On Jun 4, 2017, at 1:36 PM, David L Carlson <dcarlson at tamu.edu> wrote:
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.
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