding = enc),
> list(f = Rd,enc = encoding)))
>
I don't understand the
substitute(...())
bit (looks like an unexpected feature), but I suspect that it might also
be a good idea to read and understand the first dozen lines or so of the
lm function itself.
--
O__ ---- Peter Dalgaard ?ster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
------------------------------
Message: 26
Date: Wed, 28 Jan 2009 10:30:59 -0500
From: David Winsemius <dwinsemius at comcast.net>
Subject: Re: [R] plot slideshow
To: diego Diego <dhabbyc at gmail.com>
Cc: r-help at r-project.org
Message-ID: <68773229-5537-40A1-AFB9-2F956556487B at comcast.net>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
If you investigate how the call:
demo(graphics)
... works, you find that the first interactive event is handled by the
code at the end of the demo function, Just type:
demo
The rest of the interactive events are handled by this single line at
the beginning of the graphics.R code that creates an implicit loop:
oask <- devAskNewPage(dev.interactive(orNone = TRUE))
You could have found this by looking at the Writing R Extensions
documentation and then noting that demos are placed in demo
subdirectories of the packages. Going to a package that you knew
contained a working demo, in this cases the graphics package, you
would find a graphics.R demo script.
--
David Winsemius
On Jan 28, 2009, at 9:44 AM, diego Diego wrote:
> Dear R experts:
> I've seen that it's possible to make a sort of "slideshow" with
> several
> R-plots (each slide is activated by a click on the mouse). How can I
> put
> this on a R-script???
>
>
> Regards.
>
> D.
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 27
Date: Wed, 28 Jan 2009 15:34:47 +0000 (UTC)
From: Dieter Menne <dieter.menne at menne-biomed.de>
Subject: Re: [R] putting match.call to good use
To: r-help at stat.math.ethz.ch
Message-ID: <loom.20090128T153337-162 at post.gmane.org>
Content-Type: text/plain; charset=us-ascii
Prof Brian Ripley <ripley <at> stats.ox.ac.uk> writes:
> The way to do this is eval(substitute()). E.g. from the new Rd2HTML
>
What is Rd2HTML?
Dieter
------------------------------
Message: 28
Date: Wed, 28 Jan 2009 10:43:16 -0500
From: JLucke at ria.buffalo.edu
Subject: Re: [R] OT: Adding verbatim R code text into LaTeX documents:
texttt; verb or url?
To: "Peter Dunn" <PDunn2 at usc.edu.au>
Cc: R Help <r-help at r-project.org>, r-help-bounces at r-project.org
Message-ID:
<OFDE86EA96.6227F93B-ON8525754C.0055A1EF-8525754C.005676E8 at ria.buffalo.edu>
Content-Type: text/plain
LaTeX offers a verbatim environment.
\begin{verbatim}
This is maintained verbatim, Latex commands and environments are typeset
as written without any processing.
\end{verbatim}
Be sure to use the package verbatim.
---Joe
"Peter Dunn" <PDunn2 at usc.edu.au>
Sent by: r-help-bounces at r-project.org
01/28/2009 01:41 AM
To
"R Help" <r-help at r-project.org>
cc
Subject
[R] OT: Adding verbatim R code text into LaTeX documents: texttt; verb or
url?
Hi all
I use Sweave extensively to mix R and LaTeX, and often have R code
appearing in my LaTeX document.
Just a quick question then: What is the best way to add example of R
commands into LaTeX in-line? (That is, not using Sweave.) For example,
suppose I wish to place in my document this instruction:
...is done in R using the command \verb|lm( y ~ var.one + var.two )| as
follows:
I used \verb above, but I see three options: \verb, \url (package url),
or \texttt; there are probably others.
Here are my comments on these three:
- Using \texttt is OK, but it disappears my tildes and can hyphenate
- Using \verb is good, but it can hyphenate.
- Using \url is very good, but it:
* disappears my spaces; so for the above example, the spaces added for
clarity are gone.
* Minor: I like my verbatim text a little smaller (\small size), and
change the font size for verbatim using
\def\verbatim at font{\small\ttfamily} but \url seems to ignore this and
appears larger than if I used \text or \verb.
Also, using \url often adds line-breaks mid-variable at the dots (for
example, splitting var.one to have "var." on one line, and "one" on the
next). I'm not sure this is a problem or not; here it is just an
observation.
Ideally, one would want a LaTeX function, say \rcode{}, that displayed
in-text using non-proportional font, kept tildes, kept spacing, uses my
verb-font changes, and broke at sensible places for R. (I don't want
much, do I?)
So two questions:
* What do other people do? Maybe there is a solution I have over-looked.
* Is there an easy solution? I suppose writing such a command in LaTeX is
possible, but there is strong evidence to reject the hypothesis that I
would be able to write one. Maybe one of the above choices are easily
adopted.
If no easy solutions exist or emerge, I'm happy to run with \url.
Thanks again.
P.
Peter Dunn
Biostatistician
School of Health and Sport Science
Faculty of Science, Health and Education
University of the Sunshine Coast
Tel: +61 7 5456 5085
Fax: +61 7 5430 2896
Email: pdunn2 at usc.edu.au
www.usc.edu.au
CRICOS Provider Number: 01595D
This communication is intended for the recipient only and should not be
forwarded, distributed or otherwise read by others without express
permission. The views expressed in this email are not necessarily those of
the University of the Sunshine Coast.
--
------------------------------
Message: 29
Date: Wed, 28 Jan 2009 15:44:10 +0000 (GMT)
From: Prof Brian Ripley <ripley at stats.ox.ac.uk>
Subject: Re: [R] putting match.call to good use
To: Dieter Menne <dieter.menne at menne-biomed.de>
Cc: r-help at stat.math.ethz.ch
Message-ID: <alpine.OSX.1.00.0901281540450.97868 at tystie.local>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 28 Jan 2009, Dieter Menne wrote:
> Prof Brian Ripley <ripley <at> stats.ox.ac.uk> writes:
>
>> The way to do this is eval(substitute()). E.g. from the new Rd2HTML
>>
>
> What is Rd2HTML?
A function in the R-devel version of R (is 'new' not rather a hint?).
>From the NEWS file:
o parse_Rd(), an experimental parser for Rd files, and Rd2txt(),
Rd2HTML(), Rd2latex() and Rd2ex(), even more experimental
converters, have been added to package 'tools'.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
------------------------------
Message: 30
Date: Wed, 28 Jan 2009 21:17:35 +0530
From: venkata kirankumar <kiran4u2all at gmail.com>
Subject: [R] Grouping problem
To: r-help at r-project.org
Message-ID:
<27f678620901280747x1374203di84e0710fb5b7b9a3 at mail.gmail.com>
Content-Type: text/plain
Hi all,
I have a problem with grouping like I have to give count of employes in each
department like
if in one company there is departments like
Mechanical, Computer, Fitting, electronics and Chemical
hear I have to retreave the number of employes in each department and as
well as
I have to retreave number of John's in each department
is there any function is there which can solve my problem
i tried with subset();
but it is retreaving one department's data only
can anyone suggest what I have to do for this
thanks in advance
[[alternative HTML version deleted]]
------------------------------
Message: 31
Date: Wed, 28 Jan 2009 16:51:35 +0100
From: <mauede at alice.it>
Subject: [R] help with plot layout
To: <r-help at stat.math.ethz.ch>
Cc: gunter.berton at gene.com
Message-ID:
<6B32C438581E5D4C8A34C377C3B334A401752993 at FBCMST11V04.fbc.local>
Content-Type: text/plain; charset="iso-8859-1"
It takes a lot of sweat to generate a composite plot with R ... sigh.
I though I was almost done when I met the umpteenth hurdle. I cannot place a nice title on the 2nd plot (raw signal)
on the layout. I do not have control on where either the "main" option of "plot" function, or "title", place the text
string which keeps dysplaying chopped from above. I also tried "text", changing many times the string coordinates, but could not see any text anywhere on the canvas .
By the way, since the layout breaks the canvas into 4 parts, are the text coordinates absolute (referred to the canvas) or
relative (referred to the part) ?
Please, find attached the generated drawing. The generating script is i the following.
Thank you so much,
Maura
##################################################################
WavMaxNumCoef <- 30
setwd("C:/Documents and Settings/Monville/SpAn-Tests/16440-Raw-Dir")
xx <- read.table("Interp-Amp-PhasePlus16440.txt",header=TRUE, sep=" ")
NumCycles <- max(xx[,"cycle"])
TickPos <- vector(length=NumCycles)
TickCoord <- vector(length=NumCycles)
for(i in 1:NumCycles) {
TickPos[i] <- xx[min(which(xx[,"cycle"] == i)),1]
}
aa <- read.table( "16440-Alpha.txt" )
xaa <- seq(1:length(t(aa)))
vv <- read.table("16440-Vanishing-Moments")
vvLab <- seq(1,WavMaxNumCoef/2,1)
vvCounts <- vector(length=WavMaxNumCoef/2)
for(k in 1:(WavMaxNumCoef/2)) {
vvCounts[k] <- length(which(vv[] == k))
}
yyLab <- seq(1,length(t(vv)),2)
bb <- read.table("16440-Length")
bbLab <- seq(min(bb),max(bb),1)
bb <- sort(t(bb))
bbCounts <- as.numeric(vector(length=(max(bb)-min(bb)+1)))
for(k in 1:length(bbCounts)) {
bbCounts[k] <- length(which(bb[] == (k +min(bb) -1)))
}
zzLab <- seq(1,max(bbCounts),1)
# DEFINE LAYOUT
x11(width=22,height=14)
nf <- layout(matrix(c(1,3,2,4),2,2,byrow=TRUE), c(3,1), c(2,2),FALSE)
layout.show(nf)
# PLOT DONOHO ALPHA
par(mar=c(10,4,2,5),xaxt="n",cex.axis=1,pty="m")
plot(xaa,t(aa),type="h")
par (xaxt="s",xaxp=c(0,95.964,24),xaxs="i")
axis(1,at=TickPos,labels=as.character(TickPos),col="red",col.axis="red",font.axis=1)
# PLOT RAW SIGNAL
par(mar=c(3,4,0,5),xaxt="n",cex.axis=1,pty="m")
plot(xx[,1],xx[,2],pch=3,type="l",frame.plot=FALSE,xpd=TRUE)
title("Raw Signal 16440",cex.main=1.0,font=2)
par (xaxt="s",xaxp=c(0,95.964,24),xaxs="i")
axis(1,at=TickPos,labels=as.character(TickPos),col="red",col.axis="red", font.axis=1)
# PLOT VANISHING MOMENT DISTRIBUTION
par(mar=c(1,0,2,3),xaxt="n",yaxt="n",cex.axis=0.7,pty="m")
barplot(vvCounts,width=1,space=0,horiz=TRUE,axes=FALSE)
par(xaxt="s",yaxt="s",crt=180,srt=270,adj=1,las=3,xpd=TRUE)
text(x=25.5,y=15.3,pos=4,"Wavelet Vanishing Moments Distribution",cex=1.0,font=2)
axis(2,at=vvLab-1,labels=as.character(vvLab),col="red",col.axis="red",font.axis=1,xpd=TRUE,
cex.lab=1)
axis(3,at=yyLab-1,labels=as.character(yyLab),col="red",col.axis="red",font.axis=1,xpd=TRUE,
cex.lab=0.8,cex.axis=0.8)
# PLOT CYCLES LENGTH
par(mar=c(0,0,1,3),xaxt="n",yaxt="n",cex.axis=1)
barplot(bbCounts,width=1,axes=FALSE,space=0,horiz=TRUE)
par(xaxt="s",yaxt="s",crt=180,srt=270,adj=1,las=3,cex.lab=0.1,xpd=TRUE)
text(x=15.5,y=65.3,pos=4,"Cycles Length Distribution",cex=1.0,font=2)
axis(2,at=as.numeric(bbLab)-41,labels=bbLab,col="red",col.axis="red",font.axis=1,
lab=c(10,10,15),cex.lab=0.7,cex.axis=0.6)
axis(3,at=zzLab,labels=as.character(zzLab),col="red",col.axis="red",font.axis=1,xpd=TRUE,
cex.lab=1,cex.axis=0.8)
# cords <-locator(n=3)
e tutti i telefonini TIM!
Vai su
------------------------------
Message: 32
Date: Wed, 28 Jan 2009 09:13:51 -0500
From: "Rixon, John C." <JCRixon at wellington.com>
Subject: [R] Newbie question about "grouping"
To: <r-help at r-project.org>
Message-ID:
<761B467185125146B58FC9540C493F6A07B46196 at PROD-MSG-CLU-03.messaging.wellmanage.com>
Content-Type: text/plain
Hi folks:
I am a SQL guy who just downloaded and installed R yesterday. I am
trying to evaluate some "complex" aggregations we are currently
performing with Syncsort (and have tried in Oracle) with R. I have
loaded data in a dataframe and have performed some of the simple
aggregations on a subset of data. What I do not see how to do though,
is to "group" the aggregations on a particular key value (e.g., sum
market_value over account_id).
If you can point me in the right direction I'd very much appreciate it.
Thanks!
John
[[alternative HTML version deleted]]
------------------------------
Message: 33
Date: Wed, 28 Jan 2009 17:11:04 +0100
From: Mark Na <mtb954 at gmail.com>
Subject: [R] Logical subset of the columns in a dataframe
To: r-help at r-project.org
Message-ID:
<e40d78ce0901280811l5edc87f4lc42485ed6cbe3e33 at mail.gmail.com>
Content-Type: text/plain
Hi R-helpers,
I've been struggling with a problem for most of the day (!) so am finally
resorting to R-help.
I would like to subset the columns of my dataframe based on the frequency
with which the columns contain non-zero values. For example, let's say that
I want to retain only those columns which contain non-zero values in at
least 1% of their rows.
In Excel I would calculate a row at the bottom of my data sheet and use the
following function
=countif(range,">0")
to identify the number of non-zero cells in each column. Then, I would
divide that by the number of rows to obtain the frequency of non-zero values
in each column. Then, I would delete those columns with frequencies < 0.01.
But, I'd like to do this in R. I think the missing link is an analog to
Excel's countif function. Any ideas?
Thanks! Mark
[[alternative HTML version deleted]]
------------------------------
Message: 34
Date: Wed, 28 Jan 2009 11:11:59 -0500
From: David Winsemius <dwinsemius at comcast.net>
Subject: Re: [R] Grouping problem
To: venkata kirankumar <kiran4u2all at gmail.com>
Cc: r-help at r-project.org
Message-ID: <B0F494E3-C48F-4B55-B222-E1705451D746 at comcast.net>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
A vague answer is the best you should hope for with such a vague
question with no sample data:
?table
?xtabs
?"=="
A search on "Frequency tables from factors" should get you to the
intro to R section with that name.
--
David Winsemius
On Jan 28, 2009, at 10:47 AM, venkata kirankumar wrote:
> Hi all,
> I have a problem with grouping like I have to give count of employes
> in each
> department like
>
> if in one company there is departments like
> Mechanical, Computer, Fitting, electronics and Chemical
>
> hear I have to retreave the number of employes in each department
> and as
> well as
> I have to retreave number of John's in each department
>
> is there any function is there which can solve my problem
> i tried with subset();
> but it is retreaving one department's data only
> can anyone suggest what I have to do for this
If you had offered the code that was doing this, there may have been a
person who could explain how it could be modified to return a more
desirable value.
>
>
>
> thanks in advance
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 35
Date: Wed, 28 Jan 2009 10:12:59 -0600
From: hadley wickham <h.wickham at gmail.com>
Subject: Re: [R] Grouping problem
To: venkata kirankumar <kiran4u2all at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<f8e6ff050901280812p596c0fecr5627386123116a64 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
You might want to have a look at the plyr package,
http://had.co.nz/plyr, which includes tools for performing this sort
of grouping.
Hadley
On Wed, Jan 28, 2009 at 9:47 AM, venkata kirankumar
<kiran4u2all at gmail.com> wrote:
> Hi all,
> I have a problem with grouping like I have to give count of employes in each
> department like
>
> if in one company there is departments like
> Mechanical, Computer, Fitting, electronics and Chemical
>
> hear I have to retreave the number of employes in each department and as
> well as
> I have to retreave number of John's in each department
>
> is there any function is there which can solve my problem
> i tried with subset();
> but it is retreaving one department's data only
> can anyone suggest what I have to do for this
>
>
> thanks in advance
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
http://had.co.nz/
------------------------------
Message: 36
Date: Wed, 28 Jan 2009 11:16:30 -0500
From: David Winsemius <dwinsemius at comcast.net>
Subject: Re: [R] Newbie question about "grouping"
To: "Rixon, John C." <JCRixon at wellington.com>
Cc: r-help at r-project.org
Message-ID: <78B3843E-0007-495B-A052-8C9E3AC82788 at comcast.net>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
?by
?aggregate
?ave
Further specifics might be forthcoming if self-contained example data
and desired output were offered. The help pages will have worked
examples, of course.
--
David Winsemius
On Jan 28, 2009, at 9:13 AM, Rixon, John C. wrote:
> Hi folks:
>
> I am a SQL guy who just downloaded and installed R yesterday. I am
> trying to evaluate some "complex" aggregations we are currently
> performing with Syncsort (and have tried in Oracle) with R. I have
> loaded data in a dataframe and have performed some of the simple
> aggregations on a subset of data. What I do not see how to do though,
> is to "group" the aggregations on a particular key value (e.g., sum
> market_value over account_id).
>
> If you can point me in the right direction I'd very much appreciate
> it.
>
> Thanks!
>
> John
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 37
Date: Wed, 28 Jan 2009 08:18:15 -0800 (PST)
From: Thomas Lumley <tlumley at u.washington.edu>
Subject: Re: [R] Newbie question about "grouping"
To: "Rixon, John C." <JCRixon at wellington.com>
Cc: r-help at r-project.org
Message-ID: <Pine.LNX.4.43.0901280818150.7342 at hymn14.u.washington.edu>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
Some useful commands are:
by(), aggregate(), ave(), split().
eg
by(market_value, account_id, sum)
-thomas
On Wed, 28 Jan 2009, Rixon, John C. wrote:
> Hi folks:
>
> I am a SQL guy who just downloaded and installed R yesterday. I am
> trying to evaluate some "complex" aggregations we are currently
> performing with Syncsort (and have tried in Oracle) with R. I have
> loaded data in a dataframe and have performed some of the simple
> aggregations on a subset of data. What I do not see how to do though,
> is to "group" the aggregations on a particular key value (e.g., sum
> market_value over account_id).
>
> If you can point me in the right direction I'd very much appreciate it.
>
> Thanks!
>
> John
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Thomas Lumley Assoc. Professor, Biostatistics
tlumley at u.washington.edu University of Washington, Seattle
------------------------------
Message: 38
Date: Wed, 28 Jan 2009 10:20:15 -0600
From: hadley wickham <h.wickham at gmail.com>
Subject: Re: [R] Newbie question about "grouping"
To: "Rixon, John C." <JCRixon at wellington.com>
Cc: r-help at r-project.org
Message-ID:
<f8e6ff050901280820r3098ae20raf6b2e770a1c2d64 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
On Wed, Jan 28, 2009 at 8:13 AM, Rixon, John C. <JCRixon at wellington.com> wrote:
> Hi folks:
>
> I am a SQL guy who just downloaded and installed R yesterday. I am
> trying to evaluate some "complex" aggregations we are currently
> performing with Syncsort (and have tried in Oracle) with R. I have
> loaded data in a dataframe and have performed some of the simple
> aggregations on a subset of data. What I do not see how to do though,
> is to "group" the aggregations on a particular key value (e.g., sum
> market_value over account_id).
>
> If you can point me in the right direction I'd very much appreciate it.
Have a look at the plyr package, http://had.co.nz/plyr, and associated
documentation. If you're doing pivot table type aggregations, you
might also want to have a look at the reshape package,
http://had.co.nz/reshape.
Hadley
--
http://had.co.nz/
------------------------------
Message: 39
Date: Wed, 28 Jan 2009 16:24:11 +0000 (GMT)
From: Prof Brian Ripley <ripley at stats.ox.ac.uk>
Subject: Re: [R] Logical subset of the columns in a dataframe
To: Mark Na <mtb954 at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<alpine.LFD.2.00.0901281621030.31070 at gannet.stats.ox.ac.uk>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 28 Jan 2009, Mark Na wrote:
> Hi R-helpers,
>
> I've been struggling with a problem for most of the day (!) so am finally
> resorting to R-help.
>
> I would like to subset the columns of my dataframe based on the frequency
> with which the columns contain non-zero values. For example, let's say that
> I want to retain only those columns which contain non-zero values in at
> least 1% of their rows.
>
> In Excel I would calculate a row at the bottom of my data sheet and use the
> following function
>
> =countif(range,">0")
>
> to identify the number of non-zero cells in each column. Then, I would
> divide that by the number of rows to obtain the frequency of non-zero values
> in each column. Then, I would delete those columns with frequencies < 0.01.
>
> But, I'd like to do this in R. I think the missing link is an analog to
> Excel's countif function. Any ideas?
Use something like
DF[sapply(DF, function(x) mean(x) >= 0.01)]
Since logical values are converted to 0/1, mean() gives the frequency
(and sum() the count).
>
> Thanks! Mark
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
------------------------------
Message: 40
Date: Wed, 28 Jan 2009 11:25:55 -0500
From: Ted Byers <r.ted.byers at gmail.com>
Subject: Re: [R] Mystery Error in midnightStandard
To: Yohan Chalabi <chalabi at phys.ethz.ch>
Cc: R-help Forum <r-help at r-project.org>
Message-ID:
<4f1819890901280825w3cfa983bi8c68f1b85c22d378 at mail.gmail.com>
Content-Type: text/plain
Hi Yohan,
On Wed, Jan 28, 2009 at 10:28 AM, Yohan Chalabi <chalabi at phys.ethz.ch>wrote:
> >>>> "TB" == Ted Byers <r.ted.byers at gmail.com>
> >>>> on Wed, 28 Jan 2009 09:30:58 -0500
>
> TB> It is certain that all entries have the same format, but I'm
> TB> starting to
> TB> think that the error message is something of a red herring.
> TB> Consider this:
> TB>
> TB> > year = 2009
> TB> > week = 0
> TB> > day = 3
> TB> > datestr = sprintf(%i-%i-%i,year,week,day);datestr
> TB> [1] 2009-0-3
> TB> > date1 = timeDate(datestr, format = %Y-%U-%w);
> TB> > date1
> TB> GMT
> TB> [1] [NA]
> TB> > day = 4
> TB> > datestr = sprintf(%i-%i-%i,year,week,day);datestr
> TB> [1] 2009-0-4
> TB> > date1 = timeDate(datestr, format = %Y-%U-%w);
> TB> > date1
> TB> GMT
> TB> [1] [2009-01-01]
> TB> >
> TB> > datestr = sprintf(%i-%i-%i,year,week,3);datestr
> TB> [1] 2009-0-3
> TB> > date2 = timeDate(datestr, format = %Y-%U-%w);date2
> TB> GMT
> TB> [1] [NA]
> TB> > difftimeDate(date2,date1, units = weeks)
> TB> Error in midnightStandard(charvec, format) :
> TB> 'charvec' has non-NA entries of different number of characters
> TB> In addition: Warning messages:
> TB> 1: In min(x) : no non-missing arguments to min; returning Inf
> TB> 2: In max(x) : no non-missing arguments to max; returning -Inf
> TB>
> TB>
> TB>
> TB> The first values for year, week and day are the values on
> TB> which my loop
> TB> dies. It returns 'NA' here. It seems clear that it is
> TB> returning NA because
> TB> the date that data corresponds to is 2008-12-31.
> TB>
> TB> The error is being produced by difftimeDate rather than timeDate
> TB> (as shown
> TB> by the above session). But that represents a flaw in the
> TB> function design.
>
> This is not a flaw in timeDate. it behaves the same way as
> 'as.POSIXct'
>
That the two behave the same doesn't change the assessment that the design
is flawed. That doesn't mean that the function is wrong. It means only
that the behaviour can be made more useful. For example, in SQL, if a given
calculation returns NULL, and the result is subsequently used in another
calculation, the result that returns is also NULL. That is quite useful,
and admits algorithms that can react appropriately to NULLs when necessary.
That is arguably better than forcing the code to fail the moment a NULL is
used in a secondary calculation. In C++, OTOH, one can catch the problem
earlier using, e.g., exceptions, again allowing the program to complete even
when problems arise for certain values or combinations thereof.
As a software engineer, I understand the issues involved in creating
libraries. If I want to incorporate the functionality of a given standard
suite of functions (e.g. ANSI C standard library functions, or posix
functions), my first step would be to ensure I can duplicate how they
behave. But I would not stop there. There are, for example, serious design
flaws in many ANSI C functions that, ignored, introduce serious security
defects in applications that use them. I would therefore refactor them to
eliminate the security defects. If they can not be eliminated, I would
replace the function in question by a similar function that does not have
that security defect.
Posix is a useful, but old, standard, and I am merely suggesting that once
you have duplicated it, look beyond it to ways it can be improved upon.
There is more to the design of a function than whether or not it gives the
right result with good input. There is how it behaves when there is a
problem with the inputs and whether or not you force the calling code to die
when a problem arises or you give the calling code a way to react to such
problems. When I add functions to my own C++ or Java libraries, I normally
include more bad input data in the unit tests than good data (though the
latter is sufficient to ensure correct results are invariably obtained),
precisely so I can document how it behaves when there is a problem and give
coders who use it a variety of options to use to deal with them.
>
> strptime(datestr, format = "%Y-%U-%w")
>
> Instead of claiming that there is a flaw in the function you could have
> suggested an 'is.na' method for 'timeDate'.
>
At the time, I did not know about is.na. I have spent the past hour trying
is.na, but to no avail. I guess that is no surprise to you, but that it
would fail is not reflected in the R documentation of is.na. That mentions
S3, but not S4. As I just recently started using R, I have not yet looked
at what S3 and S4 are, so that is a few more hours of study before I get
this problem solved.
>
> I will add an 'is.na' method in the dev version of 'timeDate'.
>
>
Thanks. I'll benefit from that once it makes it into the production
release. In the mean time, I need to find a way to make something similar
now, in my script.
Thanks
Ted
[[alternative HTML version deleted]]
------------------------------
Message: 41
Date: Wed, 28 Jan 2009 11:35:19 -0500
From: David Winsemius <dwinsemius at comcast.net>
Subject: Re: [R] Logical subset of the columns in a dataframe
To: Mark Na <mtb954 at gmail.com>
Cc: r-help at r-project.org
Message-ID: <AA1C6FDC-927E-468F-B31A-0A8EA38FC2E3 at comcast.net>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
One approach to such a problem would be to use a logical vector inside
the function colSums.
?colSums
> DF <- data.frame(XX= runif(20), YY=runif(20))
> colSums(DF > 0.5)
XX YY
11 9
> colSums(DF > -Inf)
XX YY
20 20
>
> colSums(DF> 0.5)/colSums(DF > -Inf) #could have used DF >= min(DF)
in the denominator
XX YY
0.55 0.45
--
David Winsemius
On Jan 28, 2009, at 11:11 AM, Mark Na wrote:
> Hi R-helpers,
>
> I've been struggling with a problem for most of the day (!) so am
> finally
> resorting to R-help.
>
> I would like to subset the columns of my dataframe based on the
> frequency
> with which the columns contain non-zero values. For example, let's
> say that
> I want to retain only those columns which contain non-zero values in
> at
> least 1% of their rows.
>
> In Excel I would calculate a row at the bottom of my data sheet and
> use the
> following function
>
> =countif(range,">0")
>
> to identify the number of non-zero cells in each column. Then, I would
> divide that by the number of rows to obtain the frequency of non-
> zero values
> in each column. Then, I would delete those columns with frequencies
> < 0.01.
I don't think that would do what you describe unless you were only
working with single column ranges. Functions on ranges in Excel are
not calculated by column.
>
>
> But, I'd like to do this in R. I think the missing link is an analog
> to
> Excel's countif function. Any ideas?
>
> Thanks! Mark
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 42
Date: Wed, 28 Jan 2009 17:48:24 +0100
From: Yohan Chalabi <chalabi at phys.ethz.ch>
Subject: Re: [R] Mystery Error in midnightStandard
To: Ted Byers <r.ted.byers at gmail.com>
Cc: R-help Forum <r-help at r-project.org>
Message-ID: <20090128174824.561e94cb at mimi>
Content-Type: text/plain; charset=US-ASCII
>>>> "TB" == Ted Byers <r.ted.byers at gmail.com>
>>>> on Wed, 28 Jan 2009 11:25:55 -0500
TB> That the two behave the same doesn't change the assessment
TB> that the design
TB> is flawed. That doesn't mean that the function is wrong.
TB> It means only
TB> that the behaviour can be made more useful. For example,
TB> in SQL, if a given
TB> calculation returns NULL, and the result is subsequently used
TB> in another
TB> calculation, the result that returns is also NULL. That is
TB> quite useful,
TB> and admits algorithms that can react appropriately to NULLs
TB> when necessary.
TB> That is arguably better than forcing the code to fail the
TB> moment a NULL is
TB> used in a secondary calculation. In C++, OTOH, one can catch
TB> the problem
TB> earlier using, e.g., exceptions, again allowing the program
TB> to complete even
TB> when problems arise for certain values or combinations thereof.
TB>
TB> As a software engineer, I understand the issues involved
TB> in creating
TB> libraries. If I want to incorporate the functionality of a
TB> given standard
TB> suite of functions (e.g. ANSI C standard library functions,
TB> or posix
TB> functions), my first step would be to ensure I can duplicate
TB> how they
TB> behave. But I would not stop there. There are, for example,
TB> serious design
TB> flaws in many ANSI C functions that, ignored, introduce
TB> serious security
TB> defects in applications that use them. I would therefore
TB> refactor them to
TB> eliminate the security defects. If they can not be eliminated,
TB> I would
TB> replace the function in question by a similar function that
TB> does not have
TB> that security defect.
TB>
TB> Posix is a useful, but old, standard, and I am merely suggesting
TB> that once
TB> you have duplicated it, look beyond it to ways it can be
TB> improved upon.
TB> There is more to the design of a function than whether or not
TB> it gives the
TB> right result with good input. There is how it behaves when
TB> there is a
TB> problem with the inputs and whether or not you force the
TB> calling code to die
TB> when a problem arises or you give the calling code a way to
TB> react to such
TB> problems. When I add functions to my own C++ or Java libraries,
TB> I normally
TB> include more bad input data in the unit tests than good data
TB> (though the
TB> latter is sufficient to ensure correct results are invariably
TB> obtained),
TB> precisely so I can document how it behaves when there is a
TB> problem and give
TB> coders who use it a variety of options to use to deal with them.
TB>
TB>
TB> >
TB> > strptime(datestr, format = %Y-%U-%w)
TB> >
TB> > Instead of claiming that there is a flaw in the function
TB> you could have
TB> > suggested an 'is.na' method for 'timeDate'.
TB> >
TB>
TB> At the time, I did not know about is.na. I have spent the
TB> past hour trying
TB> is.na, but to no avail. I guess that is no surprise to you,
TB> but that it
TB> would fail is not reflected in the R documentation of is.na.
TB> That mentions
TB> S3, but not S4. As I just recently started using R, I have
TB> not yet looked
TB> at what S3 and S4 are, so that is a few more hours of study
TB> before I get
TB> this problem solved.
TB>
TB>
TB> >
TB> > I will add an 'is.na' method in the dev version of 'timeDate'.
TB> >
TB> >
TB> Thanks. I'll benefit from that once it makes it into the
TB> production
TB> release. In the mean time, I need to find a way to make
TB> something similar
TB> now, in my script.
setMethod("is.na", "timeDate", function(x) is.na(as.POSIXct(x)))
TB>
TB> Thanks
--
PhD student
Swiss Federal Institute of Technology
Zurich
www.ethz.ch
------------------------------
Message: 43
Date: Wed, 28 Jan 2009 16:55:40 +0000
From: June Wong <neptune545 at hotmail.com>
Subject: [R] constrainOptim
To: <r-help at r-project.org>
Message-ID: <BAY136-W82AA69673D0B171ED34178AC80 at phx.gbl>
Content-Type: text/plain
Dear R helpers
I have a question regarding the constrainOptim.
I'm coding the nested logit and would like to set a bound of rho to (0,1] as an extreme value distribution where rho = exp(lambda)/1+exp(lambda)
I wonder if I can do that directly in optim (say rho > 0 & <= 1) or need to use constrainOptim
I read the help but still don't know how to set ui and ci
Thanks,
June
_________________________________________________________________
[[alternative HTML version deleted]]
------------------------------
Message: 44
Date: Wed, 28 Jan 2009 12:07:14 -0500
From: Ravi Varadhan <rvaradhan at jhmi.edu>
Subject: Re: [R] constrainOptim
To: June Wong <neptune545 at hotmail.com>
Cc: r-help at r-project.org
Message-ID: <f6d1b5dede2.49804a72 at johnshopkins.edu>
Content-Type: text/plain; charset=us-ascii
For simple box constraints, i.e. lower and upper limits directly on the parameters themselves, you don't need ConstrOptim. You can get the job done with the "L-BFGS-B" algorithm in optim() or using nlminb() or using the spg() function in the BB package. In this case the feasible region is a hyper-rectangle (could be infinite in some dimensions).
ConstrOptim() is useful when you have more general linear inequality constraints, i.e. constraints on linear combinations of parameters. In this case the feasible region is a convex polytope.
Best,
Ravi.
____________________________________________________________________
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvaradhan at jhmi.edu
----- Original Message -----
From: June Wong <neptune545 at hotmail.com>
Date: Wednesday, January 28, 2009 11:57 am
Subject: [R] constrainOptim
To: r-help at r-project.org
> Dear R helpers
>
> I have a question regarding the constrainOptim.
> I'm coding the nested logit and would like to set a bound of rho to
> (0,1] as an extreme value distribution where rho = exp(lambda)/1+exp(lambda)
> I wonder if I can do that directly in optim (say rho > 0 & <= 1) or
> need to use constrainOptim
> I read the help but still don't know how to set ui and ci
>
> Thanks,
> June
>
> _________________________________________________________________
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
>
> PLEASE do read the posting guide
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 45
Date: Wed, 28 Jan 2009 18:07:28 +0100
From: "Strubbe Diederik" <diederik.strubbe at ua.ac.be>
Subject: [R] repeated measures design for GAM?
To: <r-help at R-project.org>
Message-ID:
<C9854550FEF14846A136100B3EC52F73B6B5AC at xmail05.ad.ua.ac.be>
Content-Type: text/plain
Dear all,
I have a question on the use of GAM with repeated measures. My dataset is as follows:
- a number of study areas where bird abundance has been determined. Counts have been performed in 3 consecutive years and there were 2 counts per year (i.e. in total 6 counts).
- a number of environmental predictors that do not change over year Xi).
When using a GLM, a repeated measures design would like: (for example)
lme(Bird_abundance = study_area + count + X1 + X2 + X3,random = ~time|cow).
However, I have found no analogue design for a GAM. For now, I have averaged my bird abundances but I wondered whether a more subtle and elegant strategy exists...?
Many thanks,
Diederik
Diederik Strubbe
Evolutionary Ecology Group
Department of Biology, University of Antwerp
Universiteitsplein 1
B-2610 Antwerp, Belgium
http://webhost.ua.ac.be/deco
tel : 32 3 820 23 85
[[alternative HTML version deleted]]
------------------------------
Message: 46
Date: Wed, 28 Jan 2009 18:09:38 +0100
From: "Strubbe Diederik" <diederik.strubbe at ua.ac.be>
Subject: [R] Repeated measures design for GAM? - corrected question...
To: <r-help at R-project.org>
Message-ID:
<C9854550FEF14846A136100B3EC52F73B6B5AD at xmail05.ad.ua.ac.be>
Content-Type: text/plain
Dear all,
I have a question on the use of GAM with repeated measures. My dataset is as follows:
- a number of study areas where bird abundance has been determined. Counts have been performed in 3 consecutive years and there were 2 counts per year (i.e. in total 6 counts).
- a number of environmental predictors that do not change over year Xi).
When using a GLM, a repeated measures design would like: (for example)
lme(Bird_abundance = study_area + count +year+ X1 + X2 + X3,random = ~count|study_area).
However, I have found no analogue design for a GAM. For now, I have averaged my bird abundances but I wondered whether a more subtle and elegant strategy exists...?
Many thanks,
Diederik
Diederik Strubbe
Evolutionary Ecology Group
Department of Biology, University of Antwerp
Universiteitsplein 1
B-2610 Antwerp, Belgium
http://webhost.ua.ac.be/deco
tel : 32 3 820 23 85
[[alternative HTML version deleted]]
------------------------------
Message: 47
Date: Wed, 28 Jan 2009 19:16:30 +0200
From: constantine <costas.magnuse at gmail.com>
Subject: [R] Sweave problem with greek text
To: r-help at r-project.org
Message-ID:
<30ddfdae0901280916i1e92e040s336ec0227305723 at mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
Dear Sweave and R aficionados,
I am using R and Latex for many years, writing texts in greek. I tried
to combine them with Sweave, but without any success.
Could you provide me with any help?
Usually my LaTeX files are like this iso-8859-7 encoded .tex file:
http://costis.name/0various/lists/R/sweave/successful.greek.tex ,
which happily produces
http://costis.name/0various/lists/R/sweave/successful.greek.pdf .
I tried using Sweave on the iso-8859-7 encoded .Rnw file:
http://costis.name/0various/lists/R/sweave/unsuccessful.sweave.Rnw ,
but I am getting misencoded greek text and also misencoded R code as
it appears in http://costis.name/0various/lists/R/sweave/unsuccessful.sweave.pdf
.
The .tex file that Sweave produces is located at
http://costis.name/0various/lists/R/sweave/unsuccessful.sweave.tex
I am also the latex R error code
http://costis.name/0various/lists/R/sweave/unsuccessful.sweave.log
In the above example I am not using the kerkis font-package. When I
am, I am getting no output at all, and a latex error of "Corrupted
NFSS tables".
I can understand that the whole problem is an encoding issue, but what
should I do in order to use Sweave with greek text flawlessly?
One solution is to edit the .tex file produced by Sweave, but this
solution is by far counter-productive.
Thank you very much in advance,
Constantine Tsardounis
http://www.costis.name
PS.: I am having no problem to run Sweave in 100% English texts.
I postscript the following files:
* unsuccessful.sweave.Rnw
* unsuccessful.sweave.tex
* successful.greek.tex
########################
unsuccessful.sweave.Rnw
########################
\documentclass[a4paper,12pt]{book}
\usepackage[greek]{babel}
\usepackage[iso-8859-7]{inputenc}
% \usepackage{kerkis}
\begin{document}
\section{\textlatin{Sweave}}
\subsection{\textlatin{in Greek}}
???? ???, ???? ????? ????????.
<<>>=
data(airquality)
library(ctest)
kruskal.test(Ozone ~ Month, data = airquality)
@
\subsection{\textlatin{in English}}
\textlatin{Hello to all, now I am writing in English.}
\end{document}
########################
unsuccessful.sweave.tex
########################
\documentclass[a4paper,12pt]{book}
\usepackage[greek]{babel}
\usepackage[iso-8859-7]{inputenc}
% \usepackage{kerkis}
\usepackage{/usr/share/R/share/texmf/Sweave}
\begin{document}
\section{\textlatin{Sweave}}
\subsection{\textlatin{in Greek}}
?????(c)?' ???'??, ???????' ?????????? ???????????(c)????.
\begin{Schunk}
\begin{Sinput}
> data(airquality)
> library(ctest)
> kruskal.test(Ozone ~ Month, data = airquality)
\end{Sinput}
\begin{Soutput}
Kruskal-Wallis rank sum test
data: Ozone by Month
Kruskal-Wallis chi-squared = 29.2666, df = 4, p-value = 6.901e-06
\end{Soutput}
\end{Schunk}
\subsection{\textlatin{in English}}
\textlatin{Hello to all, now I am writing in English.}
\end{document}
########################
successful.greek.tex
########################
\documentclass[a4paper,12pt]{book}
\usepackage[greek]{babel}
\usepackage[iso-8859-7]{inputenc}
\usepackage{kerkis}
\begin{document}
\section{\textlatin{Sweave}}
\subsection{\textlatin{in Greek}}
???? ???, ???? ????? ????????.
\subsection{\textlatin{in English}}
\textlatin{Hello to all, now I am writing in English.}
\end{document}
------------------------------
Message: 48
Date: Wed, 28 Jan 2009 12:17:05 -0500
From: Ravi Varadhan <rvaradhan at jhmi.edu>
Subject: Re: [R] StepAIC with coxph
To: Michele Santacatterina <miksanta at gmail.com>
Cc: R-help at r-project.org
Message-ID: <f6b1f699771b.49804cc1 at johnshopkins.edu>
Content-Type: text/plain; charset=iso-8859-1
Michele,
This error means that some of the variables in your formula have missing values, and hence when these terms or added/dropped you have a different sample size. Hence, the AIC cannot be compared between different models. The solution is to create a compelete-data dataframe for the largest model, i.e none of the variables in the largest model should have any missing values. Then run stepAIC on this dataframe.
Best,
Ravi.
____________________________________________________________________
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvaradhan at jhmi.edu
----- Original Message -----
From: Michele Santacatterina <miksanta at gmail.com>
Date: Wednesday, January 28, 2009 9:51 am
Subject: [R] StepAIC with coxph
To: R-help at r-project.org
> Hi,
>
> i'm trying to apply StepAIC with coxph...but i have the same error:
>
> stepAIC(fitBMT)
> Start: AIC=327.77
> Surv(TEMPO,morto==1) ? VOD + SESSO + ETA + ........
>
> Error in dropterm.default(fit,scope$drop, scale=scale,trace=max(0, :
> number of rows in use has changed: remove missing values?
>
> anybody know this error??
>
> Thanks.
>
> Michele
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
>
> PLEASE do read the posting guide
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 49
Date: Wed, 28 Jan 2009 17:30:02 +0000
From: Simon Wood <s.wood at bath.ac.uk>
Subject: Re: [R] Repeated measures design for GAM? - corrected
question...
To: r-help at r-project.org
Message-ID: <200901281730.02975.s.wood at bath.ac.uk>
Content-Type: text/plain; charset="iso-8859-1"
`gamm' in package `mgcv' will let you specify random effects as part of a
generalized additive mixed model, but I must admit that I'm a bit confused
about what `Bird_abundance' is here, and how it differs from `Count'.
best,
Simon
On Wednesday 28 January 2009 17:09, Strubbe Diederik wrote:
> Dear all,
>
> I have a question on the use of GAM with repeated measures. My dataset is
> as follows: - a number of study areas where bird abundance has been
> determined. Counts have been performed in 3 consecutive years and there
> were 2 counts per year (i.e. in total 6 counts). - a number of
> environmental predictors that do not change over year Xi). When using a
> GLM, a repeated measures design would like: (for example)
>
> lme(Bird_abundance = study_area + count +year+ X1 + X2 + X3,random =
> ~count|study_area).
>
> However, I have found no analogue design for a GAM. For now, I have
> averaged my bird abundances but I wondered whether a more subtle and
> elegant strategy exists...?
>
> Many thanks,
>
>
> Diederik
>
> Diederik Strubbe
> Evolutionary Ecology Group
> Department of Biology, University of Antwerp
> Universiteitsplein 1
> B-2610 Antwerp, Belgium
> http://webhost.ua.ac.be/deco
> tel : 32 3 820 23 85
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html and provide commented, minimal,
> self-contained, reproducible code.
--
> Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK
> +44 1225 386603 www.maths.bath.ac.uk/~sw283
------------------------------
Message: 50
Date: Wed, 28 Jan 2009 09:51:29 -0800
From: Dr Carbon <drcarbon at gmail.com>
Subject: [R] gls prediction using the correlation structure in nlme
To: r-help at r-project.org, jcp at research.bell-labs.com,
bates at stat.wisc.edu
Message-ID:
<e89bb7ac0901280951j725ef225r87965bfd08595b71 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
How does one coerce predict.gls to incorporate the fitted correlation
structure from the gls object into predictions? In the example below
the AR(1) process with phi=0.545 is not used with predict.gls. Is
there another function that does this? I'm going to want to fit a few
dozen models varying in order from AR(1) to AR(3) and would like to
look at the fits with the correlation structure included.
Thanks in advance.
-JC
PS I am including the package maintainers on this post - does this
constitute a maintainer-specific question in r-help etiquette?
# example
set.seed(123)
x <- arima.sim(list(order = c(1,0,0), ar = 0.7), n = 100)
y <-x + arima.sim(list(order = c(1,0,0), ar = 0.7), n = 100)
x <- c(x)
y <- c(y)
lm1 <- lm(y~x)
ar(residuals(lm1)) # indicates an ar1 model
cs1 <- corARMA(p=1)
fm1 <- gls(y~x,corr=cs1)
summary(fm1)
# get fits
fits <- predict(fm1)
# use coef to get fits
fits2 <- coef(fm1)[1] + (coef(fm1)[2] * x)
plot(fits,fits2)
------------------------------
Message: 51
Date: Mon, 26 Jan 2009 14:55:21 +0100
From: "ARDIA David" <david.ardia at unifr.ch>
Subject: [R] [R-pkgs] AdMit version 1-01.01
To: <r-packages at r-project.org>
Message-ID:
<2AA9A291D7760C40B00A75A149DB5D37062033 at EXCHANGE4.unifr.ch>
Content-Type: text/plain; charset="us-ascii"
Dear all,
The new version of AdMit (version 1.01-01) is now available from CRAN.
SUMMARY
The package provides functions to perform the fitting of an adaptive
mixture of Student-t distributions to a target density through its
kernel function. The mixture approximation can then be used as the importance
density in importance sampling or as the candidate density in the
Metropolis-Hastings algorithm to obtain quantities of interest for
the target density itself. We believe that this approach may be applicable in
many fields of research and hope that the R package AdMit will be
fruitful for many researchers like econometricians or applied statisticians.
MODIFICATIONS
o change in AdMit.R to deal with convergence problems for simple cases.
o the documentation file has been improved (thanks to Achim Zeilis for
comments).
o a package vignette has been added.
o a paper describing the package in detail has been published in the
Journal of Statistical Software: http://www.jstatsoft.org/v29/i03.
Abstract:
This paper presents the R package AdMit which provides functions to
approximate and sample from a certain target distribution given only a
kernel of the target density function. The core algorithm consists in
the function AdMit which fits an adaptive mixture of Student-t
distributions to the density of interest via its kernel function. Then,
importance sampling or the independence chain Metropolis-Hastings
algorithm are used to obtain quantities of interest for the target
density, using the fitted mixture as the importance or candidate
density. The estimation procedure is fully automatic and thus avoids the
time-consuming and difficult task of tuning a sampling algorithm. The
relevance of the package is shown in two examples. The first aims at
illustrating in detail the use of the functions provided by the
package in a bivariate bimodal distribution. The second shows the
relevance of the adaptive mixture procedure through the Bayesian
estimation of a mixture of ARCH model fitted to foreign exchange
log-returns data. The methodology is compared to standard cases of
importance sampling and the Metropolis-Hastings algorithm using a naive
candidate and with the Griddy-Gibbs approach.
o creation of /doc folder with AdMitJSS.txt and AdMitRnews.txt files
(the R codes used for JSS and Rnews papers).
o CITATION file simplified.
o 'coda' package is now Suggests
REFERENCES
Ardia D, Hoogerheide LF, van Dijk HK (2008). AdMit: Adaptive Mixture of
Student-t Distributions in R. R package version 1.01-01.
URL http://CRAN.R-project.org/package=AdMit.
Ardia D, Hoogerheide LF, van Dijk HK (2009). Adaptive Mixture of
Student-t Distributions as a Flexible Candidate Distribution for
Efficient Simulation: The R Package AdMit. Journal of Statistical
Software, 29(3), 1-32.
URL http://www.jstatsoft.org/v29/i03/.
Hoogerheide LF (2006). Essays on Neural Network Sampling Methods and
Instrumental Variables. Ph.D. thesis, Tinbergen Institute, Erasmus
University Rotterdam. Book nr. 379 of the Tinbergen Institute Research
Series.
Hoogerheide LF, Kaashoek JF, van Dijk HK (2007). On the Shape of
Posterior Densities and Credible Sets in Instrumental Variable
Regression Models with Reduced Rank: An Application of Flexible Sampling
Methods using Neural Networks.
Journal of Econometrics, 139(1), 154-180. doi:10.1016/j.jeconom.2006.06.009.
Hoogerheide LF, van Dijk HK (2008a). Bayesian Forecasting of Value at
Risk and Expected Shorfall Using Adaptive Importance Sampling. Technical
Report 2008-092/4, Tinbergen Institute, Erasmus University Rotterdam.
URL http://www.tinbergen.nl/ discussionpapers/08092.pdf.
Hoogerheide LF, van Dijk HK (2008b). Possibly Ill-Behaved Posteriors in
Econometric Models: On the Connection Between Model Structures,
Non-Elliptical Credible Sets and Neural Network Simulation Techniques."
Technical Report 2008-036/4, Tinbergen Institute, Erasmus University
Rotterdam.
URL http://www.tinbergen.nl/discussionpapers/08036.pdf.
Best regards,
David Ardia (package's maintainer)
Lennart F. Hoogerheide
Herman K. van Dijk
[[alternative HTML version deleted]]
_______________________________________________
R-packages mailing list
R-packages at r-project.org
https://stat.ethz.ch/mailman/listinfo/r-packages
------------------------------
Message: 52
Date: Wed, 28 Jan 2009 11:20:08 -0700
From: Greg Snow <Greg.Snow at imail.org>
Subject: Re: [R] help with plot layout
To: "mauede at alice.it" <mauede at alice.it>, "r-help at stat.math.ethz.ch"
<r-help at stat.math.ethz.ch>
Cc: "gunter.berton at gene.com" <gunter.berton at gene.com>
Message-ID:
<B37C0A15B8FB3C468B5BC7EBC7DA14CC61C939DBFA at LP-EXMBVS10.CO.IHC.COM>
Content-Type: text/plain; charset="us-ascii"
We don't have your data, so we cannot reproduce what you are doing and the plot was stripped off before we saw it (only certain types of attachments are allowed, and some e-mail programs don't give the correct information about attachments so even those types can be stripped if it is not clear what they are).
Here are some possible hints that may help (if I have guessed correctly about what you are trying to do).
Read the help page for par, looking at the information on margins and outer margins, this can be used to give you more room for your titles (also the xpd argument if you are placing things outside the plot area). Also look at the various cex arguments for controlling sizes.
Try using mtext instead of text to manually add titles or other text in the margins.
Sometimes using the outer margins works better than using the regular margins (sometimes not).
The text function uses the user coordinates of the current plot, the functions grconvertX and grconvertY can be used to convert between the different coordinate systems.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> project.org] On Behalf Of mauede at alice.it
> Sent: Wednesday, January 28, 2009 8:52 AM
> To: r-help at stat.math.ethz.ch
> Cc: gunter.berton at gene.com
> Subject: [R] help with plot layout
>
> It takes a lot of sweat to generate a composite plot with R ... sigh.
> I though I was almost done when I met the umpteenth hurdle. I cannot
> place a nice title on the 2nd plot (raw signal)
> on the layout. I do not have control on where either the "main" option
> of "plot" function, or "title", place the text
> string which keeps dysplaying chopped from above. I also tried "text",
> changing many times the string coordinates, but could not see any text
> anywhere on the canvas .
> By the way, since the layout breaks the canvas into 4 parts, are the
> text coordinates absolute (referred to the canvas) or
> relative (referred to the part) ?
> Please, find attached the generated drawing. The generating script is i
> the following.
> Thank you so much,
> Maura
>
> ##################################################################
> WavMaxNumCoef <- 30
> setwd("C:/Documents and Settings/Monville/SpAn-Tests/16440-Raw-Dir")
>
> xx <- read.table("Interp-Amp-PhasePlus16440.txt",header=TRUE, sep=" ")
> NumCycles <- max(xx[,"cycle"])
> TickPos <- vector(length=NumCycles)
> TickCoord <- vector(length=NumCycles)
> for(i in 1:NumCycles) {
> TickPos[i] <- xx[min(which(xx[,"cycle"] == i)),1]
> }
>
> aa <- read.table( "16440-Alpha.txt" )
> xaa <- seq(1:length(t(aa)))
>
> vv <- read.table("16440-Vanishing-Moments")
> vvLab <- seq(1,WavMaxNumCoef/2,1)
> vvCounts <- vector(length=WavMaxNumCoef/2)
> for(k in 1:(WavMaxNumCoef/2)) {
> vvCounts[k] <- length(which(vv[] == k))
> }
> yyLab <- seq(1,length(t(vv)),2)
>
> bb <- read.table("16440-Length")
> bbLab <- seq(min(bb),max(bb),1)
> bb <- sort(t(bb))
> bbCounts <- as.numeric(vector(length=(max(bb)-min(bb)+1)))
> for(k in 1:length(bbCounts)) {
> bbCounts[k] <- length(which(bb[] == (k +min(bb) -1)))
> }
> zzLab <- seq(1,max(bbCounts),1)
>
> # DEFINE LAYOUT
> x11(width=22,height=14)
> nf <- layout(matrix(c(1,3,2,4),2,2,byrow=TRUE), c(3,1), c(2,2),FALSE)
> layout.show(nf)
>
> # PLOT DONOHO ALPHA
> par(mar=c(10,4,2,5),xaxt="n",cex.axis=1,pty="m")
> plot(xaa,t(aa),type="h")
> par (xaxt="s",xaxp=c(0,95.964,24),xaxs="i")
>
> axis(1,at=TickPos,labels=as.character(TickPos),col="red",col.axis="red"
> ,font.axis=1)
>
> # PLOT RAW SIGNAL
> par(mar=c(3,4,0,5),xaxt="n",cex.axis=1,pty="m")
> plot(xx[,1],xx[,2],pch=3,type="l",frame.plot=FALSE,xpd=TRUE)
> title("Raw Signal 16440",cex.main=1.0,font=2)
> par (xaxt="s",xaxp=c(0,95.964,24),xaxs="i")
>
> axis(1,at=TickPos,labels=as.character(TickPos),col="red",col.axis="red"
> , font.axis=1)
>
> # PLOT VANISHING MOMENT DISTRIBUTION
> par(mar=c(1,0,2,3),xaxt="n",yaxt="n",cex.axis=0.7,pty="m")
> barplot(vvCounts,width=1,space=0,horiz=TRUE,axes=FALSE)
> par(xaxt="s",yaxt="s",crt=180,srt=270,adj=1,las=3,xpd=TRUE)
> text(x=25.5,y=15.3,pos=4,"Wavelet Vanishing Moments
> Distribution",cex=1.0,font=2)
> axis(2,at=vvLab-
> 1,labels=as.character(vvLab),col="red",col.axis="red",font.axis=1,xpd=T
> RUE,
> cex.lab=1)
> axis(3,at=yyLab-
> 1,labels=as.character(yyLab),col="red",col.axis="red",font.axis=1,xpd=T
> RUE,
> cex.lab=0.8,cex.axis=0.8)
>
> # PLOT CYCLES LENGTH
> par(mar=c(0,0,1,3),xaxt="n",yaxt="n",cex.axis=1)
> barplot(bbCounts,width=1,axes=FALSE,space=0,horiz=TRUE)
>
> par(xaxt="s",yaxt="s",crt=180,srt=270,adj=1,las=3,cex.lab=0.1,xpd=TRUE)
> text(x=15.5,y=65.3,pos=4,"Cycles Length Distribution",cex=1.0,font=2)
> axis(2,at=as.numeric(bbLab)-
> 41,labels=bbLab,col="red",col.axis="red",font.axis=1,
> lab=c(10,10,15),cex.lab=0.7,cex.axis=0.6)
>
> axis(3,at=zzLab,labels=as.character(zzLab),col="red",col.axis="red",fon
> t.axis=1,xpd=TRUE,
> cex.lab=1,cex.axis=0.8)
>
>
> # cords <-locator(n=3)
>
>
>
>
>
>
> e tutti i telefonini TIM!
> Vai su
------------------------------
Message: 53
Date: Wed, 28 Jan 2009 13:25:32 -0500 (EST)
From: Nidhi Kohli <nidhik at umd.edu>
Subject: [R] stack data sets
To: r-help at stat.math.ethz.ch, r-help at r-project.org
Message-ID: <20090128132532.BIX30409 at po3.mail.umd.edu>
Content-Type: text/plain; charset=us-ascii
Hi All,
I'm generating 10 different data sets with 1 and 0 in a matrix form and writing the output in separate files. Now I need to stack all these data sets in one vector and I know that stack only operates on list or data frame however I got these data sets by converting list to a matrix so can't go backwards now. Is there a way i can still use Stack?
Please see the program:
#Importing psych & ltm library for all the simulation related functions
library(ltm)
library(psych)
# Settting the working directory path to C:/NCME
path="C:/NCME"
setwd(path)
#IRT Data Simulation Routine#
n.exams = 500 #Sets number of examinees to be generated#
n.items = 20 #Sets number of items to be generated#
#The following intialize empty (NA) vectors or matrices#
beta.values = rep(NA,n.items)
resp.prob=matrix(rep(NA, n.exams*n.items), nrow=n.exams, ncol=n.items)
Observed_Scores=matrix(rep(NA, n.exams*n.items), nrow=n.exams, ncol=n.items)
str(Observed_Scores)
for (k in 1:10)
{
#Setting the starting point for seed
set.seed(k)
#filling item parameters into beta.values
beta.values = runif(n.items,-2,2)
#Calculating Threshold
thresh.values = .5 * beta.values
#Using the function to generate the Parallel Model CTT data
GenData <- congeneric.sim(N=500, loads = rep(.5,20), err=NULL, short = FALSE)
#Storing Observed Score in a variable
Observed_Scores = GenData[[3]]
#Exporting Observed scores to output file
ObservedScores_Data <- paste("Observed_Scores_",k,".dat")
write.table(Observed_Scores,ObservedScores_Data,row.name=FALSE,col.name=FALSE)
Zero = 0
One = 1
for (t in 1:20)
{
for (s in 1:500)
{
if (Observed_Scores[s,t]<= thresh.values[t])
resp.prob[s,t] = Zero
else
resp.prob[s,t] = One
}
}
ResponseData <- paste("ResponseMatrix_",k,".dat")
ThreshData <- paste("Threshold_",k,".dat")
write.table(resp.prob,ResponseData,row.name=FALSE,col.name=FALSE)
write.table(thresh.values,ThreshData,row.name=FALSE,col.name=FALSE)
#####STACKING ALL THE OUTPUTS#########
CommonFile <- stack(resp.prob)
######################################
#Rounding upto 2 decimal places while showing the correlation matrix
round(cor(GenData$observed),2)
#Factor Score
FactorScore=factor.pa(GenData$observed,1,scores = "TRUE")
round(cor(FactorScore$scores,GenData$latent),2)
filename_fs <- paste("FactorScore_",k,".dat")
#Exporting Factor Scores to Output file
write.table(FactorScore$scores,filename_fs,col.name=FALSE, row.name=FALSE)
}
Thank you
Nidhi
------------------------------
Message: 54
Date: Wed, 28 Jan 2009 13:25:32 -0500 (EST)
From: Nidhi Kohli <nidhik at umd.edu>
Subject: [R] stack data sets
To: r-help at stat.math.ethz.ch, r-help at r-project.org
Message-ID: <20090128132532.BIX30409 at po3.mail.umd.edu>
Content-Type: text/plain; charset=us-ascii
Hi All,
I'm generating 10 different data sets with 1 and 0 in a matrix form and writing the output in separate files. Now I need to stack all these data sets in one vector and I know that stack only operates on list or data frame however I got these data sets by converting list to a matrix so can't go backwards now. Is there a way i can still use Stack?
Please see the program:
#Importing psych & ltm library for all the simulation related functions
library(ltm)
library(psych)
# Settting the working directory path to C:/NCME
path="C:/NCME"
setwd(path)
#IRT Data Simulation Routine#
n.exams = 500 #Sets number of examinees to be generated#
n.items = 20 #Sets number of items to be generated#
#The following intialize empty (NA) vectors or matrices#
beta.values = rep(NA,n.items)
resp.prob=matrix(rep(NA, n.exams*n.items), nrow=n.exams, ncol=n.items)
Observed_Scores=matrix(rep(NA, n.exams*n.items), nrow=n.exams, ncol=n.items)
str(Observed_Scores)
for (k in 1:10)
{
#Setting the starting point for seed
set.seed(k)
#filling item parameters into beta.values
beta.values = runif(n.items,-2,2)
#Calculating Threshold
thresh.values = .5 * beta.values
#Using the function to generate the Parallel Model CTT data
GenData <- congeneric.sim(N=500, loads = rep(.5,20), err=NULL, short = FALSE)
#Storing Observed Score in a variable
Observed_Scores = GenData[[3]]
#Exporting Observed scores to output file
ObservedScores_Data <- paste("Observed_Scores_",k,".dat")
write.table(Observed_Scores,ObservedScores_Data,row.name=FALSE,col.name=FALSE)
Zero = 0
One = 1
for (t in 1:20)
{
for (s in 1:500)
{
if (Observed_Scores[s,t]<= thresh.values[t])
resp.prob[s,t] = Zero
else
resp.prob[s,t] = One
}
}
ResponseData <- paste("ResponseMatrix_",k,".dat")
ThreshData <- paste("Threshold_",k,".dat")
write.table(resp.prob,ResponseData,row.name=FALSE,col.name=FALSE)
write.table(thresh.values,ThreshData,row.name=FALSE,col.name=FALSE)
#####STACKING ALL THE OUTPUTS#########
CommonFile <- stack(resp.prob)
######################################
#Rounding upto 2 decimal places while showing the correlation matrix
round(cor(GenData$observed),2)
#Factor Score
FactorScore=factor.pa(GenData$observed,1,scores = "TRUE")
round(cor(FactorScore$scores,GenData$latent),2)
filename_fs <- paste("FactorScore_",k,".dat")
#Exporting Factor Scores to Output file
write.table(FactorScore$scores,filename_fs,col.name=FALSE, row.name=FALSE)
}
Thank you
Nidhi
------------------------------
Message: 55
Date: Wed, 28 Jan 2009 18:54:27 +0000 (UTC)
From: Ben Bolker <bolker at ufl.edu>
Subject: Re: [R] constrainOptim
To: r-help at stat.math.ethz.ch
Message-ID: <loom.20090128T184853-604 at post.gmane.org>
Content-Type: text/plain; charset=us-ascii
June Wong <neptune545 <at> hotmail.com> writes:
>
>
> Dear R helpers
>
> I have a question regarding the constrainOptim.
> I'm coding the nested logit and would like to set a bound of rho to (0,1] as
an extreme value distribution
> where rho = exp(lambda)/1+exp(lambda)
> I wonder if I can do that directly in optim (say rho > 0 & <= 1) or need to
use constrainOptim
> I read the help but still don't know how to set ui and ci
>
> Thanks,
> June
>
optim() can do box constraints (i.e., independent inequality
constraints on parameters): use method="L-BFGS-B" and
the lower and upper arguments to set the bounds for
each parameter (to -Inf and Inf if there are no bounds).
If you want to set bounds on rho you have to use rho as
the parameter in your model -- this is tricky if you
can't solve for rho, but in your case lambda=log(rho/(1-rho))
Ben Bolker
------------------------------
Message: 56
Date: Wed, 28 Jan 2009 18:26:57 +0100
From: julien cuisinier <j_cuisinier at hotmail.com>
Subject: [R] Saving plot into file without showing it
To: <r-help at r-project.org>
Message-ID: <COL102-W425B97D7E5E340366CB9128FC80 at phx.gbl>
Content-Type: text/plain
Hi List,
My apologies in advance if question is simplistic, I am quite new to R graphics capabilities and I could not find anything in past threads...
I use R 2.8.1 under Mac OS X, but I would preferrably have a cross platform answer as I use also R under Windows
I produce plots using R & save them in a file
e.g. below:
y <- rnorm(1000)
x <- rnorm(1000)
plot(x,y)
dev.copy2pdf()
Until there fine, it create a pdf file that is composed of my plot...My "issues" are the following:
1. If I want to produce the plot & save it directly in a pdf file without viewing it, how do I do that?
2. Can I create several plots in a row (without showing them in Quartz or whatever other graphic device) and save them all in separate files after creation? for example a function that would save me in separate files all what is visible through dev.list()
Let's keep the example of saving in pdf format here...I do not have target file type for saving the graphics. The point is that I would have another piece of code (HTML I guess, not developed yet) fetching all the charts and presenting it nicely.
Any feedback is appreciated
Many thanks
Julien
_________________________________________________________________
charlas.
[[alternative HTML version deleted]]
------------------------------
Message: 57
Date: Wed, 28 Jan 2009 08:48:16 -0800 (PST)
From: Frank Zhang <frankyuzhang at yahoo.com>
Subject: [R] Get median of each column
To: r-help at r-project.org
Message-ID: <724831.30195.qm at web33007.mail.mud.yahoo.com>
Content-Type: text/plain
I am new to R. How can I get column median? Thanks.Frank
[[alternative HTML version deleted]]
------------------------------
Message: 58
Date: Wed, 28 Jan 2009 18:37:58 -0000
From: "Attiglah, Mama" <Mama_Attiglah at ssga.com>
Subject: [R] R compilation
To: <r-sig-finance at stat.math.ethz.ch>
Cc: r-help at r-project.org, r-sig-hpc at r-project.org,
r-sig-db at stat.math.ethz.ch
Message-ID:
<AD34C27D4F3A0649ACE5DB999EC62E50E050BA at LCPPW1089.corp.statestr.com>
Content-Type: text/plain; charset="iso-8859-1"
Hi Mates,
I have a very long R code that needs to go to production but my portfolio managers do not use R language and the software is not supported by my bank.
Is there any way I can compile the code to an executable file and make it usable to my portfolio managers who have no knowledge at all of R?
Thanks
Mama
-----
Mama Attiglah, PhD
Quantitative Strategist
Liability Driven Investment
State Street Global Advisors
25 Bank Street, London E14 5NU
+44(0)20 7698 6290 (Direct Line)
+44 (0)207 004 2968 (Direct Fax)
Authorised and regulated by the Financial Services Authority.
State Street Global Advisors Limited, a company registered in England with company number 2509928
and VAT number 5576591 81 and whose registered office is...{{dropped:12}}
------------------------------
Message: 59
Date: Wed, 28 Jan 2009 14:38:15 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] Get median of each column
Cc: r-help at r-project.org
Message-ID:
<644e1f320901281138m7935f39cr77e92d9ccb67b35b at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
?apply
> x <- matrix(1:25,5)
> x
[,1] [,2] [,3] [,4] [,5]
[1,] 1 6 11 16 21
[2,] 2 7 12 17 22
[3,] 3 8 13 18 23
[4,] 4 9 14 19 24
[5,] 5 10 15 20 25
> apply(x, 2, median)
[1] 3 8 13 18 23
>
wrote:
> I am new to R. How can I get column median? Thanks.Frank
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 60
Date: Wed, 28 Jan 2009 14:39:40 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] Saving plot into file without showing it
To: julien cuisinier <j_cuisinier at hotmail.com>
Cc: r-help at r-project.org
Message-ID:
<644e1f320901281139h134ba472nb3673fddbfc79e05 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
pdf("yourFile.pdf")
plot(1)
plot(2)
plot(3)
.....
dev.off()
On Wed, Jan 28, 2009 at 12:26 PM, julien cuisinier
<j_cuisinier at hotmail.com> wrote:
>
> Hi List,
>
>
> My apologies in advance if question is simplistic, I am quite new to R graphics capabilities and I could not find anything in past threads...
>
> I use R 2.8.1 under Mac OS X, but I would preferrably have a cross platform answer as I use also R under Windows
>
> I produce plots using R & save them in a file
>
> e.g. below:
>
> y <- rnorm(1000)
> x <- rnorm(1000)
> plot(x,y)
> dev.copy2pdf()
>
> Until there fine, it create a pdf file that is composed of my plot...My "issues" are the following:
> 1. If I want to produce the plot & save it directly in a pdf file without viewing it, how do I do that?
> 2. Can I create several plots in a row (without showing them in Quartz or whatever other graphic device) and save them all in separate files after creation? for example a function that would save me in separate files all what is visible through dev.list()
>
> Let's keep the example of saving in pdf format here...I do not have target file type for saving the graphics. The point is that I would have another piece of code (HTML I guess, not developed yet) fetching all the charts and presenting it nicely.
>
>
> Any feedback is appreciated
>
> Many thanks
> Julien
>
> _________________________________________________________________
>
> charlas.
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 61
Date: Wed, 28 Jan 2009 20:39:09 +0100
From: Stephan Kolassa <Stephan.Kolassa at gmx.de>
Subject: Re: [R] Get median of each column
Cc: "r-help at r-project.org" <r-help at r-project.org>
Message-ID: <4980B45D.4000204 at gmx.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Assuming your data are in a data.frame called dataset,
apply(dataset,2,median)
should work. Look at
?apply
HTH,
Stephan
Frank Zhang schrieb:
> I am new to R. How can I get column median? Thanks.Frank
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
------------------------------
Message: 62
Date: Wed, 28 Jan 2009 20:40:33 +0100
From: Stephan Kolassa <Stephan.Kolassa at gmx.de>
Subject: Re: [R] Saving plot into file without showing it
To: julien cuisinier <j_cuisinier at hotmail.com>
Cc: r-help at r-project.org
Message-ID: <4980B4B1.9090504 at gmx.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Try
pdf("foo.pdf")
plot(x)
dev.off()
Other possibilities are jpeg(), tiff(), postscript() etc.
HTH,
Stephan
julien cuisinier schrieb:
> Hi List,
>
>
> My apologies in advance if question is simplistic, I am quite new to R graphics capabilities and I could not find anything in past threads...
>
> I use R 2.8.1 under Mac OS X, but I would preferrably have a cross platform answer as I use also R under Windows
>
> I produce plots using R & save them in a file
>
> e.g. below:
>
> y <- rnorm(1000)
> x <- rnorm(1000)
> plot(x,y)
> dev.copy2pdf()
>
> Until there fine, it create a pdf file that is composed of my plot...My "issues" are the following:
> 1. If I want to produce the plot & save it directly in a pdf file without viewing it, how do I do that?
> 2. Can I create several plots in a row (without showing them in Quartz or whatever other graphic device) and save them all in separate files after creation? for example a function that would save me in separate files all what is visible through dev.list()
>
> Let's keep the example of saving in pdf format here...I do not have target file type for saving the graphics. The point is that I would have another piece of code (HTML I guess, not developed yet) fetching all the charts and presenting it nicely.
>
>
> Any feedback is appreciated
>
> Many thanks
> Julien
>
> _________________________________________________________________
>
> charlas.
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
------------------------------
Message: 63
Date: Wed, 28 Jan 2009 20:44:53 +0100
From: "Strubbe Diederik" <diederik.strubbe at ua.ac.be>
Subject: Re: [R] Repeated measures design for GAM? - corrected
question...
To: <r-help at R-project.org>
Message-ID:
<C9854550FEF14846A136100B3EC52F73B6B5AF at xmail05.ad.ua.ac.be>
Content-Type: text/plain
Dear Simon,
Many thanks for pointing me to the GAMM! For clarification, Bird_abundance are breeding densities ( e.g. 1.25 BP/ha, 2.20 BP/ha,...) and count is just the actual survey(e.g. first_survey,...). The dataset looks like
Bird_abundance Study_area YEAR COUNT X1 X2 X3 X4
0.15 area_1 2004 first_survey … … … …
1.26 area_1 2004 second_survey … … … …
2.47 area_1 2005 third_survey … … … …
0.00 area_1 2005 fourth_survey … … … …
0.23 area_1 2006 fifht_survey … … … …
2.64 area_1 2006 sixth_survey … … … …
4.14 area_2 2004 first_survey … … … …
5.00 area_2 2004 second_survey … … … …
6.80 area_2 2005 third_survey … … … …
0.15 area_2 2005 fourth_survey … … … …
0.25 area_2 2006 fifht_survey … … … …
2.36 area_2 2006 sixth_survey … … … …
2.59 area_3 2004 first_survey … … … …
6.31 area_3 2004 second_survey … … … …
0.15 area_3 2005 third_survey … … … …
2.85 area_3 2005 fourth_survey … … … …
2.48 area_3 2006 fifht_survey … … … …
1.23 area_3 2006 sixth_survey … … … …
… … … … … … … …
Am I correct in assuming the following is a valid syntax for this repeated measures design?:
model <-gamm(Bird_abundance ~ YEAR + s(X1)+ s(X2)+ s(X3)+ s(X4),random=list(count=~1,park=~1))
best wishes and thanks again,
Diederik
Diederik Strubbe
Evolutionary Ecology Group
Department of Biology, University of Antwerp
Universiteitsplein 1
B-2610 Antwerp, Belgium
http://webhost.ua.ac.be/deco
tel : 32 3 820 23 85
-----Original Message-----
From: Strubbe Diederik
Sent: Wed 28-1-2009 18:09
To: r-help at R-project.org
Subject: Repeated measures design for GAM? - corrected question...
Dear all,
I have a question on the use of GAM with repeated measures. My dataset is as follows:
- a number of study areas where bird abundance has been determined. Counts have been performed in 3 consecutive years and there were 2 counts per year (i.e. in total 6 counts).
- a number of environmental predictors that do not change over year Xi).
When using a GLM, a repeated measures design would like: (for example)
lme(Bird_abundance = study_area + count +year+ X1 + X2 + X3,random = ~count|study_area).
However, I have found no analogue design for a GAM. For now, I have averaged my bird abundances but I wondered whether a more subtle and elegant strategy exists...?
Many thanks,
Diederik
Diederik Strubbe
Evolutionary Ecology Group
Department of Biology, University of Antwerp
Universiteitsplein 1
B-2610 Antwerp, Belgium
http://webhost.ua.ac.be/deco
tel : 32 3 820 23 85
[[alternative HTML version deleted]]
------------------------------
Message: 64
Date: Thu, 29 Jan 2009 08:53:44 +1300
From: Rolf Turner <r.turner at auckland.ac.nz>
Subject: Re: [R] Get median of each column
To: Stephan Kolassa <Stephan.Kolassa at gmx.de>
Cc: "r-help at r-project.org" <r-help at r-project.org>
Message-ID: <81B1D521-D4BE-451D-BC74-38B17F9D2EBE at auckland.ac.nz>
Content-Type: text/plain; charset="US-ASCII"; format=flowed
On 29/01/2009, at 8:39 AM, Stephan Kolassa wrote:
> Assuming your data are in a data.frame called dataset,
>
> apply(dataset,2,median)
>
> should work. Look at
>
> ?apply
Note that apply() works with ***matrices***. The foregoing code will
work, given that all columns of ``dataset'' are numeric, due to the
fact that apply will *coerce* a data frame to a matrix.
However it should always be remembered that
[[elided Yahoo spam]]
cheers,
Rolf Turner
######################################################################
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
------------------------------
Message: 65
Date: Wed, 28 Jan 2009 21:21:07 +0100
From: Stephan Kolassa <Stephan.Kolassa at gmx.de>
Subject: Re: [R] Power analysis for MANOVA?
To: adik at ilovebacon.org
Cc: r-help at r-project.org
Message-ID: <4980BE33.3020008 at gmx.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi Adam,
first: I really don't know much about MANOVA, so I sadly can't help you
without learning about it an Pillai's V... which I would be glad to do,
[[elided Yahoo spam]]
Second: you seem to be doing a kind of "post-hoc power analysis", "my
result isn't significant, perhaps that's due to low power? Let's look at
the power of my experiment!" My impression is that "post-hoc power
analysis" and its interpretation is, shall we say, not entirely accepted
within the statistical community, see:
Hoenig, J. M., & Heisey, D. M. (2001, February). The abuse of power: The
pervasive fallacy of power calculations for data analysis. The American
Statistician, 55 (1), 1-6
And this:
http://staff.pubhealth.ku.dk/~bxc/SDC-courses/power.pdf
However, I am sure that lots of people can discuss this more competently
than me...
Best wishes
Stephan
Adam D. I. Kramer schrieb:
>
> On Mon, 26 Jan 2009, Stephan Kolassa wrote:
>
>> My (and, judging from previous traffic on R-help about power analyses,
>> also some other people's) preferred approach is to simply simulate an
>> effect size you would like to detect a couple of thousand times, run your
>> proposed analysis and look how often you get significance. In your
>> simple
>> case, this should be quite easy.
>
> I actually don't have much experience running monte-carlo designs like
> this...so while I'd certainly prefer a bootstrapping method like this one,
> simulating the effect size given my constraints isn't something I've done
> before.
>
> The MANOVA procedure takes 5 dependent variables, and determines what
> combination of the variables best discriminates the two levels of my
> independent variable...then the discrimination rate is represented in the
> statistic (Pillai's V=.00019), which is then tested (F[5,18653] =
> 0.71). So
> coming up with a set of constraints that would produce V=.00019 given my
> data set doesn't quite sound trivial...so I'll go for the "par" library
> reference mentioned earlier before I try this. That said, if anyone can
> refer me to a tool that will help me out (or an instruction manual for
> RNG),
> I'd also be much obliged.
>
> Many thanks,
> Adam
>
>
>>
>> HTH,
>> Stephan
>>
>>
>> Adam D. I. Kramer schrieb:
>>> Hello,
>>>
>>> I have searched and failed for a program or script or method to
>>> conduct a power analysis for a MANOVA. My interest is a fairly simple
>>> case
>>> of 5 dependent variables and a single two-level categorical predictor
>>> (though the categories aren't balanced).
>>>
>>> If anybody happens to know of a script that will do this in R, I'd
>>> love to know of it! Otherwise, I'll see about writing one myself.
>>>
>>> What I currently see is this, from help.search("power"):
>>>
>>> stats::power.anova.test
>>> Power calculations for balanced one-way
>>> analysis of variance tests
>>> stats::power.prop.test
>>> Power calculations two sample test for
>>> proportions
>>> stats::power.t.test Power calculations for one and two sample t
>>> tests
>>>
>>> Any references on power in MANOVA would also be helpful, though of
>>> course I will do my own lit search for them myself.
>>>
>>> Cordially,
>>> Adam D. I. Kramer
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>
------------------------------
Message: 66
Date: Wed, 28 Jan 2009 21:28:00 +0100
From: Stephan Kolassa <Stephan.Kolassa at gmx.de>
Subject: Re: [R] Get median of each column
To: Rolf Turner <r.turner at auckland.ac.nz>
Cc: "r-help at r-project.org" <r-help at r-project.org>
Message-ID: <4980BFD0.5070301 at gmx.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Thank you, Rolf, for this well-deserved spanking :-)
I promise to amend my ways and think before I send in the future.
Best,
Stephan
Rolf Turner schrieb:
>
> On 29/01/2009, at 8:39 AM, Stephan Kolassa wrote:
>
>> Assuming your data are in a data.frame called dataset,
>>
>> apply(dataset,2,median)
>>
>> should work. Look at
>>
>> ?apply
>
> Note that apply() works with ***matrices***. The foregoing code will
> work, given that all columns of ``dataset'' are numeric, due to the
> fact that apply will *coerce* a data frame to a matrix.
>
> However it should always be remembered that
>
[[elided Yahoo spam]]
>
> cheers,
>
> Rolf Turner
>
> ######################################################################
> Attention: This e-mail message is privileged and confidential. If you
> are not the intended recipient please delete the message and notify the
> sender. Any views or opinions presented are solely those of the author.
>
> This e-mail has been scanned and cleared by MailMarshal
> www.marshalsoftware.com
> ######################################################################
>
------------------------------
Message: 67
Date: Wed, 28 Jan 2009 14:33:47 -0600
From: Kumudan <cybermails at gmail.com>
Subject: [R] Neighborhood distance calculator
To: r-help at r-project.org
Message-ID:
<d5081b140901281233v44dca65es7ced065502379055 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Hi all,
I am new to R, hence this question. I have a set of points with X and
Y coordinates. I would like to build an R code to calculate the
distances of points within a specified neighborhood (circular range)
for each point. I would like the code to be a function, so that I can
call the function, specifying parameters (data file, distance), from
another piece of code. Also, I would like the output to be as:
output = list(array)
.........code...........
........new code.....
return(output)
Something like this--
[[1496]]
[1] 1490 1491 1492 1493 1494 1495
The most important thing, however, is that I want to run the code for
100,000+ points. The current code I have calculates the neighborhood
distances for all the points with ever point, and then selects only
those points within the specified distance parameter. This code
crashes when I have over 40,000 points. Therefore, I need to figure
out a way to preclude the step where it calculates all of the paired
distances, and rather only those points within the specified distance
parameter. Please, see below for the existing code, and a sample data.
Thanks a lot for your time and effort.
Best,
Kumudan
Kumudan Grubh
EEB Graduate student
Iowa State University
-------------------------------------------------------------------------------------------------------------------------------------------------------
#--Code--
#transform coordinates in kilometers, and redefine origin
newx=(XCOORD-min(XCOORD))/1000
newy=(YCOORD-min(YCOORD))/1000
plot(newx,newy,pch=20,xlab="X coordinate (km)",ylab="Y coordinate (km)")
#set of new locations to be used from here on
tr.locs=cbind(newx,newy)
#the functions necessary for estimation are written in an external
program. All you need to do is to run this program one. To do this,
you could "source" this code
source("D:/...../Functions_distance.R")
#define distance-based neighborhood
nb.tr=dist.neighbors(tr.locs,2)
----------------------------------------------------------------------------------------------------------------------------------------------------------
#---Function_distance.R <- Function to calculate distance of pairs
of neighbors
#start by defining neighbors
library(spatstat)#need this to compute distances between all points in a data
dist.neighbors=function(data,distance)
#distance is set by user
{
#data has the x and y on the first two columns. Could have more than
two columns, or exactly two columns, it doesn't matter.
#produces a "list" object with a vector of the neighbors for each location
#distance defines the range within which we define neighbors
nd=length(data[,1])
nbmat=list(array)
dist=pairdist(data[,1:2])
for(i in 1:nd){
nbmat[[i]]=0
for(j in 1:nd) if((dist[i,j]<distance)&(i!=j)) nbmat[[i]]=c(nbmat[[i]],j)
nbmat[[i]]=nbmat[[i]][nbmat[[i]]!=0]}
return(nbmat)
}
------------------------------------------------------------------------------------------------------------------------------------
Data
-----------------------------------------------------------------------------------
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------------------------------
Message: 68
Date: Wed, 28 Jan 2009 20:41:08 +0000
From: glenn <g1enn.roberts at btinternet.com>
Subject: [R] Cor(df,method = "kendall")
To: "r-help at r-project.org" <r-help at r-project.org>
Message-ID: <C5A67364.1140%g1enn.roberts at btinternet.com>
Content-Type: text/plain
Hi All,
Does anyone know of any issues at all with using;
Cor(df,method = ³kendall²)
On a dataframe (df) 13 columns wide say?
Seems to hang my system for a while in calculating the correlation matrix
appreciate it is doing some ranking calculations so I am expecting too much
that it should return immediately ? In particular trying to use function in
Excel (reval) and I am getting OLE error boxes as RGUI hangs.
Many Thanks.
Glenn
[[alternative HTML version deleted]]
------------------------------
Message: 69
Date: Wed, 28 Jan 2009 15:46:43 -0500
From: stephen sefick <ssefick at gmail.com>
Subject: Re: [R] R compilation
To: "Attiglah, Mama" <Mama_Attiglah at ssga.com>
Cc: r-sig-finance at stat.math.ethz.ch, r-help at r-project.org,
r-sig-hpc at r-project.org, r-sig-db at stat.math.ethz.ch
Message-ID:
<c502a9e10901281246u2d14b1d7od4161f733f9bfc0f at mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
yes, first don't crosspost. It all depends on what OS .. blah, blah,
blah. You must read the posting guide because it will up your chances
of a reply.
On Wed, Jan 28, 2009 at 1:37 PM, Attiglah, Mama <Mama_Attiglah at ssga.com> wrote:
>
> Hi Mates,
> I have a very long R code that needs to go to production but my portfolio managers do not use R language and the software is not supported by my bank.
> Is there any way I can compile the code to an executable file and make it usable to my portfolio managers who have no knowledge at all of R?
>
> Thanks
>
> Mama
> -----
> Mama Attiglah, PhD
> Quantitative Strategist
> Liability Driven Investment
> State Street Global Advisors
> 25 Bank Street, London E14 5NU
> +44(0)20 7698 6290 (Direct Line)
> +44 (0)207 004 2968 (Direct Fax)
> Authorised and regulated by the Financial Services Authority.
> State Street Global Advisors Limited, a company registered in England with company number 2509928
> and VAT number 5576591 81 and whose registered office ...{{dropped:21}}
------------------------------
Message: 70
Date: Wed, 28 Jan 2009 14:51:25 -0600
From: roger koenker <rkoenker at uiuc.edu>
Subject: Re: [R] Neighborhood distance calculator
To: Kumudan <cybermails at gmail.com>
Cc: r-help <r-help at r-project.org>
Message-ID: <FEEDB763-0E6E-4E7F-8A6D-C10B94884FC3 at uiuc.edu>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
If you are willing to dig into the adjacency information returned from
tri.mesh in the tripack package, this could be done quite efficiently.
url: www.econ.uiuc.edu/~roger Roger Koenker
email rkoenker at uiuc.edu Department of Economics
vox: 217-333-4558 University of Illinois
fax: 217-244-6678 Champaign, IL 61820
On Jan 28, 2009, at 2:33 PM, Kumudan wrote:
> Hi all,
>
> I am new to R, hence this question. I have a set of points with X and
> Y coordinates. I would like to build an R code to calculate the
> distances of points within a specified neighborhood (circular range)
> for each point. I would like the code to be a function, so that I can
> call the function, specifying parameters (data file, distance), from
> another piece of code. Also, I would like the output to be as:
>
> output = list(array)
> .........code...........
> ........new code.....
> return(output)
>
> Something like this--
> [[1496]]
> [1] 1490 1491 1492 1493 1494 1495
>
> The most important thing, however, is that I want to run the code for
> 100,000+ points. The current code I have calculates the neighborhood
> distances for all the points with ever point, and then selects only
> those points within the specified distance parameter. This code
> crashes when I have over 40,000 points. Therefore, I need to figure
> out a way to preclude the step where it calculates all of the paired
> distances, and rather only those points within the specified distance
> parameter. Please, see below for the existing code, and a sample data.
> Thanks a lot for your time and effort.
>
> Best,
> Kumudan
>
> Kumudan Grubh
> EEB Graduate student
> Iowa State University
>
> -------------------------------------------------------------------------------------------------------------------------------------------------------
> #--Code--
> #transform coordinates in kilometers, and redefine origin
> newx=(XCOORD-min(XCOORD))/1000
> newy=(YCOORD-min(YCOORD))/1000
> plot(newx,newy,pch=20,xlab="X coordinate (km)",ylab="Y coordinate
> (km)")
>
> #set of new locations to be used from here on
> tr.locs=cbind(newx,newy)
>
> #the functions necessary for estimation are written in an external
> program. All you need to do is to run this program one. To do this,
> you could "source" this code
> source("D:/...../Functions_distance.R")
>
> #define distance-based neighborhood
> nb.tr=dist.neighbors(tr.locs,2)
> ----------------------------------------------------------------------------------------------------------------------------------------------------------
> #---Function_distance.R <- Function to calculate distance of pairs
> of neighbors
> #start by defining neighbors
> library(spatstat)#need this to compute distances between all points
> in a data
>
> dist.neighbors=function(data,distance)
> #distance is set by user
> {
> #data has the x and y on the first two columns. Could have more than
> two columns, or exactly two columns, it doesn't matter.
> #produces a "list" object with a vector of the neighbors for each
> location
> #distance defines the range within which we define neighbors
> nd=length(data[,1])
> nbmat=list(array)
> dist=pairdist(data[,1:2])
> for(i in 1:nd){
> nbmat[[i]]=0
> for(j in 1:nd) if((dist[i,j]<distance)&(i!=j))
> nbmat[[i]]=c(nbmat[[i]],j)
> nbmat[[i]]=nbmat[[i]][nbmat[[i]]!=0]}
> return(nbmat)
> }
> ------------------------------------------------------------------------------------------------------------------------------------
>
> Data
> -----------------------------------------------------------------------------------
> XCOORD YCOORD
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>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 71
Date: Wed, 28 Jan 2009 12:03:50 -0800 (PST)
From: pfc_ivan <pfc_ivan at hotmail.com>
Subject: [R] Newbie Question About Histograms
To: r-help at r-project.org
Message-ID: <21713626.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Hello everyone. Just have a question , cant figure out how to make this
histogram.
I have this table, that i stored in a variable name new.data2. Table looks
like this
Year GeoArea SmpNo Month Gear Maturity Length Age YearC
1989 1 362 10 22 1 225 1 1988
1991 1 267 10 10 1 191 1 1990
1991 1 267 10 10 1 202 1 1990
1992 1 305 10 8 1 162 1 1991
1992 1 305 10 8 1 165 1 1991
1992 1 305 10 8 1 166 1 1991
1992 1 305 10 8 1 167 1 1991
1992 1 305 10 8 1 167 1 1991
1992 1 305 10 8 1 169 1 1991
1992 1 305 10 8 1 170 1 1991
Now I need to make a histogram of Length vs YearC. I would guess that Length
would be on the Y-axis and YearC variable would be on X-axis. I have tried
many different combinations with command 'hist' but im always getting error
" 'x' must be numeric " ... I think im getting that error because of the
header which is not numeric. Any help would be appreciated. Thanks guys.
Ivan.
--
View this message in context: http://www.nabble.com/Newbie-Question-About-Histograms-tp21713626p21713626.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 72
Date: Wed, 28 Jan 2009 12:04:18 -0800 (PST)
From: Sea Captain 1779 <dbowen at measinc.com>
Subject: [R] Help with normal distribution in random samples...
To: r-help at r-project.org
Message-ID: <21713636.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Hi!!!
First time 'R' user looking for a little assistance. Here is what I have so
far:
practice1 = matrix ((runif(5000, min =0, max = 12)), 100)
which is creating 50 samples, for 100 cases, distributed between 0-12. What
I would like is to be able to set the mean and SD so that the data is
normally distributed around lets say 7. Any help I can get with achieving
[[elided Yahoo spam]]
-Dan
--
View this message in context: http://www.nabble.com/Help-with-normal-distribution-in-random-samples...-tp21713636p21713636.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 73
Date: Wed, 28 Jan 2009 12:37:57 -0800 (PST)
From: Alice Lin <alice.ly at gmail.com>
Subject: [R] Text data
To: r-help at r-project.org
Message-ID: <21714334.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
i have a data column of text entries:
26M_AN_C.bmp
22M_AN_C.bmp
20M_HA_O.bmp
20M_AN_C.bmp
26M_HA_O.bmp
22M_HA_O.bmp
31M_AN_C.bmp
38M_HA_O.bmp
.
.
.
.
And I would like to sort by the middle tag: AN, HA, etc.
Is there a way to parse text data in R?
In excel, I would have used the "left" and "right" function to cut out just
the middle two letters out and put into another column to sort by.
Thanks!
--
View this message in context: http://www.nabble.com/Text-data-tp21714334p21714334.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 74
Date: Thu, 29 Jan 2009 10:10:38 +1300
From: "Peter Alspach" <PAlspach at hortresearch.co.nz>
Subject: Re: [R] Newbie Question About Histograms
To: "pfc_ivan" <pfc_ivan at hotmail.com>, <r-help at r-project.org>
Message-ID:
<EC0F8FF776F3F74E9C63CE16641C9628037A486D at AKLEXB02.hort.net.nz>
Content-Type: text/plain; charset="us-ascii"
Kia ora Ivan
I think you might want a barplot.
?hist
under 'See Also:' states:
Typical plots with vertical bars are _not_ histograms. Consider
'barplot' or 'plot(*, type = "h")' for such bar plots.
[The online help in R is good.]
HTH ....
Peter Alspach
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of pfc_ivan
> Sent: Thursday, 29 January 2009 9:04 a.m.
> To: r-help at r-project.org
> Subject: [R] Newbie Question About Histograms
>
>
> Hello everyone. Just have a question , cant figure out how to
> make this histogram.
>
> I have this table, that i stored in a variable name
> new.data2. Table looks like this
>
> Year GeoArea SmpNo Month Gear Maturity Length Age YearC
> 1989 1 362 10 22 1 225 1 1988
> 1991 1 267 10 10 1 191 1 1990
> 1991 1 267 10 10 1 202 1 1990
> 1992 1 305 10 8 1 162 1 1991
> 1992 1 305 10 8 1 165 1 1991
> 1992 1 305 10 8 1 166 1 1991
> 1992 1 305 10 8 1 167 1 1991
> 1992 1 305 10 8 1 167 1 1991
> 1992 1 305 10 8 1 169 1 1991
> 1992 1 305 10 8 1 170 1 1991
>
> Now I need to make a histogram of Length vs YearC. I would
> guess that Length would be on the Y-axis and YearC variable
> would be on X-axis. I have tried many different combinations
> with command 'hist' but im always getting error " 'x' must be
> numeric " ... I think im getting that error because of the
> header which is not numeric. Any help would be appreciated.
> Thanks guys.
>
> Ivan.
> --
> View this message in context:
> http://www.nabble.com/Newbie-Question-About-Histograms-tp21713
> 626p21713626.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
The contents of this e-mail are confidential and may be subject to legal privilege.
If you are not the intended recipient you must not use, disseminate, distribute or
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e-mail in error, please notify the sender and delete all material pertaining to this
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------------------------------
Message: 75
Date: Wed, 28 Jan 2009 13:11:57 -0800 (PST)
Subject: [R] Changing histogram stack in qplot
To: R-help at r-project.org
Message-ID: <186567.94177.qm at web56006.mail.re3.yahoo.com>
Content-Type: text/plain
I've been using qplot pretty successfully to generate stacked histograms. However, it appears that I need to tweak the colors a little.
I've got three temperature variables (characters not numeric) and I need to change from the default qplot colors to the following:
Low = Blue
Middle = black
High = Red
Here is pseudo code of what I have currently:qplot(Run, data = TestData, breaks = hist_breaks, ,
fill = TestData$Temperature,
main = short_title) +
scale_x_continuous("Run, Radians") + scale_y_continuous("Frequency") +
scale_fill_discrete("Temperature")
Thanks for any advice and insights.
[[alternative HTML version deleted]]
------------------------------
Message: 76
Date: Wed, 28 Jan 2009 16:14:48 -0500
From: "Antonio, Fabio Di Narzo" <antonio.fabio at gmail.com>
Subject: Re: [R] Neighborhood distance calculator
To: roger koenker <rkoenker at uiuc.edu>
Cc: r-help <r-help at r-project.org>, Kumudan <cybermails at gmail.com>
Message-ID:
<b0808fdc0901281314j20f9f995hd3b68d345a961cc at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
The tseriesChaos package has an internal function "find_knearests"
which deals with a similar problem using a box-assisted search
strategy to alleviate time complexity explosion.
However, you will have to dig into the C sources, as these have no man pages.
HTH,
f.
2009/1/28 roger koenker <rkoenker at uiuc.edu>:
> If you are willing to dig into the adjacency information returned from
> tri.mesh in the tripack package, this could be done quite efficiently.
>
> url: www.econ.uiuc.edu/~roger Roger Koenker
> email rkoenker at uiuc.edu Department of Economics
> vox: 217-333-4558 University of Illinois
> fax: 217-244-6678 Champaign, IL 61820
>
> On Jan 28, 2009, at 2:33 PM, Kumudan wrote:
>
>> Hi all,
>>
>> I am new to R, hence this question. I have a set of points with X and
>> Y coordinates. I would like to build an R code to calculate the
>> distances of points within a specified neighborhood (circular range)
>> for each point. I would like the code to be a function, so that I can
>> call the function, specifying parameters (data file, distance), from
>> another piece of code. Also, I would like the output to be as:
>>
>> output = list(array)
>> .........code...........
>> ........new code.....
>> return(output)
>>
>> Something like this--
>> [[1496]]
>> [1] 1490 1491 1492 1493 1494 1495
>>
>> The most important thing, however, is that I want to run the code for
>> 100,000+ points. The current code I have calculates the neighborhood
>> distances for all the points with ever point, and then selects only
>> those points within the specified distance parameter. This code
>> crashes when I have over 40,000 points. Therefore, I need to figure
>> out a way to preclude the step where it calculates all of the paired
>> distances, and rather only those points within the specified distance
>> parameter. Please, see below for the existing code, and a sample data.
>> Thanks a lot for your time and effort.
>>
>> Best,
>> Kumudan
>>
>> Kumudan Grubh
>> EEB Graduate student
>> Iowa State University
>>
>>
>> -------------------------------------------------------------------------------------------------------------------------------------------------------
>> #--Code--
>> #transform coordinates in kilometers, and redefine origin
>> newx=(XCOORD-min(XCOORD))/1000
>> newy=(YCOORD-min(YCOORD))/1000
>> plot(newx,newy,pch=20,xlab="X coordinate (km)",ylab="Y coordinate (km)")
>>
>> #set of new locations to be used from here on
>> tr.locs=cbind(newx,newy)
>>
>> #the functions necessary for estimation are written in an external
>> program. All you need to do is to run this program one. To do this,
>> you could "source" this code
>> source("D:/...../Functions_distance.R")
>>
>> #define distance-based neighborhood
>> nb.tr=dist.neighbors(tr.locs,2)
>>
>> ----------------------------------------------------------------------------------------------------------------------------------------------------------
>> #---Function_distance.R <- Function to calculate distance of pairs
>> of neighbors
>> #start by defining neighbors
>> library(spatstat)#need this to compute distances between all points in a
>> data
>>
>> dist.neighbors=function(data,distance)
>> #distance is set by user
>> {
>> #data has the x and y on the first two columns. Could have more than
>> two columns, or exactly two columns, it doesn't matter.
>> #produces a "list" object with a vector of the neighbors for each location
>> #distance defines the range within which we define neighbors
>> nd=length(data[,1])
>> nbmat=list(array)
>> dist=pairdist(data[,1:2])
>> for(i in 1:nd){
>> nbmat[[i]]=0
>> for(j in 1:nd) if((dist[i,j]<distance)&(i!=j))
>> nbmat[[i]]=c(nbmat[[i]],j)
>> nbmat[[i]]=nbmat[[i]][nbmat[[i]]!=0]}
>> return(nbmat)
>> }
>>
>> ------------------------------------------------------------------------------------------------------------------------------------
>>
>> Data
>>
>> -----------------------------------------------------------------------------------
>> XCOORD YCOORD
>> 544312.87500000000 4851169.00000000000
>> 542705.87500000000 4851165.00000000000
>> 541068.87500000000 4851155.00000000000
>> 537861.81250000000 4851123.00000000000
>> 536246.81250000000 4851118.00000000000
>> 539476.87500000000 4851116.00000000000
>> 534612.81250000000 4851113.00000000000
>> 533004.75000000000 4851095.00000000000
>> 542714.87500000000 4850363.00000000000
>> 543516.87500000000 4850360.00000000000
>> 544319.93750000000 4850358.00000000000
>> 541883.87500000000 4850353.00000000000
>> 541053.87500000000 4850342.50000000000
>> 540268.87500000000 4850326.50000000000
>> 537866.81250000000 4850313.50000000000
>> 538674.87500000000 4850311.50000000000
>> 539483.87500000000 4850310.50000000000
>> 537060.81250000000 4850310.00000000000
>> 536255.81250000000 4850307.00000000000
>> 535434.81250000000 4850306.00000000000
>> 534614.81250000000 4850305.00000000000
>> 533809.81250000000 4850297.00000000000
>> 533004.81250000000 4850289.00000000000
>> 542716.87500000000 4849553.50000000000
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>> 537867.87500000000 4849507.50000000000
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>> 542718.93750000000 4848744.50000000000
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>> 540292.87500000000 4848720.50000000000
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>> 537067.87500000000 4848701.50000000000
>> 535447.81250000000 4848697.50000000000
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>> 533025.87500000000 4845443.50000000000
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>> 533027.87500000000 4844634.50000000000
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>> 529812.81250000000 4844600.50000000000
>> 544357.93750000000 4843907.50000000000
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>> 542737.93750000000 4843901.50000000000
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>> 541132.93750000000 4843883.50000000000
>> 540323.93750000000 4843875.50000000000
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>> 538708.93750000000 4843861.50000000000
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>> 537102.87500000000 4843847.50000000000
>> 536303.87500000000 4843841.50000000000
>> 535473.87500000000 4843838.50000000000
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>> 533837.87500000000 4843830.50000000000
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>> 532224.87500000000 4843813.50000000000
>> 531419.87500000000 4843801.50000000000
>> 530615.81250000000 4843800.50000000000
>> 529811.81250000000 4843799.50000000000
>> 544356.93750000000 4843106.50000000000
>> 542737.93750000000 4843094.50000000000
>> 541133.93750000000 4843075.50000000000
>> 539515.93750000000 4843065.50000000000
>> 537906.87500000000 4843047.50000000000
>> 536303.87500000000 4843031.50000000000
>> 534654.87500000000 4843025.50000000000
>> 533033.87500000000 4843015.50000000000
>> 531419.81250000000 4842993.50000000000
>> 529808.81250000000 4842989.50000000000
>> 544356.93750000000 4842305.50000000000
>> 543547.93750000000 4842296.50000000000
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>> 532228.81250000000 4842195.50000000000
>> 531419.81250000000 4842185.50000000000
>> 530612.81250000000 4842182.50000000000
>> 529806.81250000000 4842179.50000000000
>> 544356.93750000000 4841500.50000000000
>> 542738.93750000000 4841483.50000000000
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>> 539520.93750000000 4841454.50000000000
>> 537913.87500000000 4841433.50000000000
>> 536304.87500000000 4841418.50000000000
>> 534667.87500000000 4841408.50000000000
>> 533040.81250000000 4841393.50000000000
>> 531425.81250000000 4841373.50000000000
>> 529811.81250000000 4841367.50000000000
>> 544356.93750000000 4840695.50000000000
>> 543547.93750000000 4840687.50000000000
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>> 540327.93750000000 4840652.50000000000
>> 539525.87500000000 4840647.50000000000
>> 538720.87500000000 4840637.50000000000
>> 537915.87500000000 4840627.50000000000
>> 537109.87500000000 4840621.50000000000
>> 536304.87500000000 4840615.50000000000
>> 535487.87500000000 4840609.50000000000
>> 534670.87500000000 4840603.50000000000
>> 533856.81250000000 4840591.50000000000
>> 533043.81250000000 4840580.50000000000
>> 532237.81250000000 4840571.50000000000
>> 531431.81250000000 4840562.50000000000
>> 530624.81250000000 4840559.50000000000
>> 529817.81250000000 4840556.50000000000
>> 544357.93750000000 4839892.50000000000
>> 542743.93750000000 4839876.50000000000
>> 541131.93750000000 4839852.50000000000
>> 539528.87500000000 4839840.50000000000
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>> 536303.87500000000 4839804.50000000000
>> 534672.81250000000 4839794.50000000000
>> 533050.81250000000 4839778.50000000000
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>> 544358.93750000000 4839089.50000000000
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>> 541940.87500000000 4839059.50000000000
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>> 538724.87500000000 4839024.50000000000
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>> 537110.87500000000 4839004.50000000000
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>> 535489.81250000000 4838990.50000000000
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>> 537111.87500000000 4837392.50000000000
>> 536303.81250000000 4837380.50000000000
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>> 544365.87500000000 4835857.50000000000
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>> 536315.81250000000 4834974.50000000000
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>> 541984.87500000000 4832600.50000000000
>> 541179.87500000000 4832589.50000000000
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>> 546057.81250000000 4828647.50000000000
>> 547675.81250000000 4828646.50000000000
>> 581534.93750000000 4828137.50000000000
>> 580723.93750000000 4828131.50000000000
>> 579912.93750000000 4828125.50000000000
>> 579104.93750000000 4828113.50000000000
>> 578296.93750000000 4828101.50000000000
>> 577491.93750000000 4828094.50000000000
>> 576687.93750000000 4828087.50000000000
>> 575882.93750000000 4828075.50000000000
>> 575077.93750000000 4828064.50000000000
>> 574284.93750000000 4828062.50000000000
>> 573492.93750000000 4828060.50000000000
>> 572690.93750000000 4828044.50000000000
>> 571889.93750000000 4828029.50000000000
>> 571087.93750000000 4828020.50000000000
>> 568667.87500000000 4828016.50000000000
>> 569476.87500000000 4828013.50000000000
>> 570285.87500000000 4828011.50000000000
>> 567864.87500000000 4828002.50000000000
>> 567061.87500000000 4827988.50000000000
>> 566254.87500000000 4827979.50000000000
>> 565447.87500000000 4827970.50000000000
>> 564631.87500000000 4827965.50000000000
>> 563815.87500000000 4827961.50000000000
>> 563019.87500000000 4827945.50000000000
>> 562224.87500000000 4827929.50000000000
>> 561427.87500000000 4827924.50000000000
>> 560630.87500000000 4827919.50000000000
>> 559821.87500000000 4827917.50000000000
>> 559013.87500000000 4827916.50000000000
>> 558210.87500000000 4827911.50000000000
>> 557407.87500000000 4827906.50000000000
>> 555790.87500000000 4827905.50000000000
>> 556598.87500000000 4827905.50000000000
>> 554953.87500000000 4827901.50000000000
>> 554116.87500000000 4827897.50000000000
>> 553314.87500000000 4827892.50000000000
>> 552513.87500000000 4827888.50000000000
>> 551706.87500000000 4827880.50000000000
>> 550899.87500000000 4827872.50000000000
>> 550096.81250000000 4827864.50000000000
>> 549293.81250000000 4827857.50000000000
>> 548486.81250000000 4827849.50000000000
>> 547680.81250000000 4827842.50000000000
>> 546868.81250000000 4827840.50000000000
>> 546056.81250000000 4827839.50000000000
>> 545247.81250000000 4827823.50000000000
>> 544439.81250000000 4827808.50000000000
>> 581547.93750000000 4827342.50000000000
>> 579928.93750000000 4827326.50000000000
>> 578311.93750000000 4827303.50000000000
>> 576700.93750000000 4827288.50000000000
>> 575097.87500000000 4827262.50000000000
>> 573503.87500000000 4827255.50000000000
>> 571901.87500000000 4827229.50000000000
>> 570293.87500000000 4827212.50000000000
>> 568683.87500000000 4827208.50000000000
>> 567078.87500000000 4827184.50000000000
>> 565470.87500000000 4827164.50000000000
>> 563834.87500000000 4827155.50000000000
>> 562230.87500000000 4827134.50000000000
>> 560626.87500000000 4827118.50000000000
>> 559019.87500000000 4827109.50000000000
>> 557407.87500000000 4827100.50000000000
>> 555792.87500000000 4827098.50000000000
>> 554127.87500000000 4827090.50000000000
>> 552523.81250000000 4827076.50000000000
>> 550906.81250000000 4827065.50000000000
>> 549302.81250000000 4827052.50000000000
>> 547691.81250000000 4827037.50000000000
>> 546074.81250000000 4827034.50000000000
>> 581561.93750000000 4826547.50000000000
>> 580752.87500000000 4826537.50000000000
>> 579944.87500000000 4826527.50000000000
>> 579135.87500000000 4826516.50000000000
>> 578327.87500000000 4826506.50000000000
>> 577520.87500000000 4826498.50000000000
>> 576714.87500000000 4826490.50000000000
>> 575915.87500000000 4826475.50000000000
>> 575117.87500000000 4826460.50000000000
>> 574316.87500000000 4826455.50000000000
>> 573515.87500000000 4826450.50000000000
>> 572714.87500000000 4826439.50000000000
>> 571913.87500000000 4826429.50000000000
>> 571107.87500000000 4826421.50000000000
>> 570301.87500000000 4826413.50000000000
>> 569500.87500000000 4826406.50000000000
>> 568699.87500000000 4826400.50000000000
>> 567897.87500000000 4826390.50000000000
>> 567096.87500000000 4826381.50000000000
>> 566295.87500000000 4826370.50000000000
>> 565494.87500000000 4826359.50000000000
>> 564673.87500000000 4826354.50000000000
>> 563853.87500000000 4826350.50000000000
>> 563044.87500000000 4826344.50000000000
>> 562236.87500000000 4826339.50000000000
>> 561429.87500000000 4826328.50000000000
>> 560623.87500000000 4826317.50000000000
>> 559824.87500000000 4826310.50000000000
>> 559026.87500000000 4826303.50000000000
>> 558217.87500000000 4826299.50000000000
>> 557408.87500000000 4826295.50000000000
>> 556601.87500000000 4826293.50000000000
>> 555794.87500000000 4826291.50000000000
>> 554966.81250000000 4826287.50000000000
>> 554138.81250000000 4826284.50000000000
>> 553335.81250000000 4826274.50000000000
>> 552533.81250000000 4826265.50000000000
>> 551723.81250000000 4826261.50000000000
>> 550913.81250000000 4826258.50000000000
>> 550112.81250000000 4826253.50000000000
>> 549311.81250000000 4826248.50000000000
>> 548506.81250000000 4826240.50000000000
>> 547702.81250000000 4826232.50000000000
>> 546897.81250000000 4826230.50000000000
>> 546092.81250000000 4826229.50000000000
>> 545271.81250000000 4826211.50000000000
>> 544451.81250000000 4826193.50000000000
>> 581563.87500000000 4825740.50000000000
>> 579949.87500000000 4825722.50000000000
>> 578340.87500000000 4825707.50000000000
>> 576725.87500000000 4825683.50000000000
>> 575118.87500000000 4825665.50000000000
>> 573522.87500000000 4825647.50000000000
>> 571918.87500000000 4825633.50000000000
>> 570310.87500000000 4825621.50000000000
>> 568704.87500000000 4825603.50000000000
>> 567101.87500000000 4825596.50000000000
>> 565496.87500000000 4825581.50000000000
>> 563853.87500000000 4825546.50000000000
>> 562244.87500000000 4825530.50000000000
>> 560635.87500000000 4825522.50000000000
>> 559024.87500000000 4825504.50000000000
>> 557409.81250000000 4825499.50000000000
>> 555792.81250000000 4825488.50000000000
>> 554144.81250000000 4825480.50000000000
>> 552537.81250000000 4825455.50000000000
>> 549324.81250000000 4825451.50000000000
>> 546093.81250000000 4825424.50000000000
>> 547705.81250000000 4825420.50000000000
>> 581566.87500000000 4824933.50000000000
>> 580760.87500000000 4824925.50000000000
>> 579954.87500000000 4824918.50000000000
>> 579153.87500000000 4824913.50000000000
>> 578353.87500000000 4824909.50000000000
>> 577544.87500000000 4824892.50000000000
>> 576736.87500000000 4824876.50000000000
>> 575928.87500000000 4824873.50000000000
>> 575120.87500000000 4824870.50000000000
>> 574325.87500000000 4824857.50000000000
>> 573530.87500000000 4824844.50000000000
>> 572726.87500000000 4824840.50000000000
>> 571923.87500000000 4824837.50000000000
>> 571121.87500000000 4824833.50000000000
>> 570319.87500000000 4824829.50000000000
>> 569514.87500000000 4824818.50000000000
>> 567106.87500000000 4824812.50000000000
>> 567907.87500000000 4824809.50000000000
>> 566302.87500000000 4824808.50000000000
>> 568709.87500000000 4824807.50000000000
>> 565499.87500000000 4824804.50000000000
>> 564676.87500000000 4824773.50000000000
>> 563854.87500000000 4824743.50000000000
>> 563053.87500000000 4824732.50000000000
>> 560648.87500000000 4824727.50000000000
>> 561450.87500000000 4824724.50000000000
>> 562252.87500000000 4824721.50000000000
>> 559835.81250000000 4824716.50000000000
>> 559023.81250000000 4824705.50000000000
>> 557410.81250000000 4824704.50000000000
>> 558216.81250000000 4824704.50000000000
>> 556600.81250000000 4824694.50000000000
>> 555791.81250000000 4824685.50000000000
>> 554971.81250000000 4824681.50000000000
>> 554151.81250000000 4824677.50000000000
>> 553346.81250000000 4824661.50000000000
>> 549337.81250000000 4824655.50000000000
>> 552541.81250000000 4824645.50000000000
>> 548522.81250000000 4824632.50000000000
>> 546095.81250000000 4824619.50000000000
>> 546901.81250000000 4824614.50000000000
>> 547708.81250000000 4824609.50000000000
>> 581576.87500000000 4824131.50000000000
>> 579957.87500000000 4824114.50000000000
>> 578352.87500000000 4824104.50000000000
>> 576744.87500000000 4824073.50000000000
>> 575129.87500000000 4824064.50000000000
>> 573538.87500000000 4824041.50000000000
>> 571927.87500000000 4824036.50000000000
>> 570323.87500000000 4824023.50000000000
>> 567100.87500000000 4824009.50000000000
>> 568715.87500000000 4824005.50000000000
>> 565496.81250000000 4823998.50000000000
>> 563861.81250000000 4823935.50000000000
>> 562256.81250000000 4823919.50000000000
>> 560644.81250000000 4823918.50000000000
>> 557409.81250000000 4823902.50000000000
>> 559026.81250000000 4823899.50000000000
>> 581587.87500000000 4823329.5000
>
> ...
>
> [Messaggio troncato]
--
Antonio, Fabio Di Narzo
Ph.D. student at
Department of Statistical Sciences
University of Bologna, Italy
------------------------------
Message: 77
Date: Wed, 28 Jan 2009 16:18:41 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] Text data
To: Alice Lin <alice.ly at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<644e1f320901281318m7550429y763caa0c96824316 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
This will sort on those characters:
> x <- readLines(textConnection("26M_AN_C.bmp
+ 22M_AN_C.bmp
+ 20M_HA_O.bmp
+ 20M_AN_C.bmp
+ 26M_HA_O.bmp
+ 22M_HA_O.bmp
+ 31M_AN_C.bmp
+ 38M_HA_O.bmp"))
> closeAllConnections()
> # pick off characters between "_"
> sortKey <- sub(".*_(.+)_.*", "\\1", x)
> sortKey
[1] "AN" "AN" "HA" "AN" "HA" "HA" "AN" "HA"
> # output sorted list
> x[order(sortKey)]
[1] "26M_AN_C.bmp" "22M_AN_C.bmp" "20M_AN_C.bmp" "31M_AN_C.bmp"
"20M_HA_O.bmp" "26M_HA_O.bmp" "22M_HA_O.bmp" "38M_HA_O.bmp"
>
>
On Wed, Jan 28, 2009 at 3:37 PM, Alice Lin <alice.ly at gmail.com> wrote:
>
> i have a data column of text entries:
> 26M_AN_C.bmp
> 22M_AN_C.bmp
> 20M_HA_O.bmp
> 20M_AN_C.bmp
> 26M_HA_O.bmp
> 22M_HA_O.bmp
> 31M_AN_C.bmp
> 38M_HA_O.bmp
> .
> .
> .
> .
>
>
> And I would like to sort by the middle tag: AN, HA, etc.
> Is there a way to parse text data in R?
>
> In excel, I would have used the "left" and "right" function to cut out just
> the middle two letters out and put into another column to sort by.
>
> Thanks!
>
> --
> View this message in context: http://www.nabble.com/Text-data-tp21714334p21714334.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 78
Date: Wed, 28 Jan 2009 17:23:35 -0400
From: Mike Lawrence <mike at thatmike.com>
Subject: Re: [R] Help with normal distribution in random samples...
To: Sea Captain 1779 <dbowen at measinc.com>
Cc: r-help at r-project.org
Message-ID:
<8ae7763a0901281323u3e87e032gc8ceff5673246c8a at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
?rnorm
On Wed, Jan 28, 2009 at 4:04 PM, Sea Captain 1779 <dbowen at measinc.com> wrote:
>
> Hi!!!
>
> First time 'R' user looking for a little assistance. Here is what I have so
> far:
>
> practice1 = matrix ((runif(5000, min =0, max = 12)), 100)
>
> which is creating 50 samples, for 100 cases, distributed between 0-12. What
> I would like is to be able to set the mean and SD so that the data is
> normally distributed around lets say 7. Any help I can get with achieving
[[elided Yahoo spam]]
>
> -Dan
> --
> View this message in context: http://www.nabble.com/Help-with-normal-distribution-in-random-samples...-tp21713636p21713636.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Mike Lawrence
Graduate Student
Department of Psychology
Dalhousie University
www.thatmike.com
Looking to arrange a meeting? Check my public calendar:
http://www.thatmike.com/mikes-public-calendar
~ Certainty is folly... I think. ~
------------------------------
Message: 79
Date: Wed, 28 Jan 2009 16:26:58 -0500
From: "Nutter, Benjamin" <NutterB at ccf.org>
Subject: Re: [R] Text data
To: "Alice Lin" <alice.ly at gmail.com>, r-help at r-project.org
Message-ID:
<07773E68C32A644CA47651463E79E5C202476EDF at CCHSCLEXMB68.cc.ad.cchs.net>
Content-Type: text/plain; charset=us-ascii
Jim's solution is more elegant than the following (and probably more
efficient) but you could also try the following (This let's you sort by
AN/HN, and then by the number at the start of the filename):
> text <- c( "26M_AN_C.bmp", "22M_AN_C.bmp", "20M_HA_O.bmp",
"20M_AN_C.bmp", "26M_HA_O.bmp", "22M_HA_O.bmp",
"31M_AN_C.bmp", "38M_HA_O.bmp")
> split <- do.call("rbind",strsplit(text,"_"))
> o <- order(split[,2],split[,1],split[,3])
> text[o]
[1] 20M_AN_C.bmp" "22M_AN_C.bmp" "26M_AN_C.bmp" "31M_AN_C.bmp"
"20M_HA_O.bmp"
[6] "22M_HA_O.bmp" "26M_HA_O.bmp" "38M_HA_O.bmp"
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Alice Lin
Sent: Wednesday, January 28, 2009 3:38 PM
To: r-help at r-project.org
Subject: [R] Text data
i have a data column of text entries:
26M_AN_C.bmp
22M_AN_C.bmp
20M_HA_O.bmp
20M_AN_C.bmp
26M_HA_O.bmp
22M_HA_O.bmp
31M_AN_C.bmp
38M_HA_O.bmp
.
.
.
.
And I would like to sort by the middle tag: AN, HA, etc.
Is there a way to parse text data in R?
In excel, I would have used the "left" and "right" function to cut out
just
the middle two letters out and put into another column to sort by.
Thanks!
--
View this message in context:
http://www.nabble.com/Text-data-tp21714334p21714334.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
===================================
P Please consider the environment before printing this e-mail
Cleveland Clinic is ranked one of the top hospitals
in America by U.S. News & World Report (2008).
Visit us online at http://www.clevelandclinic.org for
a complete listing of our services, staff and
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Confidentiality Note: This message is intended for use\...{{dropped:13}}
------------------------------
Message: 80
Date: Wed, 28 Jan 2009 13:26:55 -0800
From: "Nordlund, Dan (DSHS/RDA)" <NordlDJ at dshs.wa.gov>
Subject: Re: [R] Help with normal distribution in random samples...
To: "Sea Captain 1779" <dbowen at measinc.com>, r-help at r-project.org
Message-ID:
<941871A13165C2418EC144ACB212BDB0BEB2A5 at dshsmxoly1504g.dshs.wa.lcl>
Content-Type: text/plain; charset=iso-8859-1
> -----Original Message-----
> From: r-help-bounces at r-project.org
> [mailto:r-help-bounces at r-project.org] On Behalf Of Sea Captain 1779
> Sent: Wednesday, January 28, 2009 12:04 PM
> To: r-help at r-project.org
> Subject: [R] Help with normal distribution in random samples...
>
>
> Hi!!!
>
> First time 'R' user looking for a little assistance. Here is
> what I have so
> far:
>
> practice1 = matrix ((runif(5000, min =0, max = 12)), 100)
>
> which is creating 50 samples, for 100 cases, distributed
> between 0-12. What
> I would like is to be able to set the mean and SD so that the data is
> normally distributed around lets say 7. Any help I can get
> with achieving
[[elided Yahoo spam]]
>
> -Dan
Use rnorm() instead of runif().
Hope this is helpful,
A different Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
------------------------------
Message: 81
Date: Wed, 28 Jan 2009 15:32:13 -0600
From: hadley wickham <h.wickham at gmail.com>
Subject: Re: [R] Changing histogram stack in qplot
Cc: R-help at r-project.org
Message-ID:
<f8e6ff050901281332r4aeee3b7jc7b8398791f3c57b at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Hi Jason,
You'll need scale_fill_manual(values = c(low = "blue", middle =
"black", high = "red"))
See http://had.co.nz/ggplot2/scale_manual.html for more examples/details.
Regards,
Hadley
wrote:
> I've been using qplot pretty successfully to generate stacked histograms. However, it appears that I need to tweak the colors a little.
>
> I've got three temperature variables (characters not numeric) and I need to change from the default qplot colors to the following:
> Low = Blue
> Middle = black
> High = Red
>
> Here is pseudo code of what I have currently:qplot(Run, data = TestData, breaks = hist_breaks, ,
> fill = TestData$Temperature,
> main = short_title) +
> scale_x_continuous("Run, Radians") + scale_y_continuous("Frequency") +
> scale_fill_discrete("Temperature")
>
> Thanks for any advice and insights.
>
>
>
>
>
> [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
--
http://had.co.nz/
------------------------------
Message: 82
Date: Wed, 28 Jan 2009 16:33:48 -0500
From: Ubuntu Diego <ubuntu.diego at gmail.com>
Subject: Re: [R] Memory issue?
To: "r-help at r-project.org" <r-help at r-project.org>
Message-ID: <1233178428.12820.38.camel at Halley>
Content-Type: text/plain
I had similar issues with memory occupancy. You should explicitly call
gc() to call the garbage collector (free memory routine) after you do
rm() of the big objects.
D.
------------------------------
Message: 83
Date: Wed, 28 Jan 2009 13:08:33 -0800 (PST)
From: pfc_ivan <pfc_ivan at hotmail.com>
Subject: Re: [R] Newbie Question About Histograms
To: r-help at r-project.org
Message-ID: <21714995.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Also I forgot to say that The Y-axis values for each YearC would be the mean
value of all the Lenghts that happen in that YearC. Basically I cant figure
out how to put the mean values of Lengths for each YearC on Y axis.
Thanks in advance!
--
View this message in context: http://www.nabble.com/Newbie-Question-About-Histograms-tp21713626p21714995.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 84
Date: Wed, 28 Jan 2009 13:14:48 -0800 (PST)
From: Cl?ment D <dudouet at gmail.com>
Subject: Re: [R] rproxy.dll
To: r-help at r-project.org
Message-ID: <21715182.post at talk.nabble.com>
Content-Type: text/plain; charset=UTF-8
Duncan Murdoch-2 wrote:
>
> On 25/10/2008 1:01 PM, Murray Eisenberg wrote:
>> Is rproxy.dll supposed to be installed as part of a Windows binary
>> installation of R? And the installer put it in R's bin subdirectory?
>>
>> If so, it seems to be missing from the R-2.8.0 patched, 2008-10-25
>> (r46779), that I installed.
>>
>
> See the CHANGES file:
>
> o Rproxy.dll is no longer part of the R distribution: it has
> been replaced by CRAN package rscproxy.
>
> Duncan Murdoch
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
Hi!
Since rproxy.dll has been replaced by rscproxy, how can I use the (R)-D COM
dll?
Cheers!
Cl?ment D
--
View this message in context: http://www.nabble.com/rproxy.dll-tp20165941p21715182.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 85
Date: Wed, 28 Jan 2009 13:59:03 -0800 (PST)
From: SnowManPaddington <wiwiana at gmail.com>
Subject: Re: [R] for/if loop
To: r-help at r-project.org
Message-ID: <21715928.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Hi ya, I've revised the code (and finally know what I m doing.. :-D)
The good news is.. I dont get any error message, but the bad news is the
following optim generate no results. I still think there is something to do
[[elided Yahoo spam]]
pp=1
rr=1
for (ii in 1:n){
if (!(panel[ii] == pp)){
hll[pp,1] == sum(lselb1[rr:ii-1])
hll[pp,2] == sum(lselb2[rr:ii-1])
rr==ii
pp==pp+1
}
if (ii==n){
hll[pp,1] == sum(lselb1[rr:ii])
hll[pp,2] == sum(lselb2[rr:ii])
rr==ii
pp==pp+1
}
ii=ii+1
}
pp=1
rr=1
for (ii in 1:n){
if (!(panel[ii] == pp)){
hll[pp,1] == sum(lselb1[rr:ii-1])
hll[pp,2] == sum(lselb2[rr:ii-1])
rr==ii
pp==pp+1
}
if (ii==n){
hll[pp,1] == sum(lselb1[rr:ii])
hll[pp,2] == sum(lselb2[rr:ii])
rr==ii
pp==pp+1
}
ii=ii+1
}
SnowManPaddington wrote:
>
> Hi, it's my first time to write a loop with R for my homework. This loop
> is part of the function. I wanna assign values for hll according to panel
> [ii,1]=pp. I didn't get any error message in this part. but then when I
> further calculate another stuff with hll, the function can't return. I
> think it must be some problem in my loop. Probably something stupid or
> easy. But I tried to look for previous posts in forum and read R language
[[elided Yahoo spam]]
>
>
>
> for (ii in 1:100){
> for (pp in 1:pp+1){
> for (rr in 1:rr+1){
> if (panel[ii,1]!=pp)
> {
> hll(pp,1)=ColSums(lselb1(rr:ii-1,1))
> hll(pp,2)=ColSums(lselb2(rr:ii-1,1))
> rr=ii
> pp=pp+1
> }
> else
> {
> hll(pp,1)=ColSums(lselb1(rr:ii,1))
> hll(pp,2)=ColSums(lselb2(rr:ii,1))
> rr=ii
> pp=pp+1}
> }
> }}}
>
>
> in fact I have the corresponding Gauss code here. But I really don't know
> how to write such loop in R.
>
> rr=1;
> ii=1;
> pp=1;
> do until ii==n+1;
> if pan[ii,1] ne pp;
> hll[pp,1]=sumc(lselb1[rr:ii-1,1]);
> hll[pp,2]=sumc(lselb2[rr:ii-1,1]);
> rr=ii;
> pp=pp+1;
> endif;
> if ii==n;
> hll[pp,1]=sumc(lselb1[rr:ii,1]);
> hll[pp,2]=sumc(lselb2[rr:ii,1]);
> rr=ii;
> pp=pp+1;
> endif;
> ii=ii+1;
> endo;
>
>
--
View this message in context: http://www.nabble.com/for-if-loop-tp21701496p21715928.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 86
Date: Wed, 28 Jan 2009 23:24:12 +0100
From: Eik Vettorazzi <E.Vettorazzi at uke.uni-hamburg.de>
Subject: Re: [R] Newbie Question About Histograms
To: pfc_ivan <pfc_ivan at hotmail.com>
Cc: r-help at r-project.org
Message-ID: <4980DB0C.10307 at uke.uni-hamburg.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
How about
> dta<-read.table("clipboard",header=T)
> means<-aggregate(dta$Length,by=list(YearC=dta$YearC),FUN=mean)
> barplot(means[,2],names.arg=means[,1])
you may have a look at ?barplot to see (lots of) options for fine tuning
the plot.
hth.
pfc_ivan schrieb:
> Also I forgot to say that The Y-axis values for each YearC would be the mean
> value of all the Lenghts that happen in that YearC. Basically I cant figure
> out how to put the mean values of Lengths for each YearC on Y axis.
>
> Thanks in advance!
>
------------------------------
Message: 87
Date: Thu, 29 Jan 2009 09:28:42 +1100
From: Gad Abraham <gabraham at csse.unimelb.edu.au>
Subject: Re: [R] Using R in a web application
To: Will Glass-Husain <wglasshusain at gmail.com>
Cc: r-help at stat.math.ethz.ch
Message-ID: <4980DC1A.8000006 at csse.unimelb.edu.au>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Will Glass-Husain wrote:
> Hi,
>
> I want to use R to do user-submitted jobs in a (java-based) webapp.
> Specifically, I want
> * users to upload R scripts
> * run the R job on user data
> * save the results to database
>
> I'm concerned about sandbox issues.
> * Is it possible to disable file read/write capability?
> * Can I prevent the user from loading packages (e.g. the database package).
>
> * Can I have users work on separate data sets while preventing access to
> other user's data?
>
> I'm trying to see if there's a secure way to let users upload their R
> scripts and run on my server.
Have a look at Rserve (http://www.rforge.net/Rserve), I've never used it
but it might be useful to you.
--
Gad Abraham
Dept. CSSE and NICTA
The University of Melbourne
Parkville 3010, Victoria, Australia
email: gabraham at csse.unimelb.edu.au
web: http://www.csse.unimelb.edu.au/~gabraham
------------------------------
Message: 88
Date: Wed, 28 Jan 2009 14:37:15 -0800 (PST)
Subject: Re: [R] Changing histogram stack in qplot
To: R-help at r-project.org
Message-ID: <370998.28760.qm at web56006.mail.re3.yahoo.com>
Content-Type: text/plain
Worked beautifully! Thank you again for providing such a flexible package.
--- On Wed, 1/28/09, hadley wickham <h.wickham at gmail.com> wrote:
From: hadley wickham <h.wickham at gmail.com>
Subject: Re: [R] Changing histogram stack in qplot
Cc: R-help at r-project.org
Date: Wednesday, January 28, 2009, 3:32 PM
Hi Jason,
You'll need scale_fill_manual(values = c(low = "blue", middle =
"black", high = "red"))
See http://had.co.nz/ggplot2/scale_manual.html for more examples/details.
Regards,
Hadley
wrote:
> I've been using qplot pretty successfully to generate stacked
histograms. However, it appears that I need to tweak the colors a little.
>
> I've got three temperature variables (characters not numeric) and I
need to change from the default qplot colors to the following:
> Low = Blue
> Middle = black
> High = Red
>
> Here is pseudo code of what I have currently:qplot(Run, data = TestData,
breaks = hist_breaks, ,
> fill = TestData$Temperature,
> main = short_title) +
> scale_x_continuous("Run, Radians") +
scale_y_continuous("Frequency") +
> scale_fill_discrete("Temperature")
>
> Thanks for any advice and insights.
>
>
>
>
>
> [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
--
http://had.co.nz/
[[alternative HTML version deleted]]
------------------------------
Message: 89
Date: Wed, 28 Jan 2009 16:39:06 -0600
From: Joseph Magagnoli <jcm331 at gmail.com>
Subject: [R] Dynamic random effects model
To: r-help at r-project.org
Message-ID:
<9729f0ae0901281439h784649d2o3d6d29ec9f0c7794 at mail.gmail.com>
Content-Type: text/plain
All R experts,
How do I fit a dynamic Random effects model with a binary dependent variable
in R
Thanks
JCM
[[alternative HTML version deleted]]
------------------------------
Message: 90
Date: Wed, 28 Jan 2009 16:41:40 -0600
From: <davidr at rhotrading.com>
Subject: Re: [R] [SPAM] - Re: for/if loop - Bayesian Filter detected
spam
To: "SnowManPaddington" <wiwiana at gmail.com>, <r-help at r-project.org>
Message-ID:
<F9F2A641C593D7408925574C05A7BE77022A612A at rhopost.rhotrading.com>
Content-Type: text/plain; charset="us-ascii"
Well, maybe you are just bad at typing then ;-)
The lines rr==ii, pp==pp+1, etc. are not setting rr and pp but comparing
them.
Probably you want rr <- ii and pp <- pp+1, etc.
And the last line of your loop 'ii=ii+1' means that,
since the for statement is already incrementing ii,
you are incrementing it twice and skipping the even indices. Omit this
line probably.
You are also forgetting(?) the operator precedence in
sum(lselb1[rr:ii-1]) and similar lines.
Note that this is equivalent to sum(lselb1[(rr-1):(ii-1)]); is that what
you meant?
Or did you want sum(lselb1[rr:(ii-1)])?
You are changing some variables but not asking R to print anything as
far as I can see.
To see the results, ask R to print hll.
HTH,
-- David
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of SnowManPaddington
Sent: Wednesday, January 28, 2009 3:59 PM
To: r-help at r-project.org
Subject: [SPAM] - Re: [R] for/if loop - Bayesian Filter detected spam
Hi ya, I've revised the code (and finally know what I m doing.. :-D)
The good news is.. I dont get any error message, but the bad news is the
following optim generate no results. I still think there is something to
do
[[elided Yahoo spam]]
pp=1
rr=1
for (ii in 1:n){
if (!(panel[ii] == pp)){
hll[pp,1] == sum(lselb1[rr:ii-1])
hll[pp,2] == sum(lselb2[rr:ii-1])
rr==ii
pp==pp+1
}
if (ii==n){
hll[pp,1] == sum(lselb1[rr:ii])
hll[pp,2] == sum(lselb2[rr:ii])
rr==ii
pp==pp+1
}
ii=ii+1
}
pp=1
rr=1
for (ii in 1:n){
if (!(panel[ii] == pp)){
hll[pp,1] == sum(lselb1[rr:ii-1])
hll[pp,2] == sum(lselb2[rr:ii-1])
rr==ii
pp==pp+1
}
if (ii==n){
hll[pp,1] == sum(lselb1[rr:ii])
hll[pp,2] == sum(lselb2[rr:ii])
rr==ii
pp==pp+1
}
ii=ii+1
}
SnowManPaddington wrote:
>
> Hi, it's my first time to write a loop with R for my homework. This
loop
> is part of the function. I wanna assign values for hll according to
panel
> [ii,1]=pp. I didn't get any error message in this part. but then when
I
> further calculate another stuff with hll, the function can't return. I
> think it must be some problem in my loop. Probably something stupid or
> easy. But I tried to look for previous posts in forum and read R
language
[[elided Yahoo spam]]
>
>
>
> for (ii in 1:100){
> for (pp in 1:pp+1){
> for (rr in 1:rr+1){
> if (panel[ii,1]!=pp)
> {
> hll(pp,1)=ColSums(lselb1(rr:ii-1,1))
> hll(pp,2)=ColSums(lselb2(rr:ii-1,1))
> rr=ii
> pp=pp+1
> }
> else
> {
> hll(pp,1)=ColSums(lselb1(rr:ii,1))
> hll(pp,2)=ColSums(lselb2(rr:ii,1))
> rr=ii
> pp=pp+1}
> }
> }}}
>
>
> in fact I have the corresponding Gauss code here. But I really don't
know
> how to write such loop in R.
>
> rr=1;
> ii=1;
> pp=1;
> do until ii==n+1;
> if pan[ii,1] ne pp;
> hll[pp,1]=sumc(lselb1[rr:ii-1,1]);
> hll[pp,2]=sumc(lselb2[rr:ii-1,1]);
> rr=ii;
> pp=pp+1;
> endif;
> if ii==n;
> hll[pp,1]=sumc(lselb1[rr:ii,1]);
> hll[pp,2]=sumc(lselb2[rr:ii,1]);
> rr=ii;
> pp=pp+1;
> endif;
> ii=ii+1;
> endo;
>
>
--
View this message in context:
http://www.nabble.com/for-if-loop-tp21701496p21715928.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 91
Date: Wed, 28 Jan 2009 17:50:03 +0800
From: Wenxia Li <ringingfeeling at gmail.com>
Subject: [R] questions about histogram
To: r-help at r-project.org
Message-ID: <D05E2017-A927-4F62-A5A2-11C4CE9281F9 at gmail.com>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
Hi all,
I'm a new R user. I have the following information about a data set
and how to make a histogram?
data number of observations
0-2 25
2-10 10
10-100 10
100-1000 5
I tried barplot(height=...,width=...,...), the output looks right but
the x-axis is missing. How to fix it?
Also can I use<hist> to draw it?
Thanks!
WX
------------------------------
Message: 92
Date: Wed, 28 Jan 2009 15:08:14 -0800
From: Henrik Bengtsson <hb at stat.berkeley.edu>
Subject: Re: [R] [SPAM] - Re: for/if loop - Bayesian Filter detected
spam
To: davidr at rhotrading.com
Cc: r-help at r-project.org, SnowManPaddington <wiwiana at gmail.com>
Message-ID:
<59d7961d0901281508w1f304b74r3c30c6b91ec03e93 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
On Wed, Jan 28, 2009 at 2:41 PM, <davidr at rhotrading.com> wrote:
> Well, maybe you are just bad at typing then ;-)
>
> The lines rr==ii, pp==pp+1, etc. are not setting rr and pp but comparing
> them.
> Probably you want rr <- ii and pp <- pp+1, etc.
> And the last line of your loop 'ii=ii+1' means that,
> since the for statement is already incrementing ii,
> you are incrementing it twice and skipping the even indices. Omit this
> line probably.
That is actually not the case (because of the scoping rules for for(),
I think). Example:
> for (ii in 1:5) { print(ii); ii <- ii + 1; }
[1] 1
[1] 2
[1] 3
[1] 4
[1] 5
Another "counter intuitive" (though it isn't) example:
for (ii in 1:3) {
cat("Outer ii:",ii,"\n");
for (ii in ii:3) {
cat(" Inner ii:",ii,"\n");
}
}
Outer ii: 1
Inner ii: 1
Inner ii: 2
Inner ii: 3
Outer ii: 2
Inner ii: 2
Inner ii: 3
Outer ii: 3
Inner ii: 3
My $.02
/Henrik
> You are also forgetting(?) the operator precedence in
> sum(lselb1[rr:ii-1]) and similar lines.
> Note that this is equivalent to sum(lselb1[(rr-1):(ii-1)]); is that what
> you meant?
> Or did you want sum(lselb1[rr:(ii-1)])?
> You are changing some variables but not asking R to print anything as
> far as I can see.
> To see the results, ask R to print hll.
>
> HTH,
> -- David
>
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
> On Behalf Of SnowManPaddington
> Sent: Wednesday, January 28, 2009 3:59 PM
> To: r-help at r-project.org
> Subject: [SPAM] - Re: [R] for/if loop - Bayesian Filter detected spam
>
>
> Hi ya, I've revised the code (and finally know what I m doing.. :-D)
>
> The good news is.. I dont get any error message, but the bad news is the
> following optim generate no results. I still think there is something to
> do
[[elided Yahoo spam]]
>
>
>
> pp=1
> rr=1
>
> for (ii in 1:n){
> if (!(panel[ii] == pp)){
> hll[pp,1] == sum(lselb1[rr:ii-1])
> hll[pp,2] == sum(lselb2[rr:ii-1])
> rr==ii
> pp==pp+1
> }
>
> if (ii==n){
> hll[pp,1] == sum(lselb1[rr:ii])
> hll[pp,2] == sum(lselb2[rr:ii])
> rr==ii
> pp==pp+1
> }
> ii=ii+1
> }
>
>
>
>
>
> pp=1
> rr=1
>
> for (ii in 1:n){
> if (!(panel[ii] == pp)){
> hll[pp,1] == sum(lselb1[rr:ii-1])
> hll[pp,2] == sum(lselb2[rr:ii-1])
> rr==ii
> pp==pp+1
> }
>
> if (ii==n){
> hll[pp,1] == sum(lselb1[rr:ii])
> hll[pp,2] == sum(lselb2[rr:ii])
> rr==ii
> pp==pp+1
> }
> ii=ii+1
> }
>
>
>
>
>
> SnowManPaddington wrote:
>>
>> Hi, it's my first time to write a loop with R for my homework. This
> loop
>> is part of the function. I wanna assign values for hll according to
> panel
>> [ii,1]=pp. I didn't get any error message in this part. but then when
> I
>> further calculate another stuff with hll, the function can't return. I
>> think it must be some problem in my loop. Probably something stupid or
>> easy. But I tried to look for previous posts in forum and read R
> language
[[elided Yahoo spam]]
>>
>>
>>
>> for (ii in 1:100){
>> for (pp in 1:pp+1){
>> for (rr in 1:rr+1){
>> if (panel[ii,1]!=pp)
>> {
>> hll(pp,1)=ColSums(lselb1(rr:ii-1,1))
>> hll(pp,2)=ColSums(lselb2(rr:ii-1,1))
>> rr=ii
>> pp=pp+1
>> }
>> else
>> {
>> hll(pp,1)=ColSums(lselb1(rr:ii,1))
>> hll(pp,2)=ColSums(lselb2(rr:ii,1))
>> rr=ii
>> pp=pp+1}
>> }
>> }}}
>>
>>
>> in fact I have the corresponding Gauss code here. But I really don't
> know
>> how to write such loop in R.
>>
>> rr=1;
>> ii=1;
>> pp=1;
>> do until ii==n+1;
>> if pan[ii,1] ne pp;
>> hll[pp,1]=sumc(lselb1[rr:ii-1,1]);
>> hll[pp,2]=sumc(lselb2[rr:ii-1,1]);
>> rr=ii;
>> pp=pp+1;
>> endif;
>> if ii==n;
>> hll[pp,1]=sumc(lselb1[rr:ii,1]);
>> hll[pp,2]=sumc(lselb2[rr:ii,1]);
>> rr=ii;
>> pp=pp+1;
>> endif;
>> ii=ii+1;
>> endo;
>>
>>
>
> --
> View this message in context:
> http://www.nabble.com/for-if-loop-tp21701496p21715928.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
------------------------------
Message: 93
Date: Wed, 28 Jan 2009 18:08:53 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] questions about histogram
To: Wenxia Li <ringingfeeling at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<644e1f320901281508t7bc7d8b2i7314b11d1da76e17 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Try this:
> x <- read.table(textConnection("0-2 25
+ 2-10 10
+ 10-100 10
+ 100-1000 5"))
>
> x
V1 V2
1 0-2 25
2 2-10 10
3 10-100 10
4 100-1000 5
> ?barplot
> x <- read.table(textConnection("0-2 25
+ 2-10 10
+ 10-100 10
+ 100-1000 5"))
> barplot(x$V2, names.arg=x$V1)
>
On Wed, Jan 28, 2009 at 4:50 AM, Wenxia Li <ringingfeeling at gmail.com> wrote:
> Hi all,
>
> I'm a new R user. I have the following information about a data set and how
> to make a histogram?
> data number of observations
> 0-2 25
> 2-10 10
> 10-100 10
> 100-1000 5
>
> I tried barplot(height=...,width=...,...), the output looks right but the
> x-axis is missing. How to fix it?
> Also can I use<hist> to draw it?
>
> Thanks!
>
> WX
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 94
Date: Wed, 28 Jan 2009 18:24:43 +0800
From: Wenxia Li <ringingfeeling at gmail.com>
Subject: Re: [R] questions about histogram
To: r-help at r-project.org
Message-ID: <9AA09DFB-A859-4ECA-B13D-BB329DF696FB at gmail.com>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
it's a frequency histogram. The total area of the bars is 1, and the
area of each bar represents its frequency, i.e.
25/50,10/50,10/50,5/50, respectively. And the width of the bars are
different.
On Jan 28, 2009, at 5:50 PM, Wenxia Li wrote:
> Hi all,
>
> I'm a new R user. I have the following information about a data set
> and how to make a histogram?
> data number of observations
> 0-2 25
> 2-10 10
> 10-100 10
> 100-1000 5
>
> I tried barplot(height=...,width=...,...), the output looks right
> but the x-axis is missing. How to fix it?
> Also can I use<hist> to draw it?
>
> Thanks!
>
> WX
>
>
------------------------------
Message: 95
Date: Wed, 28 Jan 2009 18:31:12 -0500
From: Jorge Ivan Velez <jorgeivanvelez at gmail.com>
Subject: Re: [R] questions about histogram
To: Wenxia Li <ringingfeeling at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<317737de0901281531n606d2484we61f1ff628c8747 at mail.gmail.com>
Content-Type: text/plain
Dear Wenxia,
Take a look at this post:
http://www.nabble.com/Histogram-for-grouped-data-in-R-to21624806.html#a21624806
HTH,
Jorge
On Wed, Jan 28, 2009 at 4:50 AM, Wenxia Li <ringingfeeling at gmail.com> wrote:
> Hi all,
>
> I'm a new R user. I have the following information about a data set and how
> to make a histogram?
> data number of observations
> 0-2 25
> 2-10 10
> 10-100 10
> 100-1000 5
>
> I tried barplot(height=...,width=...,...), the output looks right but the
> x-axis is missing. How to fix it?
> Also can I use<hist> to draw it?
>
> Thanks!
>
> WX
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
------------------------------
Message: 96
Date: Wed, 28 Jan 2009 21:30:04 -0500
From: Jorge Ivan Velez <jorgeivanvelez at gmail.com>
Subject: Re: [R] glm binomial loglog (NOT cloglog) link
To: William Simpson <william.a.simpson at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<317737de0901281830o468390b9occ3745a48c8cd268 at mail.gmail.com>
Content-Type: text/plain
Dear Bill,
Perhaps the "cloglog" function in the "VGAM" package might be useful for
you.
HTH,
Jorge
On Fri, Jan 23, 2009 at 11:32 AM, William Simpson <
william.a.simpson at gmail.com> wrote:
> I would like to do an R glm() with
> family = binomial(link="loglog")
>
> Right now, the cloglog link exists, which is nice when the data have a
> heavy tail to the left. I have the opposite case and the loglog link
> is what I need. Can someone suggest how to add the loglog link onto
> glm()? It would be lovely to have it there by default, and it
> certainly makes sense to have the two opposite cases cloglog and
> loglog.
>
> Thanks for any help.
>
> Bill
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
------------------------------
Message: 97
Date: Thu, 29 Jan 2009 02:31:18 +0000
From: Mark Wardle <mark at wardle.org>
Subject: Re: [R] Faced Problems with RODBC package 1.2-5 and 1.2-4 for
windows
To: Nikhil Bhide <nikhil.bhide at tcs.com>
Cc: r-help at r-project.org
Message-ID:
<b59a37130901281831m7fe573eby34b789e4f877168b at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
You need to include much more information.
Try creating a reproducible example - you may find the answer in the
process of doing this.
You may find the posting guide for the mailing list to be helpful -
you've had no replies because you haven't explained your problem well
enough.
Mark
2009/1/28 Nikhil Bhide <nikhil.bhide at tcs.com>:
> Hi,
>
> I am facing problems with RODBC package 1.2-5 and 1.2-4 built for
> windows. I am using R 2.8.1 version.
> I faced some problems when I was trying to execute sql procedure
> from R with exec/execute statement .
>
> Stored procedure contains code/statements :
> 1) Call to another procedure (R calls one procedure which itself
> calls another procedure)
> 2) Iteration (while loop)
> I created stored procedure in which I used while loop
> and while loop contains two insert statements.I executed procedure from R.
> I found that expected results are not matching with the results I got.
> Also results are not consistent.
> 3) SET QUOTED_IDENTIFIER OFF statement
>
> Please give me a solution
>
> regards,
> Nikhil Ashok Bhide
> Cell:- +919604848030
> Mailto: nikhil.bhide at tcs.com
> Website: http://www.tcs.com
> ____________________________________________
> Experience certainty. IT Services
> Business Solutions
> Outsourcing
> ____________________________________________
> ForwardSourceID:NT00001B0E
> =====-----=====-----=====
> Notice: The information contained in this e-mail
> message and/or attachments to it may contain
> confidential or privileged information. If you are
> not the intended recipient, any dissemination, use,
> review, distribution, printing or copying of the
> information contained in this e-mail message
> and/or attachments to it are strictly prohibited. If
> you have received this communication in error,
> please notify us by reply e-mail or telephone and
> immediately and permanently delete the message
> and any attachments. Thank you
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
> ______________________________________________________________________
> This email has been scanned by the MessageLabs Email Security System.
> For more information please visit http://www.messagelabs.com/email
> ______________________________________________________________________
>
--
Dr. Mark Wardle
Specialist registrar, Neurology
Cardiff, UK
------------------------------
Message: 98
Date: Wed, 28 Jan 2009 18:56:09 -0800 (PST)
From: "cameron.bracken" <cameron.bracken at gmail.com>
Subject: Re: [R] using Sweave with a master file that has several
iputted .tex files
To: r-help at r-project.org
Message-ID: <21719963.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Mark Wardle wrote:
>
> Using make means a "build" for a single chapter is cached unless the
> source file
> changes and so one can see the results of changes to one source file
> almost immediately.
>
The pgfSweave package is specifically designed for speeding up the
compilation time in large documents. It is still in development (not on
CRAN) but might be worth checking out:
https://www.rforge.net/pgfSweave/
-Cameron Bracken
--
View this message in context: http://www.nabble.com/using-Sweave-with-a-master-file-that-has-several-iputted-.tex-files-tp21690580p21719963.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 99
Date: Wed, 28 Jan 2009 21:01:56 -0600
From: Bomee Park <bombom at stanford.edu>
Subject: [R] standard error of logit parameters
To: r-help at R-project.org
Message-ID: <FD57ADE8-5658-4C1F-BD41-93454A60C0DF at stanford.edu>
Content-Type: text/plain; charset=US-ASCII; format=flowed; delsp=yes
Hi everyone.
I am now estimating the parameters for a logit model, and trying to
get the estimates by laximizing the log_likelihood.
The nlm function works nicely for maximizing the -(log_likelihood) and
returns the parameter estimates that minimize the static, and the
gradients also, but don't have any clue how I can get the standard
error for the parameters.
Any help will be greatly appreciated.
Thanks.
------------------------------
Message: 100
Date: Wed, 28 Jan 2009 16:45:41 -0800 (PST)
From: beyar <bxx at mailinator.com>
Subject: [R] Ignore text when reading data
To: r-help at r-project.org
Message-ID: <21718709.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
Hi,
I have tab delimited text files containing numerical data,
like below, but many more columns.
As you can see, the first few lines are heading and file data. I need to
skip these lines. 2 lines above where the numbers start is what I want to
use as my header rows. I then want to ignore the next line (containing
units) and start importing data.
The header row repeats. i want to ignore the blank rows and text in the
data and then continue reading. Is there an easy way to do this?
thanks
Beyar
-------------------------
main data file - file 1
by mr x
etc
Time out1
Sec mm
0.82495117 -0.020977303
1.3554688 -0.059330709
1.826416 -0.021419302
2.3295898 -0.051521059
2.8347168 -0.020661414
Time out1
Sec mm
3.8679199 -0.000439643
4.3322754 -0.063477799
4.8015137 -0.024581354
5.3286133 -0.067487299
5.8212891 -0.011978489
-----------------------------------------------
--
View this message in context: http://www.nabble.com/Ignore-text-when-reading-data-tp21718709p21718709.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 101
Date: Thu, 29 Jan 2009 14:28:05 +1100
From: Remko Duursma <remkoduursma at gmail.com>
Subject: Re: [R] Ignore text when reading data
To: beyar <bxx at mailinator.com>
Cc: r-help at r-project.org
Message-ID:
<80b45a8c0901281928x3143dbf9x3d9848011d3dce1f at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
# replace this bit, replace it with your file name
myfile <- textConnection(
"Time out1
Sec mm
0.82495117 -0.020977303
1.3554688 -0.059330709
1.826416 -0.021419302
2.3295898 -0.051521059
2.8347168 -0.020661414
Time out1
Sec mm
3.8679199 -0.000439643
4.3322754 -0.063477799
4.8015137 -0.024581354
5.3286133 -0.067487299
5.8212891 -0.011978489")
# which lines not to read
notread <- c(which(r==""),grep("Time ",r),grep("Sec ",r))
# read the data as text
mydata <- r[setdiff(1:length(r),notread)]
# make it into a dataframe (I think this can be done prettier, but whatever)
z <- paste(mydata, collapse="\n")
read.table(textConnection(z))
greetings
Remko
-------------------------------------------------
Remko Duursma
Post-Doctoral Fellow
Centre for Plant and Food Science
University of Western Sydney
Hawkesbury Campus
Richmond NSW 2753
Dept of Biological Science
Macquarie University
North Ryde NSW 2109
Australia
Mobile: +61 (0)422 096908
On Thu, Jan 29, 2009 at 11:45 AM, beyar <bxx at mailinator.com> wrote:
>
> Hi,
> I have tab delimited text files containing numerical data,
> like below, but many more columns.
>
> As you can see, the first few lines are heading and file data. I need to
> skip these lines. 2 lines above where the numbers start is what I want to
> use as my header rows. I then want to ignore the next line (containing
> units) and start importing data.
>
> The header row repeats. i want to ignore the blank rows and text in the
> data and then continue reading. Is there an easy way to do this?
>
> thanks
> Beyar
>
>
> -------------------------
> main data file - file 1
> by mr x
> etc
>
>
> Time out1
> Sec mm
> 0.82495117 -0.020977303
> 1.3554688 -0.059330709
> 1.826416 -0.021419302
> 2.3295898 -0.051521059
> 2.8347168 -0.020661414
>
>
> Time out1
> Sec mm
> 3.8679199 -0.000439643
> 4.3322754 -0.063477799
> 4.8015137 -0.024581354
> 5.3286133 -0.067487299
> 5.8212891 -0.011978489
>
> -----------------------------------------------
>
> --
> View this message in context: http://www.nabble.com/Ignore-text-when-reading-data-tp21718709p21718709.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
------------------------------
Message: 102
Date: Thu, 29 Jan 2009 14:29:01 +1100
From: Remko Duursma <remkoduursma at gmail.com>
Subject: Re: [R] Ignore text when reading data
To: beyar <bxx at mailinator.com>
Cc: r-help at r-project.org
Message-ID:
<80b45a8c0901281929x4f536cccw3cd028b8286852e4 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Sorry, forgot this line after the textConnection bit:
r <- readLines(myfile)
-------------------------------------------------
Remko Duursma
Post-Doctoral Fellow
Centre for Plant and Food Science
University of Western Sydney
Hawkesbury Campus
Richmond NSW 2753
Dept of Biological Science
Macquarie University
North Ryde NSW 2109
Australia
Mobile: +61 (0)422 096908
On Thu, Jan 29, 2009 at 2:28 PM, Remko Duursma <remkoduursma at gmail.com> wrote:
> # replace this bit, replace it with your file name
> myfile <- textConnection(
> "Time out1
> Sec mm
> 0.82495117 -0.020977303
> 1.3554688 -0.059330709
> 1.826416 -0.021419302
> 2.3295898 -0.051521059
> 2.8347168 -0.020661414
>
>
> Time out1
> Sec mm
> 3.8679199 -0.000439643
> 4.3322754 -0.063477799
> 4.8015137 -0.024581354
> 5.3286133 -0.067487299
> 5.8212891 -0.011978489")
>
> # which lines not to read
> notread <- c(which(r==""),grep("Time ",r),grep("Sec ",r))
>
> # read the data as text
> mydata <- r[setdiff(1:length(r),notread)]
>
> # make it into a dataframe (I think this can be done prettier, but whatever)
> z <- paste(mydata, collapse="\n")
> read.table(textConnection(z))
>
>
> greetings
> Remko
>
> -------------------------------------------------
> Remko Duursma
> Post-Doctoral Fellow
>
> Centre for Plant and Food Science
> University of Western Sydney
> Hawkesbury Campus
> Richmond NSW 2753
>
> Dept of Biological Science
> Macquarie University
> North Ryde NSW 2109
> Australia
>
> Mobile: +61 (0)422 096908
>
>
>
> On Thu, Jan 29, 2009 at 11:45 AM, beyar <bxx at mailinator.com> wrote:
>>
>> Hi,
>> I have tab delimited text files containing numerical data,
>> like below, but many more columns.
>>
>> As you can see, the first few lines are heading and file data. I need to
>> skip these lines. 2 lines above where the numbers start is what I want to
>> use as my header rows. I then want to ignore the next line (containing
>> units) and start importing data.
>>
>> The header row repeats. i want to ignore the blank rows and text in the
>> data and then continue reading. Is there an easy way to do this?
>>
>> thanks
>> Beyar
>>
>>
>> -------------------------
>> main data file - file 1
>> by mr x
>> etc
>>
>>
>> Time out1
>> Sec mm
>> 0.82495117 -0.020977303
>> 1.3554688 -0.059330709
>> 1.826416 -0.021419302
>> 2.3295898 -0.051521059
>> 2.8347168 -0.020661414
>>
>>
>> Time out1
>> Sec mm
>> 3.8679199 -0.000439643
>> 4.3322754 -0.063477799
>> 4.8015137 -0.024581354
>> 5.3286133 -0.067487299
>> 5.8212891 -0.011978489
>>
>> -----------------------------------------------
>>
>> --
>> View this message in context: http://www.nabble.com/Ignore-text-when-reading-data-tp21718709p21718709.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
------------------------------
Message: 103
Date: Wed, 28 Jan 2009 22:37:44 -0500
From: jim holtman <jholtman at gmail.com>
Subject: Re: [R] Ignore text when reading data
To: beyar <bxx at mailinator.com>
Cc: r-help at r-project.org
Message-ID:
<644e1f320901281937n750e9133geb9984b87ed3f59b at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Try this:
> x <- readLines(textConnection("main data file - file 1
+ by mr x
+ etc
+
+
+ Time out1
+ Sec mm
+ 0.82495117 -0.020977303
+ 1.3554688 -0.059330709
+ 1.826416 -0.021419302
+ 2.3295898 -0.051521059
+ 2.8347168 -0.020661414
+
+
+ Time out1
+ Sec mm
+ 3.8679199 -0.000439643
+ 4.3322754 -0.063477799
+ 4.8015137 -0.024581354
+ 5.3286133 -0.067487299
+ 5.8212891 -0.011978489"))
> closeAllConnections()
> # remove blanks
> x <- x[x != ""]
> # get the lines with numbers
> indx.num <- grep("^[-0-9]", x)
> header <- x[indx.num[1] - 2]
> input <- read.table(textConnection(x[indx.num]))
> names(input) <- strsplit(header, "\\s+")[[1]]
> input
Time out1
1 0.8249512 -0.020977303
2 1.3554688 -0.059330709
3 1.8264160 -0.021419302
4 2.3295898 -0.051521059
5 2.8347168 -0.020661414
6 3.8679199 -0.000439643
7 4.3322754 -0.063477799
8 4.8015137 -0.024581354
9 5.3286133 -0.067487299
10 5.8212891 -0.011978489
>
On Wed, Jan 28, 2009 at 7:45 PM, beyar <bxx at mailinator.com> wrote:
>
> Hi,
> I have tab delimited text files containing numerical data,
> like below, but many more columns.
>
> As you can see, the first few lines are heading and file data. I need to
> skip these lines. 2 lines above where the numbers start is what I want to
> use as my header rows. I then want to ignore the next line (containing
> units) and start importing data.
>
> The header row repeats. i want to ignore the blank rows and text in the
> data and then continue reading. Is there an easy way to do this?
>
> thanks
> Beyar
>
>
> -------------------------
> main data file - file 1
> by mr x
> etc
>
>
> Time out1
> Sec mm
> 0.82495117 -0.020977303
> 1.3554688 -0.059330709
> 1.826416 -0.021419302
> 2.3295898 -0.051521059
> 2.8347168 -0.020661414
>
>
> Time out1
> Sec mm
> 3.8679199 -0.000439643
> 4.3322754 -0.063477799
> 4.8015137 -0.024581354
> 5.3286133 -0.067487299
> 5.8212891 -0.011978489
>
> -----------------------------------------------
>
> --
> View this message in context: http://www.nabble.com/Ignore-text-when-reading-data-tp21718709p21718709.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
------------------------------
Message: 104
Date: Wed, 28 Jan 2009 20:09:59 -0800 (PST)
From: beyar <bxx at mailinator.com>
Subject: Re: [R] Ignore text when reading data
To: r-help at r-project.org
Message-ID: <21720588.post at talk.nabble.com>
Content-Type: text/plain; charset=us-ascii
thanks to all for the solutions. Especially to Jim H for this one which
worked perfectly...
(i only had to change the seperater on the header to /t as there are spaces
in header names)
----------------
Try this:
> x <- readLines(textConnection("main data file - file 1
+ by mr x
+ etc
+
+
+ Time out1
+ Sec mm
+ 0.82495117 -0.020977303
+ 1.3554688 -0.059330709
+ 1.826416 -0.021419302
+ 2.3295898 -0.051521059
+ 2.8347168 -0.020661414
+
+
+ Time out1
+ Sec mm
+ 3.8679199 -0.000439643
+ 4.3322754 -0.063477799
+ 4.8015137 -0.024581354
+ 5.3286133 -0.067487299
+ 5.8212891 -0.011978489"))
> closeAllConnections()
> # remove blanks
> x <- x[x != ""]
> # get the lines with numbers
> indx.num <- grep("^[-0-9]", x)
> header <- x[indx.num[1] - 2]
> input <- read.table(textConnection(x[indx.num]))
> names(input) <- strsplit(header, "\\s+")[[1]]
> input
beyar wrote:
>
> Hi,
> I have tab delimited text files containing numerical data,
> like below, but many more columns.
>
> As you can see, the first few lines are heading and file data. I need to
> skip these lines. 2 lines above where the numbers start is what I want to
> use as my header rows. I then want to ignore the next line (containing
> units) and start importing data.
>
> The header row repeats. i want to ignore the blank rows and text in the
> data and then continue reading. Is there an easy way to do this?
>
> thanks
> Beyar
>
>
> -------------------------
> main data file - file 1
> by mr x
> etc
>
>
> Time out1
> Sec mm
> 0.82495117 -0.020977303
> 1.3554688 -0.059330709
> 1.826416 -0.021419302
> 2.3295898 -0.051521059
> 2.8347168 -0.020661414
>
>
> Time out1
> Sec mm
> 3.8679199 -0.000439643
> 4.3322754 -0.063477799
> 4.8015137 -0.024581354
> 5.3286133 -0.067487299
> 5.8212891 -0.011978489
>
> -----------------------------------------------
>
>
--
View this message in context: http://www.nabble.com/Ignore-text-when-reading-data-tp21718709p21720588.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 105
Date: Thu, 29 Jan 2009 07:52:37 +0100
From: "Weiss, Bernd " <bernd.weiss at uni-koeln.de>
Subject: [R] Question about collapse/aggregate and avoidance of loops
To: r-help at r-project.org
Message-ID: <49815235.2000303 at uni-koeln.de>
Content-Type: text/plain; charset=ISO-8859-15; format=flowed
Dear all,
given the following data
## original data
id <- c(1,1,1,2,2,3)
author <- c("A","B","C","D","E","F")
tmp <- data.frame(id,author)
tmp
> tmp
id author
1 1 A
2 1 B
3 1 C
4 2 D
5 2 E
6 3 F
What is the best (most efficient/vectorized/avoiding loops) approach to
obtain the following data frame?
id author
1 "A, B, C"
2 "D, E"
3 "F"
Thanks for your help,
Bernd
> version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status Patched
major 2
minor 8.1
year 2008
month 12
day 22
svn rev 47296
language R
version.string R version 2.8.1 Patched (2008-12-22 r47296)
------------------------------
Message: 106
Date: Thu, 29 Jan 2009 06:54:49 +0000 (GMT)
From: justin bem <justin_bem at yahoo.fr>
Subject: [R] Re : standard error of logit parameters
To: R Maillist <r-help at stat.math.ethz.ch>
Message-ID: <609242.98892.qm at web23208.mail.ird.yahoo.com>
Content-Type: text/plain
Run
outfit<-nlm(..., hessian=T) and then standards error are
se<-diag(solve(outfit$hessian))
Â
Justin BEM
BP 1917 Yaoundé
Tél (237) 76043774
Â
________________________________
De : Bomee Park <bombom at stanford.edu>
À : r-help at r-project.org
Envoyé le : Jeudi, 29 Janvier 2009, 4h01mn 56s
Objet : [R] standard error of logit parameters
Hi everyone.
I am now estimating the parameters for a logit model, and trying to get the estimates by laximizing the log_likelihood.
The nlm function works nicely for maximizing the -(log_likelihood) and returns the parameter estimates that minimize the static, and the gradients also, but don't have any clue how I can get the standard error for the parameters.
Any help will be greatly appreciated.
Thanks.
______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
------------------------------
Message: 107
Date: Thu, 29 Jan 2009 08:17:48 +0100
From: Petr PIKAL <petr.pikal at precheza.cz>
Subject: Re: [R] t.test in a loop
To: Michael Pearmain <mpearmain at google.com>
Cc: r-help at r-project.org
Message-ID:
<OF1FE4CBEF.AEE8B85E-ONC125754D.0027723A-C125754D.0028054F at precheza.cz>
Content-Type: text/plain; charset="US-ASCII"
Hi
r-help-bounces at r-project.org napsal dne 28.01.2009 12:57:55:
> On Wed, 28 Jan 2009, Michael Pearmain wrote:
>
> > Hi All,
> > I've been having a little trouble with creating a loop that will run a
a
> > series of t.tests for inspection,
> > Below is the code i've tried, and some checks i've looked at.
> >
> > I've used the get(paste()) idea as i was told previously that the use
of the
> > eval should try and be avoided.
> >
> > I've run a single syntax to check that my systax is correct and works
> > without any problems
> >> t.test(channel.data.train$News~channel.data.train$power)
> >
> > Can anyone offer any advice?
>
> There's the additional problem that if your code worked it would do 16
t-tests
> but only report the last one.
>
> Assuming you want them printed
>
> for(v in names(channel.data.train)[1:16]) {
> print(v)
> print(t.test(channel.data.train[[v]]~channel.data.train$power)
> }
>
> or
> for(v in names(channel.data.train)[1:16]){
> test <- eval(bquote(.(v)~power, data=channel.data.train)
> print(eval(test))
> }
>
> This sort of use of eval is fairly harmless.
Another option is to use lapply
lapply(channel.data.train[, 1:16], function(x)
t.test((x)~channel.data.train$power)
Regards
Petr
>
> -thomas
> > Many thanks
> >
> > Mike
> >
> >> str(channel.data.train$power)
> > num [1:9913] 0 0 0 0 0 0 0 0 0 0 ...
> >> summary(channel.data.train$power)
> > Min. 1st Qu. Median Mean 3rd Qu. Max.
> > 0.0000 0.0000 0.0000 0.2368 0.0000 1.0000
> >> names(channel.data.train)
> > [1] "News" "Entertainment" "Communicate"
> > [4] "Lifestyle" "Games" "Music"
> > [7] "Money" "Celebrity" "Shopping"
> > [10] "Sport" "Film" "Travel"
> > [13] "Cars" "Property" "Chat"
> > [16] "Bet.Play.Win" "config" "exposed"
> > [19] "site" "referrer" "started"
> > [22] "last_viewed" "num_views" "secs_since_viewed"
> > [25] "register" "secs.na" "power"
> > [28] "tt"
> >> for(i in names(channel.data.train[,c(1:16)])){
> > +
> >
t.test(get(paste("channel.data.train$",i,"~channel.data.train$power",sep="")))
> > + }
> > Error in get(paste("channel.data.train$", i,
"~channel.data.train$power",
> > :
> > variable "channel.data.train$News~channel.data.train$power" was not
found
> >
> >
> >
> > --
> > Michael Pearmain
> > Senior Analytics Research Specialist
> >
> >
> > Google UK Ltd
> > Belgrave House
> > 76 Buckingham Palace Road
> > London SW1W 9TQ
> > United Kingdom
> > t +44 (0) 2032191684
> > mpearmain at google.com
> >
> > If you received this communication by mistake, please don't forward it
to
> > anyone else (it may contain confidential or privileged information),
please
> > erase all copies of it, including all attachments, and please let the
sender
> > know it went to the wrong person. Thanks.
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help at r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> Thomas Lumley Assoc. Professor, Biostatistics
> tlumley at u.washington.edu University of Washington, Seattle
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 108
Date: Thu, 29 Jan 2009 01:29:48 -0600 (CST)
From: markleeds at verizon.net
Subject: Re: [R] Re : standard error of logit parameters
To: justin bem <justin_bem at yahoo.fr>
Cc: R Maillist <r-help at stat.math.ethz.ch>
Message-ID:
<1961203354.65345681233214188971.JavaMail.javamailuser at localhost>
Content-Type: text/plain; charset=UTF-8; format=flowed; delsp=no
I'm sure below is fine but john fox's CAR book has some nice examples
of how to compute the logit parameters and variances from scratch using
iteratively weighted least squares.
On Thu, Jan 29, 2009 at 1:54 AM, justin bem wrote:
> Run
>
> outfit<-nlm(..., hessian=T) and then standards error are
> se<-diag(solve(outfit$hessian))
>
>
> ??
> Justin BEM
> BP 1917 Yaound??
> T??l (237) 76043774
>
> ??
>
>
>
> ________________________________
> De : Bomee Park <bombom at stanford.edu>
> ?? : r-help at r-project.org
> Envoy?? le : Jeudi, 29 Janvier 2009, 4h01mn 56s
> Objet??: [R] standard error of logit parameters
>
> Hi everyone.
>
> I am now estimating the parameters for a logit model, and trying to
> get the estimates by laximizing the log_likelihood.
> The nlm function works nicely for maximizing the -(log_likelihood) and
> returns the parameter estimates that minimize the static, and the
> gradients also, but don't have any clue how I can get the standard
> error for the parameters.
>
> Any help will be greatly appreciated.
> Thanks.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
> [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 109
Date: Thu, 29 Jan 2009 08:38:57 +0100
From: Petr PIKAL <petr.pikal at precheza.cz>
Subject: [R] Odp: Question about collapse/aggregate and avoidance of
loops
To: "Weiss, Bernd " <bernd.weiss at uni-koeln.de>
Cc: r-help at r-project.org
Message-ID:
<OFCB718E52.0A552D54-ONC125754D.0029CE12-C125754D.0029F514 at precheza.cz>
Content-Type: text/plain; charset="US-ASCII"
Hi
r-help-bounces at r-project.org napsal dne 29.01.2009 07:52:37:
> Dear all,
>
> given the following data
>
> ## original data
> id <- c(1,1,1,2,2,3)
> author <- c("A","B","C","D","E","F")
> tmp <- data.frame(id,author)
> tmp
>
>
> > tmp
> id author
> 1 1 A
> 2 1 B
> 3 1 C
> 4 2 D
> 5 2 E
> 6 3 F
>
> What is the best (most efficient/vectorized/avoiding loops) approach to
> obtain the following data frame?
>
> id author
> 1 "A, B, C"
> 2 "D, E"
> 3 "F"
Not sure if it is most efficient but
aggregate(tmp$author, list(tmp$id), function(x) paste(x, collapse=","))
can do the trick
Regards
Petr
>
>
> Thanks for your help,
>
> Bernd
>
>
>
>
>
> > version
> _
> platform i386-pc-mingw32
> arch i386
> os mingw32
> system i386, mingw32
> status Patched
> major 2
> minor 8.1
> year 2008
> month 12
> day 22
> svn rev 47296
> language R
> version.string R version 2.8.1 Patched (2008-12-22 r47296)
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 110
Date: Thu, 29 Jan 2009 08:50:03 +0100
From: Petr PIKAL <petr.pikal at precheza.cz>
Subject: [R] Odp: stack data sets
To: Nidhi Kohli <nidhik at umd.edu>, r-help at stat.math.ethz.ch
Message-ID:
<OF8ECC66A1.290C3232-ONC125754D.002A39DA-C125754D.002AF935 at precheza.cz>
Content-Type: text/plain; charset="US-ASCII"
Hi
r-help-bounces at r-project.org napsal dne 28.01.2009 19:25:32:
> Hi All,
>
> I'm generating 10 different data sets with 1 and 0 in a matrix form and
> writing the output in separate files. Now I need to stack all these data
sets
> in one vector and I know that stack only operates on list or data frame
> however I got these data sets by converting list to a matrix so can't go
> backwards now. Is there a way i can still use Stack?
It is rather difficult to understand what you want to do. The code you
provide do not work as I presume only you have C:/NCME path and your data.
Stack works with data frames as you point out. However if your matrix has
appropriate form and names you can easily transform it to data frame and
use stack.
pg <- unstack(PlantGrowth)
pg<-as.matrix(pg)
pg
ctrl trt1 trt2
[1,] 4.17 4.81 6.31
[2,] 5.58 4.17 5.12
...
[9,] 5.33 4.32 5.80
[10,] 5.14 4.69 5.26
> stack(pg)
Error in rep.int(names(x), lapply(x, length)) : invalid 'times' value
> as.data.frame(pg)
ctrl trt1 trt2
1 4.17 4.81 6.31
2 5.58 4.17 5.12
...
9 5.33 4.32 5.80
10 5.14 4.69 5.26
> stack(as.data.frame(pg))
values ind
1 4.17 ctrl
2 5.58 ctrl
3 5.18 ctrl
4 6.11 ctrl
5 4.50 ctrl
6 4.61 ctrl
....
Regards
Petr
>
> Please see the program:
>
> #Importing psych & ltm library for all the simulation related functions
> library(ltm)
> library(psych)
> # Settting the working directory path to C:/NCME
> path="C:/NCME"
> setwd(path)
> #IRT Data Simulation Routine#
> n.exams = 500 #Sets number of examinees to be generated#
> n.items = 20 #Sets number of items to be generated#
> #The following intialize empty (NA) vectors or matrices#
> beta.values = rep(NA,n.items)
> resp.prob=matrix(rep(NA, n.exams*n.items), nrow=n.exams, ncol=n.items)
> Observed_Scores=matrix(rep(NA, n.exams*n.items), nrow=n.exams,
ncol=n.items)
> str(Observed_Scores)
> for (k in 1:10)
> {
> #Setting the starting point for seed
> set.seed(k)
> #filling item parameters into beta.values
> beta.values = runif(n.items,-2,2)
> #Calculating Threshold
> thresh.values = .5 * beta.values
>
> #Using the function to generate the Parallel Model CTT data
> GenData <- congeneric.sim(N=500, loads = rep(.5,20), err=NULL, short =
FALSE)
>
> #Storing Observed Score in a variable
> Observed_Scores = GenData[[3]]
> #Exporting Observed scores to output file
> ObservedScores_Data <- paste("Observed_Scores_",k,".dat")
>
write.table(Observed_Scores,ObservedScores_Data,row.name=FALSE,col.name=FALSE)
> Zero = 0
> One = 1
> for (t in 1:20)
> {
> for (s in 1:500)
> {
> if (Observed_Scores[s,t]<= thresh.values[t])
> resp.prob[s,t] = Zero
> else
> resp.prob[s,t] = One
>
> }
> }
> ResponseData <- paste("ResponseMatrix_",k,".dat")
> ThreshData <- paste("Threshold_",k,".dat")
> write.table(resp.prob,ResponseData,row.name=FALSE,col.name=FALSE)
> write.table(thresh.values,ThreshData,row.name=FALSE,col.name=FALSE)
>
> #####STACKING ALL THE OUTPUTS#########
> CommonFile <- stack(resp.prob)
> ######################################
>
> #Rounding upto 2 decimal places while showing the correlation matrix
> round(cor(GenData$observed),2)
> #Factor Score
> FactorScore=factor.pa(GenData$observed,1,scores = "TRUE")
> round(cor(FactorScore$scores,GenData$latent),2)
> filename_fs <- paste("FactorScore_",k,".dat")
> #Exporting Factor Scores to Output file
> write.table(FactorScore$scores,filename_fs,col.name=FALSE,
row.name=FALSE)
> }
>
>
> Thank you
> Nidhi
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 111
Date: Thu, 29 Jan 2009 10:08:47 +0200
From: Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>
Subject: Re: [R] Character SNP data to binary MAF data
To: Jorge Ivan Velez <jorgeivanvelez at gmail.com>
Cc: r-help at r-project.org
Message-ID:
<db80b30d0901290008j2f25ab58mcbe87677c55c1c52 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Hi
An example is as follows. Consider the character 3x6 matrix:
a A a T A t
G g t T T t
A a C C c c
For each row I would like to identify the most frequent letter and
assign a 1 to it and 0
to the less frequent character. That is, in row 1 the most frequent
letter is A (I do not differentiate between capital and non-capital
letters), in row 2 T and in row 3 C. After the binary conversion
the resulting matrix would look like that:
1 1 1 0 1 0
0 0 1 1 1 1
0 0 1 1 1 1
Any suggestions on how to do that (and I am sure I am not the first
one to try this).
Thanks
Hadassa
On Thu, Jan 29, 2009 at 1:50 AM, Jorge Ivan Velez
<jorgeivanvelez at gmail.com> wrote:
>
> Hi Hadassa,
> Do you have a sample of your data and the output you want? It might be
> useful for us in order to provide any help to you.
> Regards,
>
> Jorge
>
>
> On Wed, Jan 28, 2009 at 8:36 AM, Hadassa Brunschwig
> <hadassa.brunschwig at mail.huji.ac.il> wrote:
>>
>> Hi
>>
>> I am sure there is a function out there already but I couldn't find it.
>> I have SNP data, that is, a matrix which contains in each row two
>> characters (they are different in each row) and I would like to
>> convert this matrix to a binary one according to the minor allele
>> frequency. For non-geneticists: I want to have a binary matrix
>> for which in each row the 0 stands for the less frequent character
>> and 1 for the more frequent character.
>>
>> Thanks for any suggestions.
>> Hadassa
>>
>> --
>> Hadassa Brunschwig
>> PhD Student
>> Department of Statistics
>> The Hebrew University of Jerusalem
>> http://www.stat.huji.ac.il
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
--
Hadassa Brunschwig
PhD Student
Department of Statistics
The Hebrew University of Jerusalem
http://www.stat.huji.ac.il
------------------------------
Message: 112
Date: Thu, 29 Jan 2009 09:14:49 +0100
From: Bernd Weiss <bernd.weiss at uni-koeln.de>
Subject: Re: [R] Odp: Question about collapse/aggregate and avoidance
of loops
To: r-help at r-project.org, petr.pikal at precheza.cz,
markleeds at verizon.net
Message-ID: <49816579.2000102 at uni-koeln.de>
Content-Type: text/plain; charset=UTF-8; format=flowed
markleeds at verizon.net schrieb:
> Thanks Petr because I sent Bernd a solution offline but yours is MUCH
> NICER. it's not worth showing you because it was pretty ugly.
>
Dear Mark & Petr,
Thank your very much! I like both solutions. Petr's is the more obvious
one but Mark's solution is good for rethinking my understanding of
lapply :-)
Again, thank you very much.
Bernd
------------------------------
Message: 113
Date: Thu, 29 Jan 2009 08:28:44 +0000
From: Barry Rowlingson <b.rowlingson at lancaster.ac.uk>
Subject: Re: [R] Character SNP data to binary MAF data
To: Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>
Cc: r-help at r-project.org
Message-ID:
<d8ad40b50901290028j3b97129bo34100a6606467217 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
2009/1/29 Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>:
> Hi
>
> An example is as follows. Consider the character 3x6 matrix:
>
> a A a T A t
> G g t T T t
> A a C C c c
>
> For each row I would like to identify the most frequent letter and
> assign a 1 to it and 0
> to the less frequent character. That is, in row 1 the most frequent
> letter is A (I do not differentiate between capital and non-capital
> letters), in row 2 T and in row 3 C. After the binary conversion
> the resulting matrix would look like that:
>
> 1 1 1 0 1 0
> 0 0 1 1 1 1
> 0 0 1 1 1 1
What if there's a tie for most frequent? Do you want 1s for all the
most frequent characters? Or choose one randomly? Or zeroes?
Examples: what do the following become:
A A C C T G
A A C C T T
A A A A A A
Or are such cases not possible?
Some hints for you to work on this yourself:
help('table') - the table function works out counts of elements of vectors
help('tolower') - for changing upper to lower case
help('apply') - for working on rows of data frames
then check out any basic R tutorial on subscripting and replacement,
and you may need to work out how to loop over things with 'for'. You
should be able to make a working solution in a dozen or so lines of R.
Don't be surprised if some R guru on here does it in 2 or 3 lines of
[[elided Yahoo spam]]
Barry
------------------------------
Message: 114
Date: Thu, 29 Jan 2009 00:33:31 -0800 (PST)
From: Thomas Lumley <tlumley at u.washington.edu>
Subject: Re: [R] Character SNP data to binary MAF data
To: Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>
Cc: r-help at r-project.org
Message-ID:
<Pine.LNX.4.43.0901290033310.11961 at hymn32.u.washington.edu>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
The first step is to convert your data to all uppercase with toupper().
Then it depends on how tidy the data are: are there missing data, are some SNPs monomorphic in your sample, etc.
If there are no missing data you can use
N<-ncol(the_data)
halfN <- N/2
maf_one_row <-function(arow) {
rval<-numeric(N)
if (sum(i<-arow=="A")>halfN) {
rval[]<-1
} else if (sum(i<-arow=="C")>halfN){
rval[i]<-1
} else if (sum(i<-arow=="T"))>halfN){
rval[i]<-1
} else if (sum(i<-arow=="G")>halfN){
rval[i]<-1
}
rval
}
apply(the_data, 1, maf_one_row)
YOu could also use table() to find the two alleles, but you have to make sure that the code still works when there is only one allele.
-thomas
On Thu, 29 Jan 2009, Hadassa Brunschwig wrote:
> Hi
>
> An example is as follows. Consider the character 3x6 matrix:
>
> a A a T A t
> G g t T T t
> A a C C c c
>
> For each row I would like to identify the most frequent letter and
> assign a 1 to it and 0
> to the less frequent character. That is, in row 1 the most frequent
> letter is A (I do not differentiate between capital and non-capital
> letters), in row 2 T and in row 3 C. After the binary conversion
> the resulting matrix would look like that:
>
> 1 1 1 0 1 0
> 0 0 1 1 1 1
> 0 0 1 1 1 1
>
> Any suggestions on how to do that (and I am sure I am not the first
> one to try this).
>
> Thanks
> Hadassa
>
>
> On Thu, Jan 29, 2009 at 1:50 AM, Jorge Ivan Velez
> <jorgeivanvelez at gmail.com> wrote:
>>
>> Hi Hadassa,
>> Do you have a sample of your data and the output you want? It might be
>> useful for us in order to provide any help to you.
>> Regards,
>>
>> Jorge
>>
>>
>> On Wed, Jan 28, 2009 at 8:36 AM, Hadassa Brunschwig
>> <hadassa.brunschwig at mail.huji.ac.il> wrote:
>>>
>>> Hi
>>>
>>> I am sure there is a function out there already but I couldn't find it.
>>> I have SNP data, that is, a matrix which contains in each row two
>>> characters (they are different in each row) and I would like to
>>> convert this matrix to a binary one according to the minor allele
>>> frequency. For non-geneticists: I want to have a binary matrix
>>> for which in each row the 0 stands for the less frequent character
>>> and 1 for the more frequent character.
>>>
>>> Thanks for any suggestions.
>>> Hadassa
>>>
>>> --
>>> Hadassa Brunschwig
>>> PhD Student
>>> Department of Statistics
>>> The Hebrew University of Jerusalem
>>> http://www.stat.huji.ac.il
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
>
>
>
> --
> Hadassa Brunschwig
> PhD Student
> Department of Statistics
> The Hebrew University of Jerusalem
> http://www.stat.huji.ac.il
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Thomas Lumley Assoc. Professor, Biostatistics
tlumley at u.washington.edu University of Washington, Seattle
------------------------------
Message: 115
Date: Thu, 29 Jan 2009 00:41:09 -0800
From: Patrick Aboyoun <paboyoun at fhcrc.org>
Subject: Re: [R] Character SNP data to binary MAF data
To: r-help at r-project.org
Cc: Thomas Lumley <tlumley at u.washington.edu>, Hadassa Brunschwig
<hadassa.brunschwig at mail.huji.ac.il>
Message-ID: <20090129004109.s7162gkn2uc4gsko at webmail.fhcrc.org>
Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes";
format="flowed"
Hadassa,
You may want to check out the snpMatrix package in Bioconductor
http://bioconductor.org/packages/2.3/bioc/html/snpMatrix.html
http://bioconductor.org/packages/2.4/bioc/html/snpMatrix.html
It contains classes that manage this type of information and should
minimize your coding effort.
Patrick
Quoting Thomas Lumley <tlumley at u.washington.edu>:
>
> The first step is to convert your data to all uppercase with toupper().
>
> Then it depends on how tidy the data are: are there missing data, are
> some SNPs monomorphic in your sample, etc.
>
> If there are no missing data you can use
>
> N<-ncol(the_data)
> halfN <- N/2
>
> maf_one_row <-function(arow) {
> rval<-numeric(N)
> if (sum(i<-arow=="A")>halfN) {
> rval[]<-1
> } else if (sum(i<-arow=="C")>halfN){
> rval[i]<-1
> } else if (sum(i<-arow=="T"))>halfN){
> rval[i]<-1
> } else if (sum(i<-arow=="G")>halfN){
> rval[i]<-1
> }
> rval
> }
>
> apply(the_data, 1, maf_one_row)
>
> YOu could also use table() to find the two alleles, but you have to
> make sure that the code still works when there is only one allele.
>
> -thomas
>
> On Thu, 29 Jan 2009, Hadassa Brunschwig wrote:
>
>> Hi
>>
>> An example is as follows. Consider the character 3x6 matrix:
>>
>> a A a T A t
>> G g t T T t
>> A a C C c c
>>
>> For each row I would like to identify the most frequent letter and
>> assign a 1 to it and 0
>> to the less frequent character. That is, in row 1 the most frequent
>> letter is A (I do not differentiate between capital and non-capital
>> letters), in row 2 T and in row 3 C. After the binary conversion
>> the resulting matrix would look like that:
>>
>> 1 1 1 0 1 0
>> 0 0 1 1 1 1
>> 0 0 1 1 1 1
>>
>> Any suggestions on how to do that (and I am sure I am not the first
>> one to try this).
>>
>> Thanks
>> Hadassa
>>
>>
>> On Thu, Jan 29, 2009 at 1:50 AM, Jorge Ivan Velez
>> <jorgeivanvelez at gmail.com> wrote:
>>>
>>> Hi Hadassa,
>>> Do you have a sample of your data and the output you want? It might be
>>> useful for us in order to provide any help to you.
>>> Regards,
>>>
>>> Jorge
>>>
>>>
>>> On Wed, Jan 28, 2009 at 8:36 AM, Hadassa Brunschwig
>>> <hadassa.brunschwig at mail.huji.ac.il> wrote:
>>>>
>>>> Hi
>>>>
>>>> I am sure there is a function out there already but I couldn't find it.
>>>> I have SNP data, that is, a matrix which contains in each row two
>>>> characters (they are different in each row) and I would like to
>>>> convert this matrix to a binary one according to the minor allele
>>>> frequency. For non-geneticists: I want to have a binary matrix
>>>> for which in each row the 0 stands for the less frequent character
>>>> and 1 for the more frequent character.
>>>>
>>>> Thanks for any suggestions.
>>>> Hadassa
>>>>
>>>> --
>>>> Hadassa Brunschwig
>>>> PhD Student
>>>> Department of Statistics
>>>> The Hebrew University of Jerusalem
>>>> http://www.stat.huji.ac.il
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>
>>
>>
>> --
>> Hadassa Brunschwig
>> PhD Student
>> Department of Statistics
>> The Hebrew University of Jerusalem
>> http://www.stat.huji.ac.il
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> Thomas Lumley Assoc. Professor, Biostatistics
> tlumley at u.washington.edu University of Washington, Seattle
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 116
Date: Thu, 29 Jan 2009 08:46:01 +0000
From: Patrick Burns <pburns at pburns.seanet.com>
Subject: Re: [R] Question about collapse/aggregate and avoidance of
loops
To: "Weiss, Bernd " <bernd.weiss at uni-koeln.de>
Cc: r-help at r-project.org
Message-ID: <49816CC9.9060905 at pburns.seanet.com>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
I think you are looking for
split(author, id)
Patrick Burns
patrick at burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of "The R Inferno" and "A Guide for the Unwilling S User")
Weiss, Bernd wrote:
> Dear all,
>
> given the following data
>
> ## original data
> id <- c(1,1,1,2,2,3)
> author <- c("A","B","C","D","E","F")
> tmp <- data.frame(id,author)
> tmp
>
>
> > tmp
> id author
> 1 1 A
> 2 1 B
> 3 1 C
> 4 2 D
> 5 2 E
> 6 3 F
>
> What is the best (most efficient/vectorized/avoiding loops) approach
> to obtain the following data frame?
>
> id author
> 1 "A, B, C"
> 2 "D, E"
> 3 "F"
>
>
> Thanks for your help,
>
> Bernd
>
>
>
>
>
> > version
> _
> platform i386-pc-mingw32
> arch i386
> os mingw32
> system i386, mingw32
> status Patched
> major 2
> minor 8.1
> year 2008
> month 12
> day 22
> svn rev 47296
> language R
> version.string R version 2.8.1 Patched (2008-12-22 r47296)
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
------------------------------
Message: 117
Date: Thu, 29 Jan 2009 08:53:05 +0000
From: Barry Rowlingson <b.rowlingson at lancaster.ac.uk>
Subject: Re: [R] Character SNP data to binary MAF data
To: Patrick Aboyoun <paboyoun at fhcrc.org>
Cc: r-help at r-project.org, Thomas Lumley <tlumley at u.washington.edu>,
Hadassa Brunschwig <hadassa.brunschwig at mail.huji.ac.il>
Message-ID:
<d8ad40b50901290053m2f937ddft8923db40c1a1b7c3 at mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
2009/1/29 Patrick Aboyoun <paboyoun at fhcrc.org>:
> Hadassa,
> You may want to check out the snpMatrix package in Bioconductor
>
> http://bioconductor.org/packages/2.3/bioc/html/snpMatrix.html
> http://bioconductor.org/packages/2.4/bioc/html/snpMatrix.html
>
> It contains classes that manage this type of information and should minimize
> your coding effort.
It's not that much effort - this code turns all ties into 1s:
snp2maf=function(m){
m=toupper(m)
return(t(apply(m,1,makeBin)))
}
makeBin = function(chars){
tc = table(chars)
maxV = names(tc[tc==max(tc)])
matches = match(chars,maxV)
r=as.integer(!is.na(matches))
return(r)
}
then:
> m
[,1] [,2] [,3] [,4] [,5]
[1,] "t" "g" "g" "g" "t"
[2,] "a" "G" "a" "C" "c"
[3,] "A" "T" "c" "c" "C"
[4,] "g" "T" "c" "A" "C"
[5,] "G" "C" "G" "g" "G"
[6,] "G" "t" "T" "a" "C"
[7,] "A" "G" "T" "g" "T"
[8,] "T" "a" "C" "a" "T"
[9,] "t" "g" "g" "c" "T"
[10,] "A" "t" "t" "c" "A"
> snp2maf(m)
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 1 1 0
[2,] 1 0 1 1 1
[3,] 0 0 1 1 1
[4,] 0 0 1 0 1
[5,] 1 0 1 1 1
[6,] 0 1 1 0 0
[7,] 0 1 1 1 1
[8,] 1 1 0 1 1
[9,] 1 1 1 0 1
[10,] 1 1 1 0 1
>
Barry
------------------------------
Message: 118
Date: Thu, 29 Jan 2009 10:19:37 +0100
From: "Gerit Offermann" <gerit.offermann at gmx.de>
Subject: [R] Multiple tables
To: R-help at r-project.org
Message-ID: <20090129091937.122580 at gmx.net>
Content-Type: text/plain; charset="iso-8859-1"
Dear list,
I have a set of 100+ variables. I would like to have one by one crosstables for each variable. I started with
table(variable1, variable2)
table(variable1, variable3)
table(variable1, variable4)
...
table(variable2, variable3)
table(variable2, variable4)
...
It seems rather tedious.
Any better ideas around?
Thanks for any help!
Gerit
--
NUR NOCH BIS 31.01.! GMX FreeDSL - Telefonanschluss + DSL
f?r nur 16,37 EURO/mtl.!* http://dsl.gmx.de/?ac=OM.AD.PD003K11308T4569a
------------------------------
Message: 119
Date: Thu, 29 Jan 2009 01:25:13 -0800 (PST)
From: joe1985 <johannes at dsr.life.ku.dk>
Subject: [R] Text in a character vector to indicate "ifelse" argument
To: r-help at r-project.org
Message-ID: <21722983.post at talk.nabble.com>
Content-Type: text/plain; charset=UTF-8
Hello
I have a data set that looks like this;
> b2
dato chr status PRRSvac
PRRSsanVac PRRSsanDk PRRSdk
33 2007-12-03 090432 R?d SPF
34 2007-02-09 090432 R?d SPF+sanDK
35 2002-12-17 090432 R?d SPF+DK
36 2002-11-27 090432 R?d SPF+sanDK
37 2002-07-23 090432 R?d SPF+DK
38 2001-08-23 090432 R?d SPF
39 2000-01-01 090432 SPF-X, PRRS-neg.
40 1999-05-01 090432 MS-X, PRRS-neg.
81 2001-08-23 022458 R?d SPF
82 1999-01-22 022458 SPF-X, PRRS-neg.
130 2008-10-16 080385 R?d SPF+Myc+Ap2+Nys+DK+Vac
131 2003-03-18 080385 R?d SPF+Myc+Ap2+DK+Vac
132 2002-11-01 080385 R?d SPF+Myc+DK+Vac
133 2002-02-07 080385 R?d SPF+Myc+Vac
134 2000-09-19 080385 MS-X, PRRS-pos VAC
135 1999-01-22 080385 MS-X, PRRS-neg
176 2008-10-28 013168 R?d SPF+Myc+Ap2+Nys+DK+Vac
177 2003-05-23 013168 R?d SPF+Myc+Ap2+DK+Vac
178 2002-11-01 013168 R?d SPF+Myc+DK+Vac
179 2001-07-03 013168 R?d SPF+Myc+Vac
180 2000-09-01 013168 MS-X, PRRS-pos VAC
181 2000-06-02 013168 MS-X, PRRS-neg
182 2000-04-03 013168 SKM-X, +Ap2, PRRS-neg
183 1999-01-22 013168 MS-X, PRRS-neg
Where I have used;
b2$PRRSvac <- ifelse(b2$status=='PRRS-pos VAC' | b2$status=='Vac',1,0)
b2$PRRSdk <- ifelse(b2$status=='PRRS-pos DK' | b2$status=='DK',1,0)
b2$PRRSsanVac <- ifelse(b2$status=='sanVac',1,0)
b2$PRRSsanDk <- ifelse(b2$status=='sanDK',1,0)
to creat the last four variables, but it wont work!!! The variable status
has class "character".
Can anyone help me?
--
View this message in context: http://www.nabble.com/Text-in-a-character-vector-to-indicate-%22ifelse%22-argument-tp21722983p21722983.html
Sent from the R help mailing list archive at Nabble.com.
------------------------------
Message: 120
Date: Thu, 29 Jan 2009 20:43:25 +1100
From: Jim Lemon <jim at bitwrit.com.au>
Subject: Re: [R] help with plot layout
To: mauede at alice.it
Cc: r-help at stat.math.ethz.ch, gunter.berton at gene.com
Message-ID: <49817A3D.6060702 at bitwrit.com.au>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
mauede at alice.it wrote:
> It takes a lot of sweat to generate a composite plot with R ... sigh.
> I though I was almost done when I met the umpteenth hurdle. I cannot place a nice title on the 2nd plot (raw signal)
> on the layout. I do not have control on where either the "main" option of "plot" function, or "title", place the text
> string which keeps dysplaying chopped from above. I also tried "text", changing many times the string coordinates, but could not see any text anywhere on the canvas .
> By the way, since the layout breaks the canvas into 4 parts, are the text coordinates absolute (referred to the canvas) or
> relative (referred to the part) ?
> Please, find attached the generated drawing. The generating script is i the following.
>
Hi Maura,
You may find tab.title in the plotrix package helpful.
Jim
------------------------------
Message: 121
Date: Thu, 29 Jan 2009 08:20:30 +0000
From: geert aarts <geert_aarts at hotmail.com>
Subject: Re: [R] convergence problem gamm / lme
To: <r-help at r-project.org>
Message-ID: <BLU128-W2832C2C6A471A6C725C85684C90 at phx.gbl>
Content-Type: text/plain; charset="utf-8"
Simon, thanks for your reply and your suggestions.
I fitted the following glmm's
gamm3<-try(glmmPQL(count~offset(offsetter)+poly(lon,3)*poly(lat,3),random=list(code_tripnr=~1),family="poisson"))
Which worked OK (see summary below)
I also fitted a model using quasipoisson, but that didn't help. I actually also thought that glmmPQL and gamm estimate the dispersion parameter and hence assumes a quasipoisson distribution, even if you specify poisson. Is that correct?
Finally I tried fitting a model to less data, and sometimes gamm managed to converge (see summary below).
So would it be possible to use the parameter estimates from the model fitted to less data as starting values for the gamm fitted to the full data set?
Or do you have any other suggestions?
Thanks.
Cheers Geert
>
gamm3<-try(glmmPQL(count~offset(offsetter)+poly(lon,3)*poly(lat,3),random=list(code_tripnr=~1),f
amily="poisson"))
iteration
1
iteration
2
iteration
3
> detach(Disc_age)
>
summary(gamm3)
Linear
mixed-effects model fit by maximum likelihood
Data: NULL
AIC BIC logLik
NA NA
NA
Random
effects:
Formula: ~1 | code_tripnr
(Intercept) Residual
StdDev:
0.001391914 231.9744
Variance
function:
Structure: fixed weights
Formula: ~invwt
Fixed
effects: count ~ offset(offsetter) + poly(lon, 3) * poly(lat, 3)
Value
Std.Error DF t-value p-value
(Intercept) -1.582 11.96 2024 -0.13232174 0.8947
poly(lon,
3)1 -4.048 1397.33 2024 -0.00289673 0.9977
poly(lon,
3)2 -22.013 699.71 2024 -0.03145996 0.9749
poly(lon,
3)3 -8.538 593.87 2024 -0.01437683 0.9885
poly(lat,
3)1 -109.624 666.05 2024 -0.16458856 0.8693
poly(lat,
3)2 -104.179 381.37 2024 -0.27316977 0.7848
poly(lat,
3)3 -10.661 221.93 2024 -0.04803585 0.9617
poly(lon,
3)1:poly(lat, 3)1 4290.737 61369.98 2024
0.06991589 0.9443
poly(lon,
3)2:poly(lat, 3)1 1853.559 36835.63 2024
0.05031972 0.9599
poly(lon,
3)3:poly(lat, 3)1 -240.521 25771.80 2024 -0.00933272 0.9926
poly(lon,
3)1:poly(lat, 3)2 2540.147 41378.38 2024
0.06138826 0.9511
poly(lon,
3)1:poly(lat, 3)2 2540.147 41378.38 2024
0.06138826 0.9511
poly(lon,
3)2:poly(lat, 3)2 -1803.911 21522.17
2024 -0.08381643 0.9332
poly(lon,
3)3:poly(lat, 3)2 1040.858 16352.56 2024
0.06365109 0.9493
poly(lon,
3)1:poly(lat, 3)3 632.587 12180.28 2024
0.05193535 0.9586
poly(lon,
3)2:poly(lat, 3)3 -394.339 13088.72 2024 -0.03012818 0.9760
poly(lon,
3)3:poly(lat, 3)3 -543.502 6221.71 2024 -0.08735569 0.9304
Correlation:
(Intr) ply(ln,3)1
ply(ln,3)2 ply(ln,3)3 ply(lt,3)1
poly(lon,
3)1 0.889
poly(lon,
3)2 0.938 0.878
poly(lon,
3)3 0.843 0.981
0.792
poly(lat,
3)1 -0.829 -0.949 -0.906
-0.882
poly(lat,
3)2 0.859 0.578 0.742
0.538 -0.474
poly(lat,
3)3 -0.552 -0.783 -0.579
-0.756 0.837
poly(lon,
3)1:poly(lat, 3)1 -0.947 -0.974
-0.940 -0.940 0.925
poly(lon,
3)2:poly(lat, 3)1 -0.934 -0.950
-0.857 -0.929 0.881
poly(lon,
3)3:poly(lat, 3)1 -0.818 -0.963
-0.866 -0.945 0.931
poly(lon,
3)1:poly(lat, 3)2 0.808 0.975
0.784 0.968 -0.928
poly(lon,
3)2:poly(lat, 3)2 0.737 0.575
0.853 0.465 -0.659
poly(lon,
3)3:poly(lat, 3)2 0.735 0.896
0.647 0.938 -0.765
poly(lon,
3)1:poly(lat, 3)3 -0.794 -0.592
-0.823 -0.518 0.591
poly(lon,
3)2:poly(lat, 3)3 -0.542 -0.737
-0.419 -0.781 0.635
poly(lon,
3)3:poly(lat, 3)3 -0.398 -0.383
-0.534 -0.334 0.425
ply(lt,3)2
ply(lt,3)3 p(,3)1:(,3)1 p(,3)2:(,3)1
poly(lon,
3)1
poly(lon,
3)2
poly(lon,
3)3
poly(lat,
3)1
poly(lat,
3)2
poly(lat,
3)3 -0.136
poly(lon,
3)1:poly(lat, 3)1 -0.708 0.690
poly(lon,
3)2:poly(lat, 3)1 -0.701 0.710 0.933
poly(lon,
3)3:poly(lat, 3)1 -0.499 0.738 0.956 0.849
poly(lon,
3)1:poly(lat, 3)2 0.458 -0.845
-0.915 -0.934
poly(lon,
3)2:poly(lat, 3)2 0.683 -0.344
-0.719 -0.522
poly(lon,
3)2:poly(lat, 3)2 0.683 -0.344
-0.719 -0.522
poly(lon,
3)3:poly(lat, 3)2 0.464 -0.655
-0.834 -0.884
poly(lon,
3)1:poly(lat, 3)3 -0.823 0.241 0.752 0.594
poly(lon,
3)2:poly(lat, 3)3 -0.300 0.707 0.612 0.788
poly(lon,
3)3:poly(lat, 3)3 -0.266 0.148 0.493 0.250
p(,3)3:(,3)1
p(,3)1:(,3)2 p(,3)2:(,3)2 p(,3)3:(,3)2
poly(lon,
3)1
poly(lon,
3)2
poly(lon,
3)3
poly(lat,
3)1
poly(lat,
3)2
poly(lat,
3)3
poly(lon,
3)1:poly(lat, 3)1
poly(lon,
3)2:poly(lat, 3)1
poly(lon,
3)3:poly(lat, 3)1
poly(lon,
3)1:poly(lat, 3)2 -0.928
poly(lon,
3)2:poly(lat, 3)2 -0.637 0.432
poly(lon,
3)3:poly(lat, 3)2 -0.851
0.935 0.245
poly(lon,
3)1:poly(lat, 3)3 0.642 -0.482 -0.894 -0.410
poly(lon,
3)2:poly(lat, 3)3 0.609 -0.822 0.007 -0.847
poly(lon,
3)3:poly(lat, 3)3 0.551 -0.327 -0.637 -0.291
p(,3)1:(,3)3
p(,3)2:(,3)3
poly(lon,
3)1
poly(lon,
3)2
poly(lon,
3)3
poly(lat,
3)1
poly(lat,
3)2
poly(lat,
3)3
poly(lon,
3)1:poly(lat, 3)1
poly(lon,
3)2:poly(lat, 3)1
poly(lon,
3)3:poly(lat, 3)1
poly(lon,
3)1:poly(lat, 3)2
poly(lon,
3)2:poly(lat, 3)2
poly(lon,
3)3:poly(lat, 3)2
poly(lon,
3)1:poly(lat, 3)3
poly(lon,
3)3:poly(lat, 3)1
poly(lon,
3)1:poly(lat, 3)2
poly(lon,
3)2:poly(lat, 3)2
poly(lon,
3)3:poly(lat, 3)2
poly(lon,
3)1:poly(lat, 3)3
poly(lon,
3)2:poly(lat, 3)3 0.080
poly(lon,
3)3:poly(lat, 3)3 0.684 -0.180
Standardized
Within-Group Residuals:
Min Q1 Med Q3 Max
-0.504980771 -0.000866948
0.028470924 0.078583094
33.247831244
Number
of Observations: 2113
Number
of Groups: 74
gamm3<-try(gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),family="quasipoisson",
niterPQL=200))
>
summary(gamm3$gam)
Family:
quasipoisson
Link
function: log
Formula:
count
~ offset(offsetter) + s(lon, lat)
Parametric
coefficients:
Estimate Std. Error t value Pr(>|t|)
X 1.31370
0.09854 13.33
>
summary(gamm3$lme)
Linear
mixed-effects model fit by maximum likelihood
Data: data
AIC
BIC logLik
2808.398 2837.845 -1398.199
Random
effects:
Formula: ~Xr.1 - 1 | g.1
Structure: pdIdnot
Xr.11 Xr.12
Xr.13 Xr.14 Xr.15
Xr.16 Xr.17 Xr.18
StdDev:
12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623
Xr.19 Xr.110
Xr.111 Xr.112 Xr.113
Xr.114 Xr.115 Xr.116
StdDev:
12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623
Xr.117 Xr.118
Xr.119 Xr.120 Xr.121
Xr.122 Xr.123 Xr.124
StdDev:
12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623 12.49623
Xr.125 Xr.126
Xr.127
StdDev:
12.49623 12.49623 12.49623
Formula: ~1 | code_tripnr %in% g.1
(Intercept) Residual
StdDev: 0.8132693 5.077804
Variance
function:
Structure: fixed weights
Formula: ~invwt
Fixed
effects: list(fixed)
Value Std.Error
DF t-value p-value
XX 1.3137042 0.09863463 923
13.318894 0.0000
Xs(lon,lat)Fx1
-0.4406352 0.23114503 923 -1.906315
0.0569
Xs(lon,lat)Fx2
-0.6217519 0.24918031 923 -2.495189 0.0128
Correlation:
XX X(,)F1
Xs(lon,lat)Fx1 0.015
Xs(lon,lat)Fx2
-0.009 -0.148
Standardized
Within-Group Residuals:
Min Q1 Med Q3 Max
-3.42951750 -0.37448354
0.06432438 0.53690322 8.62026552
Number
of Observations: 1000
Number
of Groups:
g.1 code_tripnr %in% g.1
1 75
>
----------------------------------------
> From: s.wood at bath.ac.uk
> To: r-help at r-project.org
> Date: Fri, 23 Jan 2009 11:32:21 +0000
> Subject: Re: [R] convergence problem gamm / lme
>
> Geert,
>
> Can you get a simpler model with, say, a quadratic dependence on lon, lat to
> converge, using glmmPQL? The answer might give a clue about whether the issue
> is related to using a smoother, or is something more basic.
>
> How confident are you that the Poisson assumption is reasonable?
>
> Can the model be fitted to a random subsample of the data, or does it always
> fail? PQL can fail to converge, but it's usually not as obstinate as it seems
> to be in this case, if the model structure is reasonable and identifiable.
>
> best,
> Simon
>
>
>
>
>
> On Thursday 22 January 2009 15:52, geert aarts wrote:
>> Hope one of you could help with the following question/problem:
>> We would like to explain the spatial
>> distribution of juvenile fish. We have 2135 records, from 75 vessels
>> (code_tripnr) and 7 to 39 observations for each vessel, hence the random
>> effect for code_tripnr. The offset (?offsetter?) accounts for the haul
>> duration and sub sampling factor. There are no extreme outliers in lat/lon.
>> The model we try to fit is:
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>> We tried several things. We added some
>> noise to lon and lat, modelled the density instead of using a count with
>> model offset, and we normalized the explanatory variables. We also changed
>> several settings (see models below).
>>
>>
>>
>> Interestingly, we do manage to fit a more
>> complex model:
>>
>> gamm2<-gamm(count~offset(offsetter)+
>> s(lat,lon,year,dayofyear), random=list(code_tripnr=~1),family="poisson",
>> correlation = corGaus(0.1, form=~lat + lon))
>>
>>
>>
>> The models are fitted using mgcv 1.4-1 and
>> R 2.7.1 on a 64Bits Debian OS.
>>
>>
>>
>> So there seems to be a convergence problem, correct? And does someone have
>> an idea what might cause this? Secondly are there some tricks/solutions.
>> E.g. perhaps we could use the results from the more complex model (gamm2
>> above), but I do not know exactly how. All help/advice would be greatly
>> appreciated.
>>
>>
>>
>> Kind regards, Geert
>>
>>
>>
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),
>> random=list(code_tripnr=~1),family="poisson", correlation = corExp(1,
>> form=~X + Y),nite
>>
>> rPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in recalc.corSpatial(object[[i]],
>> conLin) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat,k=c(1,1)),random=list(code_
>>>tripnr=~1),family="poisson",
>>
>> niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in lme.formula(fixed = fixed, random
>> = random, data = data, correlation = correlation, :
>>
>> nlminb
>> problem, convergence error code = 1
>>
>>
>> message = false convergence (8)
>>
>> In addition: Warning messages:
>>
>> 1: In if (k < M + 1) { :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>>
>>
>>
>>
>> .Options$mgcv.vc.logrange=0.001 # we also
>> tried higher settings
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200, control=lmeControl(opt="optim"))
>>
>>
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in optim(c(coef(lmeSt)),
>> function(lmePars) -logLik(lmeSt, lmePars),
>>
>>
>>
>> initial value in 'vmmin' is not finite
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200,control=lmeControl(minAbsParApV
>>
>> ar=0.0000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in recalc.corSpatial(object[[i]],
>> conLin) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat,k=c(1,1)),random=list(code_tr
>>ipnr=~1),family="poisson", niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in lme.formula(fixed = fixed, random
>> = random, data = data, correlation = correlation, :
>>
>>
>> nlminb problem, convergence
>> error code = 1
>>
>>
>> message = false convergence (8)
>>
>> In addition: Warning messages:
>>
>> 1: In if (k < M + 1) { :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>> 2: In smooth.construct.tp.smooth.spec(object,
>> dk$data, dk$knots) :
>>
>>
>> basis dimension, k, increased to minimum possible
>>
>>
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat,k=c(8,8)),random=list(code_tr
>>ipnr=~1),family="poisson", niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in lme.formula(fixed = fixed, random
>> = random, data = data, correlation = correlation, :
>>
>>
>> nlminb problem, convergence
>> error code = 1
>>
>>
>> message = false convergence (8)
>>
>> In addition: Warning messages:
>>
>> 1: In if (k < M + 1) { :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>> 2: In 1:UZ.len : numerical expression has 2
>> elements: only the first used
>>
>> 3: In if (p.rank> ncol(XZ)) p.rank
>> <- ncol(XZ) :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>> 4: In 1:p.rank : numerical expression has 2
>> elements: only the first used
>>
>> 5: In if (p.rank < k - j) Xf <- XZU[,
>> (p.rank + 1):(k - j), drop = FALSE] else Xf <- matrix(0, :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>> 6: In (p.rank + 1):(k - j) :
>>
>>
>> numerical expression has 2 elements: only the first used
>>
>> 7: In 1:p.rank : numerical expression has 2
>> elements: only the first used
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat,k=c(4,4),fx=T),random=list(co
>>de_tripnr=~1),family="poisson", niterPQL=200)
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>> In addition: Warning messages:
>>
>> 1: In if (k < M + 1) { :
>>
>> the
>> condition has length> 1 and only the first element will be used
>>
>> 2: In 1:UZ.len : numerical expression has 2
>> elements: only the first used
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+te(lon,lat),random=list(code_tripnr=~1)
>>,family="poisson", niterPQL=200,control=lmeControl(opt="opti
>>
>> m"))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in optim(c(coef(lmeSt)),
>> function(lmePars) -logLik(lmeSt, lmePars),
>>
>>
>>
>> initial value in 'vmmin' is not finite
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200,control=lmeControl(tolerance=
>>
>> 0.00000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1
>>>),family="poisson",
>>
>> niterPQL=200,control=lmeControl(niterEM=200))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson",
>> niterPQL=200,control=lmeControl(msTol=0.00000000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson",
>> niterPQL=200,control=lmeControl(.relStep=0.00000000000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson",
>> niterPQL=200,control=lmeControl(nlmStepMax=0.00000000000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson",
>> niterPQL=200,control=lmeControl(minAbsParApVar=0.0000000000001))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> NA/NaN/Inf in foreign function call (arg 1)
>>
>>
>>
>>
>> gamm3<-gamm(count~offset(offsetter)+s(lon,lat),random=list(code_tripnr=~1),
>>family="poisson", niterPQL=200, control=lmeControl(returnObject=T))
>>
>> Maximum number of PQL iterations: 200
>>
>> iteration 1
>>
>> iteration 2
>>
>> Error in MEestimate(lmeSt, grps) :
>>
>>
>> Singularity in backsolve at level 0, block 1
>>
>> In addition: Warning messages:
>>
>> 1: In logLik.reStruct(object, conLin) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>> 2: In logLik.reStruct(object, conLin) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>> 3: In logLik.reStruct(object, conLin) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>> 4: In logLik.reStruct(object, conLin) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>> 5: In logLik.reStruct(object, conLin) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>> 6: In MEestimate(lmeSt, grps) :
>>
>>
>> Singular precision matrix in level -1, block 1
>>
>>
>> _________________________________________________________________
>>
>>
>> [[alternative HTML version deleted]]
>
> --
>> Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK
>> +44 1225 386603 www.maths.bath.ac.uk/~sw283
>
_________________________________________________________________
[[elided Yahoo spam]]
http://video.msn.com/video.aspx?mkt=nl-nl
------------------------------
Message: 122
Date: Thu, 29 Jan 2009 11:09:53 +0100
From: Petr PIKAL <petr.pikal at precheza.cz>
Subject: [R] Odp: Text in a character vector to indicate "ifelse"
argument
To: joe1985 <johannes at dsr.life.ku.dk>
Cc: r-help at r-project.org
Message-ID:
<OFFF65FBD8.BEBD7F83-ONC125754D.00363477-C125754D.0037C6AF at precheza.cz>
Content-Type: text/plain; charset="ISO-8859-1"
Hi
r-help-bounces at r-project.org napsal dne 29.01.2009 10:25:13:
>
> Hello
>
> I have a data set that looks like this;
>
> > b2
> dato chr status PRRSvac
> PRRSsanVac PRRSsanDk PRRSdk
> 33 2007-12-03 090432 R?d SPF
> 34 2007-02-09 090432 R?d SPF+sanDK
> 35 2002-12-17 090432 R?d SPF+DK
> 36 2002-11-27 090432 R?d SPF+sanDK
> 37 2002-07-23 090432 R?d SPF+DK
> 38 2001-08-23 090432 R?d SPF
> 39 2000-01-01 090432 SPF-X, PRRS-neg.
> 40 1999-05-01 090432 MS-X, PRRS-neg.
> 81 2001-08-23 022458 R?d SPF
> 82 1999-01-22 022458 SPF-X, PRRS-neg.
> 130 2008-10-16 080385 R?d SPF+Myc+Ap2+Nys+DK+Vac
> 131 2003-03-18 080385 R?d SPF+Myc+Ap2+DK+Vac
> 132 2002-11-01 080385 R?d SPF+Myc+DK+Vac
> 133 2002-02-07 080385 R?d SPF+Myc+Vac
> 134 2000-09-19 080385 MS-X, PRRS-pos VAC
> 135 1999-01-22 080385 MS-X, PRRS-neg
> 176 2008-10-28 013168 R?d SPF+Myc+Ap2+Nys+DK+Vac
> 177 2003-05-23 013168 R?d SPF+Myc+Ap2+DK+Vac
> 178 2002-11-01 013168 R?d SPF+Myc+DK+Vac
> 179 2001-07-03 013168 R?d SPF+Myc+Vac
> 180 2000-09-01 013168 MS-X, PRRS-pos VAC
> 181 2000-06-02 013168 MS-X, PRRS-neg
> 182 2000-04-03 013168 SKM-X, +Ap2, PRRS-neg
> 183 1999-01-22 013168 MS-X, PRRS-neg
>
> Where I have used;
>
> b2$PRRSvac <- ifelse(b2$status=='PRRS-pos VAC' | b2$status=='Vac',1,0)
> b2$PRRSdk <- ifelse(b2$status=='PRRS-pos DK' | b2$status=='DK',1,0)
> b2$PRRSsanVac <- ifelse(b2$status=='sanVac',1,0)
> b2$PRRSsanDk <- ifelse(b2$status=='sanDK',1,0)
>
> to creat the last four variables, but it wont work!!! The variable
status
> has class "character".
What "wont work!!!" means
> zdrzeni
sklon ot doba typ spolf spol.f
1 35 3.00 70.00 stand 35.stand 35.stand
2 50 20.00 9.50 stand 50.stand 50.stand
3 50 5.00 29.50 stand 50.stand 50.stand
4 50 15.00 13.00 stand 50.stand 50.stand
....
> zdrzeni$v1<-ifelse(zdrzeni$typ=="stand"|zdrzeni$typ=="not", 1,0)
> zdrzeni
sklon ot doba typ spolf spol.f v1
1 35 3.00 70.00 stand 35.stand 35.stand 1
2 50 20.00 9.50 stand 50.stand 50.stand 1
3 50 5.00 29.50 stand 50.stand 50.stand 1
4 50 15.00 13.00 stand 50.stand 50.stand 1
....
obviously "works", so the problem is in your data and your use of "==".
ifelse(x=="abc", 1,0)
means that the result is 1 if x is exactly equivalent to "abc" and 0 if x
is " abc", "def-abc" or in any other variation. You probably need to use
grep and regexpr expression for test but it is not my cup of tea.
Regards
Petr
>
> Can anyone help me?
>
> --
> View this message in context: http://www.nabble.com/Text-in-a-character-
> vector-to-indicate-%22ifelse%22-argument-tp21722983p21722983.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------
Message: 123
Date: Thu, 29 Jan 2009 11:42:02 +0100
From: Sigbert Klinke <sigbert at wiwi.hu-berlin.de>
Subject: [R] Graphic device & graphics primitives
To: r-help at r-project.org
Message-ID: <498187FA.2080509 at wiwi.hu-berlin.de>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi,
I know that some graphics devices in R store graphics primitives such
that a redraw is possible (e.g. when resizing the window). Is it
possible to get the current number of stored graphic primitives?
Thanks in advance
Sigbert Klinke
------------------------------
_______________________________________________
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
End of R-help Digest, Vol 71, Issue 29
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