[R] New var

David L Carlson dcarlson at tamu.edu
Sun Jun 4 19:36:01 CEST 2017


Since the number of choices is small (6), how about this?

Starting with Jeff's initial DFM:

DFM <- structure(list(obs = 1:6, start = structure(c(16467, 14710, 13152, 
13787, 15126, 12696), class = "Date"), end = structure(c(17167, 
14975, 13636, 13879, 15340, 12753), class = "Date"), D = c(700, 
265, 484, 92, 214, 57), bin = structure(c(6L, 3L, 5L, 1L, 3L, 
1L), .Label = c("[0,100)", "[100,200)", "[200,300)", "[300,400)", 
"[400,500)", "[500,Inf)"), class = c("ordered", "factor"))), .Names = c("obs", 
"start", "end", "D", "bin"), row.names = c(NA, -6L), class = "data.frame")

Construct a matrix of the six alternatives:

tvals <- c(1, -1, -1, -1, -1, 0, 1, -1, -1, -1, 0, 0, 1, -1, -1, 0, 0, 
    0, 1, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0)
tmat <- matrix(tvals, 6, 5, byrow=TRUE)
colnames(tmat) <- paste0("t", 1:5)
tmat
#      t1 t2 t3 t4 t5
# [1,]  1 -1 -1 -1 -1
# [2,]  0  1 -1 -1 -1
# [3,]  0  0  1 -1 -1
# [4,]  0  0  0  1 -1
# [5,]  0  0  0  0  1
# [6,]  0  0  0  0  0

idx <-as.numeric(DFM$bin)
(DFM <- data.frame(DFM, tmat[idx, ]))
#    obs      start        end   D       bin t1 t2 t3 t4 t5
# 1   1 2015-02-01 2017-01-01 700 [500,Inf)  0  0  0  0  0
# 2   2 2010-04-11 2011-01-01 265 [200,300)  0  0  1 -1 -1
# 3   3 2006-01-04 2007-05-03 484 [400,500)  0  0  0  0  1
# 4   4 2007-10-01 2008-01-01  92   [0,100)  1 -1 -1 -1 -1
# 5   5 2011-06-01 2012-01-01 214 [200,300)  0  0  1 -1 -1
# 6   6 2004-10-05 2004-12-01  57   [0,100)  1 -1 -1 -1 -1


David L. Carlson
Department of Anthropology
Texas A&M University

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Val
Sent: Sunday, June 4, 2017 11:31 AM
To: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
Cc: r-help at R-project.org
Subject: Re: [R] New var

Thank you Jeff and All,

Within a given time period (say 700 days, from the start day),  I am
expecting measurements taken at each time interval;. In this case "0" means
measurement taken, "1"  not taken (stopped or opted out  and " -1"  don't
consider that time period for that individual. This will be compared with
the actual measurements taken (Observed- expected)  within each time
interval.




On Sat, Jun 3, 2017 at 9:50 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
wrote:

> # read.table is NOT part of the data.table package
> #library(data.table)
> DFM <- read.table( text=
> 'obs start end
> 1 2/1/2015   1/1/2017
> 2 4/11/2010  1/1/2011
> 3 1/4/2006   5/3/2007
> 4 10/1/2007  1/1/2008
> 5 6/1/2011   1/1/2012
> 6 10/5/2004 12/1/2004
> ',header = TRUE, stringsAsFactors = FALSE)
> # cleaner way to compute D
> DFM$start <- as.Date( DFM$start, format="%m/%d/%Y" )
> DFM$end <- as.Date( DFM$end, format="%m/%d/%Y" )
> DFM$D <- as.numeric( DFM$end - DFM$start, units="days" )
> # categorize your data into groups
> DFM$bin <- cut( DFM$D
>               , breaks=c( seq( 0, 500, 100 ), Inf )
>               , right=FALSE # do not include the right edge
>               , ordered_result = TRUE
>               )
> # brute force method you should have been able to figure out to show us
> some work
> DFM$t1 <- ifelse( DFM$D < 100, 1, 0 )
> DFM$t2 <- ifelse( 100 <= DFM$D & DFM$D < 200, 1, ifelse( DFM$D < 100, -1,
> 0 ) )
> DFM$t3 <- ifelse( 200 <= DFM$D & DFM$D < 300, 1, ifelse( DFM$D < 200, -1,
> 0 ) )
> DFM$t4 <- ifelse( 300 <= DFM$D & DFM$D < 400, 1, ifelse( DFM$D < 300, -1,
> 0 ) )
> DFM$t5 <- ifelse( 400 <= DFM$D & DFM$D < 500, 1, ifelse( DFM$D < 400, -1,
> 0 ) )
> # brute force method with ordered factor
> DFM$tf1 <- ifelse( "[0,100)" == DFM$bin, 1, 0 )
> DFM$tf2 <- ifelse( "[100,200)" == DFM$bin, 1, ifelse( "[100,200)" <
> DFM$bin, 0, -1 ) )
> DFM$tf3 <- ifelse( "[200,300)" == DFM$bin, 1, ifelse( "[200,300)" <
> DFM$bin, 0, -1 ) )
> DFM$tf4 <- ifelse( "[300,400)" == DFM$bin, 1, ifelse( "[300,400)" <
> DFM$bin, 0, -1 ) )
> DFM$tf5 <- ifelse( "[400,500)" == DFM$bin, 1, ifelse( "[400,500)" <
> DFM$bin, 0, -1 ) )
> # less obvious approach using the fact that factors are integers
> # and using the outer function to find all combinations of elements of two
> vectors
> # and the sign function
> DFM[ , paste0( "tm", 1:5 )] <- outer( as.integer( DFM$bin )
>                                     , 1:5
>                                     , FUN = function(x,y) {
>                                           z <- sign(y-x)+1L
>                                           ifelse( 2 == z, -1L, z )
>                                       }
>                                     )
>
> # my result, provided using dput for precise representation
> DFMresult <- structure(list(obs = 1:6, start = structure(c(16467, 14710,
> 13152, 13787, 15126, 12696), class = "Date"), end = structure(c(17167,
> 14975, 13636, 13879, 15340, 12753), class = "Date"), D = c(700,
> 265, 484, 92, 214, 57), bin = structure(c(6L, 3L, 5L, 1L, 3L,
> 1L), .Label = c("[0,100)", "[100,200)", "[200,300)", "[300,400)",
> "[400,500)", "[500,Inf)"), class = c("ordered", "factor")), t1 = c(0,
> 0, 0, 1, 0, 1), t2 = c(0, 0, 0, -1, 0, -1), t3 = c(0, 1, 0, -1,
> 1, -1), t4 = c(0, -1, 0, -1, -1, -1), t5 = c(0, -1, 1, -1, -1,
> -1), tf1 = c(0, 0, 0, 1, 0, 1), tf2 = c(0, 0, 0, -1, 0, -1),
>     tf3 = c(0, 1, 0, -1, 1, -1), tf4 = c(0, -1, 0, -1, -1, -1
>     ), tf5 = c(0, -1, 1, -1, -1, -1), tm1 = c(0, 0, 0, 1, 0,
>     1), tm2 = c(0, 0, 0, -1, 0, -1), tm3 = c(0, 1, 0, -1, 1,
>     -1), tm4 = c(0, -1, 0, -1, -1, -1), tm5 = c(0, -1, 1, -1,
>     -1, -1)), row.names = c(NA, -6L), .Names = c("obs", "start",
> "end", "D", "bin", "t1", "t2", "t3", "t4", "t5", "tf1", "tf2",
> "tf3", "tf4", "tf5", "tm1", "tm2", "tm3", "tm4", "tm5"), class =
> "data.frame")
>
> You did not address Bert's request for some context, but I am curious how
> he or Peter would have approached this problem, so I encourage you do
> provide some insight on the list as to why you are doing this.
>
>
> On Sat, 3 Jun 2017, Val wrote:
>
> Thank you all for the useful suggestion. I did some of my homework.
>>
>> library(data.table)
>> DFM <- read.table(header=TRUE, text='obs start end
>> 1 2/1/2015   1/1/2017
>> 2 4/11/2010  1/1/2011
>> 3 1/4/2006   5/3/2007
>> 4 10/1/2007  1/1/2008
>> 5 6/1/2011   1/1/2012
>> 6 10/5/2004 12/1/2004',stringsAsFactors = FALSE)
>> DFM
>>
>> DFM$D =as.numeric(difftime(as.Date(DFM$end,format="%m/%d/%Y"),
>> as.Date(DFM$start,format="%m/%d/%Y"), units = "days"))
>> DFM
>>
>> output.
>>     obs     start       end   D
>> 1   1  2/1/2015  1/1/2017 700
>> 2   2 4/11/2010  1/1/2011 265
>> 3   3  1/4/2006  5/3/2007 484
>> 4   4 10/1/2007  1/1/2008  92
>> 5   5  6/1/2011  1/1/2012 214
>> 6   6 10/5/2004 12/1/2004  57
>>
>> My problem is how do I get the other new variables
>>
>> obs     start       end   D  t1,t2,t3,t4, t5
>> 1, 2/1/2015,  1/1/2017, 700,0,0,0,0,0
>> 2, 4/11/2010, 1/1/2011, 265,0,0,1,-1,-1
>> 3, 1/4/2006,  5/3/2007, 484,0,0,0,0,1
>> 4, 10/1/2007, 1/1/2008, 92,1,-1,-1,-1,-1
>> 5, 6/1/2011,  1/1/2012, 214,0,0,1,-1,-1
>> 6, 10/15/2004,12/1/2004,47,1,-1,-1,-1,-1
>>
>> Thank you again.
>>
>>
>>
>> On Sat, Jun 3, 2017 at 12:13 AM, Bert Gunter <bgunter.4567 at gmail.com>
>> wrote:
>>
>>> Ii is difficult to provide useful help, because you have failed to
>>> read and follow the posting guide. In particular:
>>>
>>> 1. Plain text, not HTML.
>>> 2. Use dput() or provide code to create your example. Text printouts
>>> such as that which you gave require some work to wrangle into into an
>>> example that we can test.
>>>
>>> Specifically:
>>>
>>> 3. Have you gone through any R tutorials?-- it sure doesn't look like
>>> it. We do expect some effort to learn R before posting.
>>>
>>> 4. What is the format of your date columns? character, factors,
>>> POSIX,...? See ?date-time for details. Note particularly the
>>> "difftime" link to obtain intervals.
>>>
>>> 5. ?ifelse  for vectorized conditionals.
>>>
>>> Also, you might want to explain the context of what you are trying to
>>> do. I strongly suspect you shouldn't be doing it at all, but that is
>>> just a guess.
>>>
>>> Be sure to cc your reply to the list, not just to me.
>>>
>>> Cheers,
>>> Bert
>>>
>>>
>>> Bert Gunter
>>>
>>> "The trouble with having an open mind is that people keep coming along
>>> and sticking things into it."
>>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>>
>>>
>>> On Fri, Jun 2, 2017 at 8:49 PM, Val <valkremk at gmail.com> wrote:
>>>
>>>> Hi all,
>>>>
>>>> I have a data set with time interval and depending on the interval I
>>>> want
>>>> to create 5 more variables . Sample data below
>>>>
>>>> obs,   Start,   End
>>>> 1,2/1/2015,  1/1/2017
>>>> 2,4/11/2010, 1/1/2011
>>>> 3,1/4/2006,  5/3/2007
>>>> 4,10/1/2007, 1/1/2008
>>>> 5,6/1/2011,  1/1/2012
>>>> 6,10/15/2004,12/1/2004
>>>>
>>>> First, I want get  interval between the start date and end dates
>>>> (End-start).
>>>>
>>>>  obs,  Start , end, datediff
>>>> 1,2/1/2015,  1/1/2017, 700
>>>> 2,4/11/2010, 1/1/2011, 265
>>>> 3,1/4/2006,  5/3/2007, 484
>>>> 4,10/1/2007, 1/1/2008, 92
>>>> 5,6/1/2011,  1/1/2012, 214
>>>> 6,10/15/2004,12/1/2004,47
>>>>
>>>> Second. I want create 5 more variables  t1, t2, t3, t4 and  t5
>>>> The value of each variable is defined as follows
>>>> if datediff <   100 then  t1=1,  t2=t3=t4=t5=-1.
>>>> if datediff >= 100 and  < 200 then  t1=0, t2=1,t3=t4=t5=-1,
>>>> if datediff >= 200 and  < 300 then  t1=0, t2=0,t3=1,t4=t5=-1,
>>>> if datediff >= 300 and  < 400 then  t1=0, t2=0,t3=0,t4=1,t5=-1,
>>>> if datediff >= 400 and  < 500 then  t1=0, t2=0,t3=0,t4=0,t5=1,
>>>> if datediff >= 500 then  t1=0, t2=0,t3=0,t4=0,t5=0
>>>>
>>>> The complete out put looks like as follow.
>>>> obs, start,         end,    datediff,   t1, t2, t3, t4, t5
>>>> 1,    2/1/2015,   1/1/2017,    700, 0,  0,  0,  0,  0
>>>> 2,  4/11/2010,   1/1/2011,    265, 0,  0,  1, -1,  -1
>>>> 3,    1/4/2006,   5/3/2007,    484, 0,  0,  0, 0,   1
>>>> 4,   10/1/2007,  1/1/2008,      92, 1, -1, -1,-1,  -1
>>>> 5 ,    6/1/2011,    1/1/2012,  214,  0,  0,  1,-1,  -1
>>>> 6, 10/15/2004, 12/1/2004,     47, 1, -1, -1, -1, -1
>>>>
>>>> Thank you.
>>>>
>>>>         [[alternative HTML version deleted]]
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide http://www.R-project.org/posti
>>>> ng-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posti
>> ng-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>>
> ------------------------------------------------------------
> ---------------
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