Hi all,

I am a real novice to R. :)

I am struggling with a problem for generating colored dendrogram. I have
searched the R list and complied/collected a R code which can generated a
colored dendrogram based on the rainbow color and 4x4 similarity matrix (say
matrix:m).

In this dendrogram, each leaf is colored differently. But, I do not want the
leaf colored on a random basis. I want to assign color for leaf on basis of
some criterion.

My actual problem: How to generate a dendrogram with the leaf colored
according to the values in the second matrix (which is 4x1 dim, say matrix
c)? Meaning, the leaf 1 and 2 should be colored in neighboring spectra of
color (say different shade of red) and leaf 3 and 4 in a different color
(say different shade of blue)

Below is R code and data matrix

Any help would be highly appreciated.

Thanks
Om Prakash
Bioinformatics Center
University of Pune
India

---------------------------
Matrix m
"" "V1" "V2" "V3" "V4"
"1" 1 0.96 0.86 0.68
"2" 0.96 1 0.75 0.84
"3" 0.86 0.75 1 0.58
"4" 0.68 0.84 0.58 1
Matrix c
"" "V1"
"1" 7.450000
"2" 5.962963
"3" 0.473462
"4" 0.682300

R code:

hc<-hcluster(m, link="ave")
dhc<-as.dendrogram(hc)
dendrapply(dhc, function(n) str(attributes(n)))

local({
colLab <<- function(n){
if(is.leaf(n)){
a <- attributes(n)
i <<- i+1
attr(n, "nodePar") <- c(a$nodePar, list(lab.col = mycols[i], lab.font=i%%3))
}
n
}
#mycols <- grDevices::rainbow(attr(dhc,"members"))
i <- 0
})
dL <- dendrapply(dhc, colLab)
plot(dL)
-------------------------------------------------------------


On 12/26/06, r-help-request@stat.math.ethz.ch <
r-help-request@stat.math.ethz.ch> wrote:
>
> Send R-help mailing list submissions to
>        r-help@stat.math.ethz.ch
>
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> or, via email, send a message with subject or body 'help' to
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>
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>
> When replying, please edit your Subject line so it is more specific
> than "Re: Contents of R-help digest..."
>
>
> Today's Topics:
>
>   1. Re: how to 'get' an object that is part of a list
>      (Gabor Grothendieck)
>   2. Re: extend summary.lm for hccm? (Achim Zeileis)
>   3. Re: extend summary.lm for hccm? (John Fox)
>   4. Hmisc - some latex problems (steve)
>   5. Re: Hmisc - some latex problems (Frank E Harrell Jr)
>   6. Problem to generate training data set and test data set
>      (Aimin Yan)
>   7. No fonts in graphics using GDD (Miguel Tremblay)
>   8. Re: how to 'get' an object that is part of a list
>      (Christos Hatzis)
>   9. Bayesian data mining (Frank Grex)
> 10. Bayesian data mining (fwd) (Frank Grex)
> 11. defining color sequence in image() (Milton Cezar Ribeiro)
> 12. Re: Problem to generate training data set and test data set
>      (Jim Lemon)
> 13. Re: Higher Dimensional Matrices (Bill.Venables@csiro.au)
> 14. sequential row selection in dataframe (Pedro Mardones)
> 15. Re: Higher Dimensional Matrices (downunder)
> 16. Merry Xmas (justin bem)
> 17. Re: defining color sequence in image() (talepanda)
> 18. HOW TO: vectorize an iterative process. (Geoffrey Zhu)
> 19. writing to S-PLUS .dat file (Sebastian Michalski)
> 20. Re: extend summary.lm for hccm? (Achim Zeileis)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Mon, 25 Dec 2006 08:33:54 -0500
> From: "Gabor Grothendieck" <ggrothendieck@gmail.com>
> Subject: Re: [R] how to 'get' an object that is part of a list
> To: christos@nuverabio.com
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>        <971536df0612250533n6ba6e8f4o75975036512154ae@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> my.length.2 also has the advantage of eliminating the eval.
>
> On 12/25/06, Christos Hatzis <christos@nuverabio.com> wrote:
> > Thank you Gabor.  Very interesting solution.
> > If I get it right, the first argument in function f is just a
> placeholder to
> > help extract the right element out of the list(...) that is passed to
> > length.  Very smart trick.
> >
> > Jim's solution appears a bit simpler, at least along the lines that I
> was
> > thinking:
> >
> > my.length <- function(...) {
> >        names <- as.character(substitute(list(...)))[-1]
> >        sapply(names, function(x){y <- eval(parse(text=x)); length(y)})
> > }
> >
> > -Christos
> >
> > -----Original Message-----
> > From: Gabor Grothendieck [mailto:ggrothendieck@gmail.com]
> > Sent: Sunday, December 24, 2006 7:41 AM
> > To: jim holtman
> > Cc: christos@nuverabio.com;
> > Subject: Re: [R] how to 'get' an object that is part of a list
> >
> > Is this what you are looking for:
> >
> > > my.length.2 <- function(...) {
> > +    f <- function(nm, val) length(val)
> > +    mapply(f, make.names(as.list(match.call()[-1])), list(...)) }
> > > my.length.2(xx, xx$b)
> >  xx xx.b
> >   2    5
> >
> > On 12/24/06, jim holtman <jholtman@gmail.com> wrote:
> > > use 'eval' and 'parse'
> > >
> > >
> > > > xx
> > > $a
> > > [1] 1 2 3 4 5
> > >
> > > $b
> > > [1] "a" "b" "c" "d" "e"
> > >
> > > > eval(parse(text='xx$a'))
> > > [1] 1 2 3 4 5
> > > >
> > >
> > > So that way you can pass in the character string and then 'parse' it.
> > >
> > >
> > >
> > > On 12/24/06, Christos Hatzis <christos@nuverabio.com> wrote:
> > > >
> > > > This might be an trivial thing but I am stuck.
> > > >
> > > > Consider:
> > > >
> > > > xx <- list(a=1:5, b=letters[1:5])
> > > >
> > > > Although object xx is accessible through its name, how can object
> > > > xx$b be accessed similarly through its name?
> > > >
> > > > > get("xx")
> > > > $a
> > > > [1] 1 2 3 4 5
> > > >
> > > > $b
> > > > [1] "a" "b" "c" "d" "e"
> > > >
> > > > > get("xx$b")
> > > > Error in get(x, envir, mode, inherits) : variable "xx$b" was not
> > > > found
> > > >
> > > > get("xx")$b will not work in my case because it will probably
> > > > require parsing to make it work within a function. E.g.
> > > >
> > > > my.length <- function(...) {
> > > >        names <- as.character(substitute(list(...)))[-1]
> > > >        sapply(names, FUN=function(x){y <- get(x); length(y)}) }
> > > > > my.length(xx)
> > > > xx
> > > > 2
> > > > > my.length(xx$a)
> > > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > > found
> > > > > my.length(xx$a, xx$b)
> > > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > > found
> > > >
> > > > Thank you.
> > > >
> > > > Christos Hatzis, Ph.D.
> > > > Nuvera Biosciences, Inc.
> > > > 400 West Cummings Park
> > > > Suite 5350
> > > > Woburn, MA 01801
> > > > Tel: 781-938-3830
> > > > www.nuverabio.com
> > > >
> > > > ______________________________________________
> > > > R-help@stat.math.ethz.ch 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 you are trying to solve?
>
>
>
> ------------------------------
>
> Message: 2
> Date: Mon, 25 Dec 2006 14:38:07 +0100 (CET)
> From: Achim Zeileis <Achim.Zeileis@wu-wien.ac.at>
> Subject: Re: [R] extend summary.lm for hccm?
> To: John Fox <jfox@mcmaster.ca>
> Cc: r-help@stat.math.ethz.ch, 'ivo welch' <ivowel@gmail.com>,   'Dirk
>        Eddelbuettel' <edd@debian.org>
> Message-ID:
>        <Pine.LNX.4.44.0612251406500.8229-100000@disco.wu-wien.ac.at>
> Content-Type: TEXT/PLAIN; charset=US-ASCII
>
> On Sun, 24 Dec 2006, John Fox wrote:
>
> > Dear Dirk and Ivo,
> >
> > It's true that the sandwich package has more extensive facilities for
> > sandwich variance estimation than the hccm function in the car package,
> but
> > I think that the thrust of Ivo's question is to get a model summary that
> > automatically uses the adjusted standard errors rather than having to
> > compute them and use them "manually."
>
> I've written the function coeftest() in package "lmtest" for this purpose.
> With this you can
> coeftest(obj, vcov = sandwich)
> or you can put this into a specialized summary() function as John
> suggested (where you probably would want the F statistic to use the other
> vcov as well). See also function waldtest() in "lmtest",
> linear.hypothesis() in "car" and
> vignette("sandwich", package = "sandwich")
>
> Although this works, it is still a nuisance to use a different function
> and not summary() directly. In addition, it would also be nice to plug in
> different vcovs into confint() or predict() methods. Of course, one could
> write different generics or overload the methods in certain packages.
> However, I guess that many practitioners want to use different vcov
> estimators - especially in the social and political scieneces, and
> econometrics etc. - so that this might justify that the original "lm" (and
> "glm") methods are extended to allow for plugging in different vcov
> matrices. Maybe we could try to convince R-core to include somthing like
> this?
>
> Best wishes,
> Z
>
>
>
> ------------------------------
>
> Message: 3
> Date: Mon, 25 Dec 2006 10:01:59 -0500
> From: "John Fox" <jfox@mcmaster.ca>
> Subject: Re: [R] extend summary.lm for hccm?
> To: "'Achim Zeileis'" <Achim.Zeileis@wu-wien.ac.at>
> Cc: r-help@stat.math.ethz.ch, 'ivo welch' <ivowel@gmail.com>,   'Dirk
>        Eddelbuettel' <edd@debian.org>
> Message-ID:
>        <
> 20061225150156.ZBJX17401.tomts10-srv.bellnexxia.net@JohnDesktop8300>
> Content-Type: text/plain;       charset="us-ascii"
>
> Dear Achim,
>
> > -----Original Message-----
> > From: Achim Zeileis [mailto:Achim.Zeileis@wu-wien.ac.at]
> > Sent: Monday, December 25, 2006 8:38 AM
> > To: John Fox
> > Cc: 'Dirk Eddelbuettel'; 'ivo welch'; r-help@stat.math.ethz.ch
> > Subject: Re: [R] extend summary.lm for hccm?
> >
> > On Sun, 24 Dec 2006, John Fox wrote:
> >
> > > Dear Dirk and Ivo,
> > >
> > > It's true that the sandwich package has more extensive
> > facilities for
> > > sandwich variance estimation than the hccm function in the car
> > > package, but I think that the thrust of Ivo's question is to get a
> > > model summary that automatically uses the adjusted standard errors
> > > rather than having to compute them and use them "manually."
> >
> > I've written the function coeftest() in package "lmtest" for
> > this purpose.
> > With this you can
> >   coeftest(obj, vcov = sandwich)
>
> Since coeftest() already exists in a package, I think that this is a
> preferable solution.
>
> > or you can put this into a specialized summary() function as
> > John suggested (where you probably would want the F statistic
> > to use the other vcov as well).
>
> Good point. Here's a modified version that also fixes up the F-test:
>
> summaryHCCM.lm <- function(model, type=c("hc3", "hc0", "hc1", "hc2",
> "hc4"),
>
>    ...){
>    if (!require(car)) stop("Required car package is missing.")
>    type <- match.arg(type)
>    V <- hccm(model, type=type)
>    sumry <- summary(model)
>    table <- coef(sumry)
>    table[,2] <- sqrt(diag(V))
>    table[,3] <- table[,1]/table[,2]
>    table[,4] <- 2*pt(abs(table[,3]), df.residual(model), lower.tail=FALSE)
>    sumry$coefficients <- table
>    p <- nrow(table)
>    hyp <- if (has.intercept(model)) cbind(0, diag(p - 1)) else diag(p)
>    sumry$fstatistic[1] <- linear.hypothesis(model, hyp,
> white.adjust=type)[2,"F"]
>    print(sumry)
>    cat("Note: Heteroscedasticity-consistant standard errors using
> adjustment",
>        type, "\n")
>    }
>
> Regards,
> John
>
> > See also function waldtest()
> > in "lmtest",
> > linear.hypothesis() in "car" and
> >   vignette("sandwich", package = "sandwich")
> >
> > Although this works, it is still a nuisance to use a
> > different function and not summary() directly. In addition,
> > it would also be nice to plug in different vcovs into
> > confint() or predict() methods. Of course, one could write
> > different generics or overload the methods in certain packages.
> > However, I guess that many practitioners want to use
> > different vcov estimators - especially in the social and
> > political scieneces, and econometrics etc. - so that this
> > might justify that the original "lm" (and
> > "glm") methods are extended to allow for plugging in
> > different vcov matrices. Maybe we could try to convince
> > R-core to include somthing like this?
> >
> > Best wishes,
> > Z
> >
> >
> >
>
>
>
> ------------------------------
>
> Message: 4
> Date: Mon, 25 Dec 2006 10:21:55 -0500
> From: steve <fisk@bowdoin.edu>
> Subject: [R] Hmisc - some latex problems
> To: r-help@stat.math.ethz.ch
> Message-ID: <emoqak$6di$1@sea.gmane.org>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> If I use latex with describe:
> (faithful is the Old faithful data)
>
> latex(describe(faithful),file="describe.tex")
>
> then the first few lines of describe.tex are
>
> \begin{spacing}{0.7}
> \begin{center} \bf faithful \\ 2 Variables~~~~~ 272 ~Observations
> \end{center}
>
> I have two problems. First, I don't know what package
> the environment "spacing" comes from. (There is also a command \smaller,
> but that is from the relsize package. Perhaps the latex man page could
> list the latex packages that are necessary to create correct output.)
>
> Second, I use the memoir class, and it flags \bf as an error -
> \textbf should be used instead. A correct version would be
>
>
> \begin{center}\textbf{  faithful \\ 2 Variables~~~~~ 272
> ~Observations}\end{center}
>
> thank you,
>
> Steve
>
>
>
> ------------------------------
>
> Message: 5
> Date: Mon, 25 Dec 2006 09:50:08 -0600
> From: Frank E Harrell Jr <f.harrell@vanderbilt.edu>
> Subject: Re: [R] Hmisc - some latex problems
> To: steve <fisk@bowdoin.edu>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <458FF330.606@vanderbilt.edu>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> steve wrote:
> > If I use latex with describe:
> > (faithful is the Old faithful data)
> >
> > latex(describe(faithful),file="describe.tex")
> >
> > then the first few lines of describe.tex are
> >
> > \begin{spacing}{0.7}
> > \begin{center} \bf faithful \\ 2 Variables~~~~~ 272 ~Observations
> > \end{center}
> >
> > I have two problems. First, I don't know what package
> > the environment "spacing" comes from. (There is also a command \smaller,
> > but that is from the relsize package. Perhaps the latex man page could
> > list the latex packages that are necessary to create correct output.)
>
> The latex object created by latex.describe lists the necessary styles -
> setspace and relsize - but I'll make that clear in the documentation.
> Note for debian linux users - these styles are included in the package
> tetex-extra.
>
> >
> > Second, I use the memoir class, and it flags \bf as an error -
> > \textbf should be used instead. A correct version would be
> >
> >
> > \begin{center}\textbf{  faithful \\ 2 Variables~~~~~ 272
> > ~Observations}\end{center}
>
> Thanks - will fix in next release
>
> Frank
>
> >
> > thank you,
> >
> > Steve
>
>
>
> ------------------------------
>
> Message: 6
> Date: Mon, 25 Dec 2006 11:35:17 -0600
> From: Aimin Yan <aiminy@iastate.edu>
> Subject: [R] Problem to generate training data set and test data set
> To: r-help@stat.math.ethz.ch
> Message-ID:
>        <6.1.2.0.2.20061225112056.01c5c4e0@aiminy.mail.iastate.edu>
> Content-Type: text/plain; charset="us-ascii"; format=flowed
>
> I have a full data set like this:
>
>    aa bas    aas bms   ams bcu        acu     omega       y
> 1 ALA   0 127.71   0 69.99   0 -0.2498560  79.91470 outward
> 2 PRO   0  68.55   0 55.44   0 -0.0949008  76.60380 outward
> 3 ALA   0  52.72   0 47.82   0 -0.0396550  52.19970 outward
> 4 PHE   0  22.62   0 31.21   0  0.1270330 169.52500  inward
> 5 SER   0  71.32   0 52.84   0 -0.1312380   7.47528 outward
> 6 VAL   0  12.92   0 22.40   0  0.1728390 149.09400  inward
>
> ......................................................................................
>
>
> aa have 19 levels, and there are different number of observation for each
> levels.
> I want to pick 75% of observations of each levels randomly to generate a
> training set,
> and 25% of observation of each levels to generate a testing set.
>
> Does anyone know to do this?
>
> Thanks
>
> Aimin Yan
>
>
>
> ------------------------------
>
> Message: 7
> Date: Mon, 25 Dec 2006 15:10:23 -0500 (EST)
> From: Miguel Tremblay <miguel.tremblay@ptaff.ca>
> Subject: [R] No fonts in graphics using GDD
> To: r-help@stat.math.ethz.ch
> Message-ID: <Pine.LNX.4.64.0612251508440.14703@octet.ca>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
> I have problem to have fonts in graphics using GDD.
>
> R version: R 2.2.1 (2005-12-20).
> GDD version: GDD_0.1-8.tar.gz
> Platform: Linux 2.6.17-gentoo-r8 #2 SMP Sat Nov 4 21:16:47 EST 2006 i686
> Intel(R) Pentium(R) 4 CPU 3.00GHz GenuineIntel GNU/Linux
>
>
> I have configured the basefont.mapping file accordingly to the path of the
> font on my computer:
> base.norm:/usr/share/fonts/corefonts/arial.ttf
> base.ital:/usr/share/fonts/corefonts/ariali.ttf
> base.bold:/usr/share/fonts/corefonts/arialbd.ttf
> base.bita:/usr/share/fonts/corefonts/arialbi.ttf
>
> Those files exists and are present at the right place:
> ls /usr/share/fonts/corefonts/arial*.ttf
> /usr/share/fonts/corefonts/arial.ttf
> /usr/share/fonts/corefonts/arialbi.ttf
> /usr/share/fonts/corefonts/arialbd.ttf
> /usr/share/fonts/corefonts/ariali.ttf
>
> In R, when I import the GDD library and look at the font path, they are
> correct:
> > library('GDD')
> > .Call("gdd_look_up_font", NULL)
> [1] "/usr/share/fonts/corefonts/arial.ttf"
> [2] "/usr/share/fonts/corefonts/arialbd.ttf"
> [3] "/usr/share/fonts/corefonts/ariali.ttf"
> [4] "/usr/share/fonts/corefonts/arialbi.ttf"
> [5] NA
>
>
> When I create a graphic, there is no font:
> > library(GDD)
> > GDD()
> > plot(rnorm)
> > dev.off()
>
> I have went through the following threads and it did'nt help:
>
> http://mailman.rz.uni-augsburg.de/pipermail/stats-rosuda-devel/2006q1/000197.html
> http://tolstoy.newcastle.edu.au/R/help/06/04/25159.html
>
>
> The ouptut of my GDD installation is:
> R CMD INSTALL GDD_0.1-8.tar.gz
> * Installing *source* package 'GDD' ...
> checking for gcc... i686-pc-linux-gnu-gcc
> checking for C compiler default output file name... a.out
> checking whether the C compiler works... yes
> checking whether we are cross compiling... no
> checking for suffix of executables...
> checking for suffix of object files... o
> checking whether we are using the GNU C compiler... yes
> checking whether i686-pc-linux-gnu-gcc accepts -g... yes
> checking for i686-pc-linux-gnu-gcc option to accept ANSI C... none needed
> checking how to run the C preprocessor... i686-pc-linux-gnu-gcc -E
> checking for egrep... grep -E
> checking for ANSI C header files... yes
> checking for sys/wait.h that is POSIX.1 compatible... yes
> checking for sys/types.h... yes
> checking for sys/stat.h... yes
> checking for stdlib.h... yes
> checking for string.h... yes
> checking for memory.h... yes
> checking for strings.h... yes
> checking for inttypes.h... yes
> checking for stdint.h... yes
> checking for unistd.h... yes
> checking for string.h... (cached) yes
> checking sys/time.h usability... yes
> checking sys/time.h presence... yes
> checking for sys/time.h... yes
> checking for unistd.h... (cached) yes
> checking for an ANSI C-conforming const... yes
> checking whether time.h and sys/time.h may both be included... yes
> checking for stdlib.h... (cached) yes
> checking for GNU libc compatible malloc... yes
> checking return type of signal handlers... void
> checking for memset... yes
> checking for mkdir... yes
> checking for rmdir... yes
> checking for gdlib-config... /usr/bin/gdlib-config
> libgd-flags according to gdlib-config:
>    -I/usr/include -L/usr/lib -ljpeg -lfontconfig -lpng12 -lz -lm
> if that is not correct, fix your gd installation
> checking gd.h usability... yes
> checking gd.h presence... yes
> checking for gd.h... yes
> checking for gdImageCreateFromPng in -lgd... yes
> checking usability of FreeType in GD... yes
> checking whether GD programs can be compiled... yes
> configure: creating ./config.status
> config.status: creating src/Makevars
> config.status: creating src/gddconfig.h
> ** libs
> i686-pc-linux-gnu-gcc -I/usr/lib/R/include -I/usr/include -I/usr/include
> -DHAVE_GDDCONFIG_H -I. -Iinclude -I/usr/local/include
> -D_FILE_OFFSET_BITS=64 -D_LARGEFILE64_SOURCE -D_LARGEFILE_SOURCE -O2
> -march=pentium4 -fomit-frame-pointer -pipe -fPIC  -D_FILE_OFFSET_BITS=64
> -D_LARGEFILE64_SOURCE -D_LARGEFILE_SOURCE -O2 -march=pentium4
> -fomit-frame-pointer -pipe -c GDD.c -o GDD.o
> i686-pc-linux-gnu-gcc -I/usr/lib/R/include -I/usr/include -I/usr/include
> -DHAVE_GDDCONFIG_H -I. -Iinclude -I/usr/local/include
> -D_FILE_OFFSET_BITS=64 -D_LARGEFILE64_SOURCE -D_LARGEFILE_SOURCE -O2
> -march=pentium4 -fomit-frame-pointer -pipe -fPIC  -D_FILE_OFFSET_BITS=64
> -D_LARGEFILE64_SOURCE -D_LARGEFILE_SOURCE -O2 -march=pentium4
> -fomit-frame-pointer -pipe -c GDDtalk.c -o GDDtalk.o
> i686-pc-linux-gnu-gcc -shared  -o GDD.so GDD.o GDDtalk.o -L/usr/lib
> -L/usr/lib -lgd  -ljpeg -lfontconfig -lpng12 -lz -lm  -L/usr/lib/R/lib -lR
> ** R
> ** inst
> ** help
> >>> Building/Updating help pages for package 'GDD'
>      Formats: text html latex example
>   GDD                               text    html    latex   example
>   GDDfont                           text    html    latex   example
> ** building package indices ...
> * DONE (GDD)
>
>
>
> Anyone can help me with this?
>
>
> Miguel Tremblay
> http://ptaff.ca/miguel/
>
>
>
> ------------------------------
>
> Message: 8
> Date: Mon, 25 Dec 2006 17:08:26 -0500
> From: "Christos Hatzis" <christos@nuverabio.com>
> Subject: Re: [R] how to 'get' an object that is part of a list
> To: "'Gabor Grothendieck'" <ggrothendieck@gmail.com>
> Cc: r-help@stat.math.ethz.ch
> Message-ID:
>        <001401c72871$34007850$0202a8c0@headquarters.silicoinsights>
> Content-Type: text/plain;       charset="us-ascii"
>
> True.  Thanks again.
>
> -----Original Message-----
> From: Gabor Grothendieck [mailto:ggrothendieck@gmail.com]
> Sent: Monday, December 25, 2006 8:34 AM
> To: christos@nuverabio.com
> Cc: r-help@hypatia.math.ethz.ch
> Subject: Re: [R] how to 'get' an object that is part of a list
>
> my.length.2 also has the advantage of eliminating the eval.
>
> On 12/25/06, Christos Hatzis <christos@nuverabio.com> wrote:
> > Thank you Gabor.  Very interesting solution.
> > If I get it right, the first argument in function f is just a
> > placeholder to help extract the right element out of the list(...)
> > that is passed to length.  Very smart trick.
> >
> > Jim's solution appears a bit simpler, at least along the lines that I
> > was
> > thinking:
> >
> > my.length <- function(...) {
> >        names <- as.character(substitute(list(...)))[-1]
> >        sapply(names, function(x){y <- eval(parse(text=x)); length(y)})
> > }
> >
> > -Christos
> >
> > -----Original Message-----
> > From: Gabor Grothendieck [mailto:ggrothendieck@gmail.com]
> > Sent: Sunday, December 24, 2006 7:41 AM
> > To: jim holtman
> > Cc: christos@nuverabio.com;
> > Subject: Re: [R] how to 'get' an object that is part of a list
> >
> > Is this what you are looking for:
> >
> > > my.length.2 <- function(...) {
> > +    f <- function(nm, val) length(val)
> > +    mapply(f, make.names(as.list(match.call()[-1])), list(...)) }
> > > my.length.2(xx, xx$b)
> >  xx xx.b
> >   2    5
> >
> > On 12/24/06, jim holtman <jholtman@gmail.com> wrote:
> > > use 'eval' and 'parse'
> > >
> > >
> > > > xx
> > > $a
> > > [1] 1 2 3 4 5
> > >
> > > $b
> > > [1] "a" "b" "c" "d" "e"
> > >
> > > > eval(parse(text='xx$a'))
> > > [1] 1 2 3 4 5
> > > >
> > >
> > > So that way you can pass in the character string and then 'parse' it.
> > >
> > >
> > >
> > > On 12/24/06, Christos Hatzis <christos@nuverabio.com> wrote:
> > > >
> > > > This might be an trivial thing but I am stuck.
> > > >
> > > > Consider:
> > > >
> > > > xx <- list(a=1:5, b=letters[1:5])
> > > >
> > > > Although object xx is accessible through its name, how can object
> > > > xx$b be accessed similarly through its name?
> > > >
> > > > > get("xx")
> > > > $a
> > > > [1] 1 2 3 4 5
> > > >
> > > > $b
> > > > [1] "a" "b" "c" "d" "e"
> > > >
> > > > > get("xx$b")
> > > > Error in get(x, envir, mode, inherits) : variable "xx$b" was not
> > > > found
> > > >
> > > > get("xx")$b will not work in my case because it will probably
> > > > require parsing to make it work within a function. E.g.
> > > >
> > > > my.length <- function(...) {
> > > >        names <- as.character(substitute(list(...)))[-1]
> > > >        sapply(names, FUN=function(x){y <- get(x); length(y)}) }
> > > > > my.length(xx)
> > > > xx
> > > > 2
> > > > > my.length(xx$a)
> > > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > > found
> > > > > my.length(xx$a, xx$b)
> > > > Error in get(x, envir, mode, inherits) : variable "xx$a" was not
> > > > found
> > > >
> > > > Thank you.
> > > >
> > > > Christos Hatzis, Ph.D.
> > > > Nuvera Biosciences, Inc.
> > > > 400 West Cummings Park
> > > > Suite 5350
> > > > Woburn, MA 01801
> > > > Tel: 781-938-3830
> > > > www.nuverabio.com
> > > >
> > > > ______________________________________________
> > > > R-help@stat.math.ethz.ch 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 you are trying to solve?
>
>
>
> ------------------------------
>
> Message: 9
> Date: Mon, 25 Dec 2006 18:25:33 -0500 (EST)
> From: Frank Grex <franco@grex.cyberspace.org>
> Subject: [R] Bayesian data mining
> To: r-help@stat.math.ethz.ch
> Message-ID: <Pine.BSO.4.63.0612251817440.31320@grex.cyberspace.org>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
> Hi, I need a help to know whether I can perform the following in R:
> I have a set of observations (Ns) and each observation is drawn from a
> poisson distribution with an unkown mean, lambda. The set of lambdas in
> their turn are drawn from a common prior distribution which is supposed to
> be a a mixture of two gamma distributions.
> Is there a way to determine the poisson means in R, given the Ns and the
> probabilities?
> And how can one determine the two alphas and the two betas of the gamma
> mixtures? I am assuming there will be an MLE somewhere.
> This is to help me understand and apply William DuMouchel concept of
> datamining especially in his article: "Bayesian data mining in large
> frequency tables". Thanks
>
>
>
> ------------------------------
>
> Message: 10
> Date: Mon, 25 Dec 2006 18:31:33 -0500 (EST)
> From: Frank Grex <franco@grex.cyberspace.org>
> Subject: [R] Bayesian data mining (fwd)
> To: r-help@stat.math.ethz.ch
> Message-ID: <Pine.BSO.4.63.0612251830460.3501@grex.cyberspace.org>
> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
>
> Hi, I need a help to know whether I can perform the following in R:
> I have a set of observations (Ns) and each observation is drawn from a
> poisson distribution with an unkown mean, lambda. The set of lambdas in
> their turn are drawn from a common prior distribution which is supposed to
> be a a mixture of two gamma distributions.
> Is there a way to determine the poisson means in R, given the Ns and the
> probabilities?
> And how can one determine the two alphas and the two betas of the gamma
> mixtures? I am assuming there will be an MLE somewhere.
> This is to help me understand and apply William DuMouchel concept of
> datamining especially in his article: "Bayesian data mining in large
> frequency tables". Thanks
>
> ______________________________________________
> R-help@stat.math.ethz.ch 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: 11
> Date: Tue, 26 Dec 2006 00:03:44 +0000 (GMT)
> From: Milton Cezar Ribeiro <milton_ruser@yahoo.com.br>
> Subject: [R] defining color sequence in image()
> To: R-help <r-help@stat.math.ethz.ch>
> Message-ID: <973009.27835.qm@web56608.mail.re3.yahoo.com>
> Content-Type: text/plain
>
> Dear All,
>
> How can I define a color sequence for each image value? I have several
> images with values varying from 1 to 5, and I would like to assing a fixed
> color for each value (green for 1, yellow for 2...). I used somethink like:
>
>      > image(img,col=c("green","yellow",...))
>
> but unfortunately whem I have absent values, the color that I defined for
> each values change.
>
> Thanks a lot,
>
> Miltinho
> Brazil
>
> PS : Merry Christmas!!!
>
> __________________________________________________
>
>
>        [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 12
> Date: Tue, 26 Dec 2006 11:16:20 +1100
> From: Jim Lemon <jim@bitwrit.com.au>
> Subject: Re: [R] Problem to generate training data set and test data
>        set
> To: Aimin Yan <aiminy@iastate.edu>
> Cc: r-help@stat.math.ethz.ch
> Message-ID: <459069D4.8080103@bitwrit.com.au>
> Content-Type: text/plain; charset=us-ascii; format=flowed
>
> Aimin Yan wrote:
> > I have a full data set like this:
> >
> >     aa bas    aas bms   ams bcu        acu     omega       y
> > 1 ALA   0 127.71   0 69.99   0 -0.2498560  79.91470 outward
> > 2 PRO   0  68.55   0 55.44   0 -0.0949008  76.60380 outward
> > 3 ALA   0  52.72   0 47.82   0 -0.0396550  52.19970 outward
> > 4 PHE   0  22.62   0 31.21   0  0.1270330 169.52500  inward
> > 5 SER   0  71.32   0 52.84   0 -0.1312380   7.47528 outward
> > 6 VAL   0  12.92   0 22.40   0  0.1728390 149.09400  inward
> >
> ......................................................................................
> >
> >
> > aa have 19 levels, and there are different number of observation for
> each
> > levels.
> > I want to pick 75% of observations of each levels randomly to generate a
> > training set,
> > and 25% of observation of each levels to generate a testing set.
> >
> Hi Aimin,
> I haven't tested this exhaustively, but I think it does what you want.
>
> get.prob.sample<-function(x,prob=0.5) {
> xlevels<-levels(as.factor(x))
> xlength<-length(x)
> xsamp<-rep(FALSE,xlength)
> for(i in xlevels) {
>   lengthi<-length(x[x == i])
>   xsamp[sample(which(x == i),lengthi*prob)]<-TRUE
> }
> return(xsamp)
> }
>
> get.prob.sample(mydata$aa,0.75)
>
> Jim
>
>
>
> ------------------------------
>
> Message: 13
> Date: Tue, 26 Dec 2006 10:59:45 +1000
> From: <Bill.Venables@csiro.au>
> Subject: Re: [R] Higher Dimensional Matrices
> To: <larsfromspace@web.de>, <r-help@stat.math.ethz.ch>
> Message-ID:
>        <B998A44C8986644EA8029CFE6396A924840A8E@exqld2-bne.qld.csiro.au>
> Content-Type: text/plain;       charset="us-ascii"
>
> Lars from space [sic] asks:
> > -----Original Message-----
>
> > From: r-help-bounces@stat.math.ethz.ch
> > [mailto:r-help-bounces@stat.math.ethz.ch] On Behalf Of downunder
> > Sent: Monday, 25 December 2006 12:31 PM To: r-help@stat.math.ethz.ch
> > Subject: [R] Higher Dimensional Matrices
> >
> >
> > Hi all.
> >
> > I want to calculate partial correlations while controlling for one
> > or more variables. That works already fine when I control for
> > example just for x[,1] and x[,2] that gives me one single
> > correlation matrix and i have to to it for x [,1]...x[,10]. That
> > would give me out 10 matrices. Controlling for 2 Variables 100
> > matrices. how can I run a loop to get f.e the 10 or 100 matrices at
> > once? I appreciate for every hint. have nice holidays.
>
> I don't quite understand this.  You have 10 variables and you want to
> find the partial correlations controlling for two of them at a time.
> If you take each possible set of two variables to condition upon at a
> time, this would give you choose(10, 2) = 45 matrices, wouldn't it?
> Where do you get '10 or 100' matrices from?
>
> >
> > greetings lars
> >
> > x <- read.table("C:/.....dat")
> > dim(x) #200x10
> > a <- matrix(0,200,10)
> > for (i in 1:10)
> > a[,i] <- residuals(lm(x[,i]~1+x[,1]+x[,2]))
> > b <- matrix(0,200,10)
> > for (i in 1:10)
> > b[,i] <- residuals(lm(x[,i]~1+x[,1]+x[,2]))
> > #a=round(a,5)
> > #b=round(b,5)
> > d <- cor(a,b)
> > d
>
> But a and b are the same, aren't they?  Why do you need to compute
> them twice?  Why not just use cor(a, a) [which is the same as cor(a),
> of course]?
>
> There is a more serious problem here, though.  Your residuals are
> after regression on x[, 1:2] so if you *select* x[, 1:2] as the
> y-variables in your regression then the residuals are going to be
> zero, but in practice round-off error.  so the first two rows and
> columns of d will be correlations with round-off error,
> i.e. misleading junk.  It doesn't make sense to include the
> conditioning variables in the correlation matrix *conditioning on
> them*.  Only the 8 x 8 matrix of the others among themselves is
> defined, really.
>
> So how do we do it?  Here are a few pointers.
>
> To start, here is a function that uses a somewhat more compact way of
> finding the partial correlations than your method.  Sorting out how it
> works should be a useful exercise.
>
> partialCorr <- function (cond, mat)
>        cor(qr.resid(qr(cbind(1, mat[, cond])), mat[, -cond]))
>
> To find the matrix of partial correlations conditioning on x[, 1:2]
> you would use
>
> d <- partialCorr(c(1,2), x)
>
> So how to do it for all possible conditioning pairs of variables.
> Well you could do it in an obvious loop:
>
> cmats <- array(0, c(8,8,45))
> k <- 0
> for(i in 1:9) for(j in (i+1):10) {
>    k <- k+1
>    cmats[, , k] <- partialCorr(c(i, j), x)
> }
>
> Now the results are in a 3-way array, but without any kind of labels.
> Perhaps you should think about how to fix that one yourself...
>
> With more than two conditioning variables you should probably use a
> function to generate the subsets of the appropriate size rather than
> trying to write ever more deeply nested loops.  There are plenty of
> functions around to do this.
>
> Bill Venables.
>
>
>
> ------------------------------
>
> Message: 14
> Date: Tue, 26 Dec 2006 00:07:57 -0500
> From: "Pedro Mardones" <mardones.p@gmail.com>
> Subject: [R] sequential row selection in dataframe
> To: R-help@stat.math.ethz.ch
> Message-ID:
>        <83dca7860612252107j60f4836aq6cf3debc1dec4045@mail.gmail.com>
> Content-Type: text/plain; charset=UTF-8; format=flowed
>
> Dear all;
>
> I'm wondering if there is any 'efficient' approach for selecting a
> sample of 'every nth rows'  from a dataframe. For example, let's use
> the dataframe GAGurine in MASS library:
>
> > length(GAGurine[,1])
> [1] 314
>
> # select an 75% of the dataset, i.e. = 236 rows, every 2 rows starting
> from row 1
> > test<-GAGurine[seq(1,314,2),]
> > length(test[,1])
> [1] 157
>
> # so, I still need another 79 rows, one way could be:
> test2<-GAGurine[-seq(1,314,2),]
> > length(test2[,1])
> [1] 157
> > test3<-test2[seq(1,157,2),]
>
> # and then
> final<-rbind(test2,test3)
> > length(final[,1])
> [1] 236
>
> Does anyone have a better idea to get the same results but without
> creating different datasets like test2 and test3?
>
> Thanks
> PM
>
>
>
> ------------------------------
>
> Message: 15
> Date: Mon, 25 Dec 2006 22:15:08 -0800 (PST)
> From: downunder <larsfromspace@web.de>
> Subject: Re: [R] Higher Dimensional Matrices
> To: r-help@stat.math.ethz.ch
> Message-ID: <8051407.post@talk.nabble.com>
> Content-Type: text/plain; charset=us-ascii
>
>
> Hi Bill. Thanks for your extensive hints especially with arrays.
> That solves my problem now and I am also able to control for every
> combination of variables.
>
> Have a merry christmas. lars
>
>
> >
> > x <- read.table("C:/.....dat")
> > dim(x) #200x10
> > a <- matrix(0,200,10)
> > for (i in 1:10)
> > a[,i] <- residuals(lm(x[,i]~1+x[,1]+x[,2]))
> > b <- matrix(0,200,10)
> > for (i in 1:10)
> > b[,i] <- residuals(lm(x[,i]~1+x[,1]+x[,2]))
> > #a=round(a,5)
> > #b=round(b,5)
> > d <- cor(a,b)
> > d
>
> But a and b are the same, aren't they?  Why do you need to compute
> them twice?  Why not just use cor(a, a) [which is the same as cor(a),
> of course]?
>
> There is a more serious problem here, though.  Your residuals are
> after regression on x[, 1:2] so if you *select* x[, 1:2] as the
> y-variables in your regression then the residuals are going to be
> zero, but in practice round-off error.  so the first two rows and
> columns of d will be correlations with round-off error,
> i.e. misleading junk.  It doesn't make sense to include the
> conditioning variables in the correlation matrix *conditioning on
> them*.  Only the 8 x 8 matrix of the others among themselves is
> defined, really.
>
> So how do we do it?  Here are a few pointers.
>
> To start, here is a function that uses a somewhat more compact way of
> finding the partial correlations than your method.  Sorting out how it
> works should be a useful exercise.
>
> partialCorr <- function (cond, mat)
>        cor(qr.resid(qr(cbind(1, mat[, cond])), mat[, -cond]))
>
> To find the matrix of partial correlations conditioning on x[, 1:2]
> you would use
>
> d <- partialCorr(c(1,2), x)
>
> So how to do it for all possible conditioning pairs of variables.
> Well you could do it in an obvious loop:
>
> cmats <- array(0, c(8,8,45))
> k <- 0
> for(i in 1:9) for(j in (i+1):10) {
>    k <- k+1
>    cmats[, , k] <- partialCorr(c(i, j), x)
> }
>
> Now the results are in a 3-way array, but without any kind of labels.
> Perhaps you should think about how to fix that one yourself...
>
> With more than two conditioning variables you should probably use a
> function to generate the subsets of the appropriate size rather than
> trying to write ever more deeply nested loops.  There are plenty of
> functions around to do this.
>
> Bill Venables.
>
> ______________________________________________
> R-help@stat.math.ethz.ch 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.
>
>
>
> --
> View this message in context:
> http://www.nabble.com/Re%3A--R--Higher-Dimensional-Matrices-tf2880928.html#a8051407
> Sent from the R help mailing list archive at Nabble.com.
>
>
>
> ------------------------------
>
> Message: 16
> Date: Tue, 26 Dec 2006 06:41:16 +0000 (GMT)
> From: justin bem <justin_bem@yahoo.fr>
> Subject: [R] Merry Xmas
> To: R Maillist <r-help@stat.math.ethz.ch>
> Message-ID: <20061226064116.53756.qmail@web23014.mail.ird.yahoo.com>
> Content-Type: text/plain
>
> Merry Xmas to U all and happy New year 2007.
>
> Justin BEM
> El�ve Ing�nieur Statisticien Economiste
> BP 294 Yaound�.
> T�l (00237)9597295.
>
>
>
>
>
>
>
>
>
> ___________________________________________________________________________
>
> interface r�volutionnaire.
>
>        [[alternative HTML version deleted]]
>
>
>
> ------------------------------
>
> Message: 17
> Date: Tue, 26 Dec 2006 17:42:18 +0900
> From: talepanda <talepanda@gmail.com>
> Subject: Re: [R] defining color sequence in image()
> To: R-help <r-help@stat.math.ethz.ch>
> Message-ID:
>        <3948d9e50612260042l49c52176teb4996fdb6ebe585@mail.gmail.com>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
> one solution is:
>
> img1<-matrix(1:5)
> img2<-matrix(2:5)
> col<-1:5 # col<-c("green","yellow",...)
> image(img1,col=col[sort(unique(img1))])
> image(img2,col=col[sort(unique(img2))])
>
> On 12/26/06, Milton Cezar Ribeiro <milton_ruser@yahoo.com.br> wrote:
> > Dear All,
> >
> >   How can I define a color sequence for each image value? I have several
> > images with values varying from 1 to 5, and I would like to assing a
> fixed
> > color for each value (green for 1, yellow for 2...). I used somethink
> like:
> >
> >       > image(img,col=c("green","yellow",...))
> >
> >   but unfortunately whem I have absent values, the color that I defined
> for
> > each values change.
> >
> >   Thanks a lot,
> >
> >   Miltinho
> >   Brazil
> >
> >   PS : Merry Christmas!!!
> >
> >  __________________________________________________
> >
> >
> >       [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@stat.math.ethz.ch 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: 18
> Date: Mon, 25 Dec 2006 17:13:46 -0600
> From: "Geoffrey Zhu" <gzhu@peak6.com>
> Subject: [R] HOW TO: vectorize an iterative process.
> To: <r-help@stat.math.ethz.ch>
> Message-ID:
>        <99F81FFD0EA54E4DA8D4F1BFE272F3410821BE@ppi-mail1.chicago.peak6.net
> >
> Content-Type: text/plain;       charset="utf-8"
>
> SGkgRXZlcnlvbmUsDQogDQpJIGFtIHN0dWNrIHdpdGggYSBzaW1wbGUgcHJvYmxlbS4gU3Vw
> cG9zZSBJIGhhdmUgYSB2ZWN0b3IgeCwgYW5kIEkgd2FudCB0byBjYWxjdWxhdGUgeVtpXT14
> W2krMV0teFtpXSwgaXQgaXMgdmVyeSBlYXN5LiBJIGp1c3QgbmVlZCB0byB3cml0ZSB5PC14
> WzI6bGVuZ3RoKHgpXS14WzE6bGVuZ3RoKHgpLTFdLiANCiANCk5vdyBpZiBJIGtub3cgeSwg
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>
>
>
> ------------------------------
>
> Message: 19
> Date: Mon, 25 Dec 2006 13:36:50 +0100
> From: "Sebastian Michalski" <s_michalski@o2.pl>
> Subject: [R] writing to S-PLUS .dat file
> To: <r-help@stat.math.ethz.ch>
> Message-ID: <005001c72821$59e96bd0$51cf1251@smicha>
> Content-Type: text/plain; format=flowed; charset="iso-8859-2";
>        reply-type=original
>
> Dear Users,
> I am new to R. I use write() to write my data in .txt format. I'd like
> to write to a disc any kind of data in a .dat S-PLUS format.
> Please help.
> SM
>
>
>
> ------------------------------
>
> Message: 20
> Date: Mon, 25 Dec 2006 14:57:04 +0100 (CET)
> From: Achim Zeileis <Achim.Zeileis@wu-wien.ac.at>
> Subject: Re: [R] extend summary.lm for hccm?
> To: John Fox <jfox@mcmaster.ca>
> Cc: r-help@stat.math.ethz.ch, 'ivo welch' <ivowel@gmail.com>,   'Dirk
>        Eddelbuettel' <edd@debian.org>
> Message-ID:
>        <Pine.LNX.4.44.0612251439200.8229-100000@disco.wu-wien.ac.at>
> Content-Type: TEXT/PLAIN; charset=US-ASCII
>
> On Sun, 24 Dec 2006, John Fox wrote:
>
> > If I remember Freedman's recent paper correctly, he argues that sandwich
> > variance estimator, though problematic in general, is not problematic in
> the
> > case that White described -- an otherwise correctly specified linear
> model
> > with heteroscedasticity estimated by least-squares.
>
> More generally, sandwich-type estimators are valid (i.e., estimate
> the right quantity, although not necessarily precisely, as Frank pointed
> out) in situations where the estimating functions are correctly specified
> but remaining aspets of the likelihood (not captured in the estimating
> functions) are potentially not.
>
> In linear models, it is easy to see what this means: the mean function has
> to be correctly specified (i.e., the errors have zero mean) but the
> correlation structure of the errors (i.e., their (co-)variances) might
> differ from the usual assumptions. In GLMs, in particular logistic
> regression, it is much more difficult to see against which types of
> misspecification sandwich-based inference is robust.
>
> Freedman's paper stresses the point that many model misspecifications also
> imply misspecified estimating functions (and in his example this is rather
> obvious) so that consequently the sandwich-type estimators estimate the
> wrong quantity.
>
> Best wishes,
> Z
>
>
>
> ------------------------------
>
> _______________________________________________
> R-help@stat.math.ethz.ch 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.
>
> End of R-help Digest, Vol 46, Issue 26
> **************************************
>



-- 
Om Prakash
Research Student,
Bioinformatics Center
University of Pune
Pune-411007
India

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