Message-ID: <E66794E69CFDE04D9A70842786030B930CE2B1A6@PA-MBX04.na.tibco.com>
Date: 2012-11-04T01:21:23Z
From: William Dunlap
Subject: Replacing NAs in long format
In-Reply-To: <E66794E69CFDE04D9A70842786030B930CE2B16D@PA-MBX04.na.tibco.com>
Or, even simpler,
> flag <- with(dat2, ave(schyear<=5 & year==0, idr, FUN=any))
> data.frame(dat2, flag)
idr schyear year flag
1 1 4 -1 TRUE
2 1 5 0 TRUE
3 1 6 1 TRUE
4 1 7 2 TRUE
5 2 9 0 FALSE
6 2 10 1 FALSE
7 2 11 2 FALSE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> Of William Dunlap
> Sent: Saturday, November 03, 2012 5:38 PM
> To: arun; Christopher Desjardins
> Cc: R help
> Subject: Re: [R] Replacing NAs in long format
>
> ave() or split<-() can make that easier to write, although it
> may take some time to internalize the idiom. E.g.,
>
> > flag <- rep(NA, nrow(dat2)) # add as.integer if you prefer 1,0 over TRUE,FALSE
> > split(flag, dat2$idr) <- lapply(split(dat2, dat2$idr), function(d)with(d, any(schyear<=5 &
> year==0)))
> > data.frame(dat2, flag)
> idr schyear year flag
> 1 1 4 -1 TRUE
> 2 1 5 0 TRUE
> 3 1 6 1 TRUE
> 4 1 7 2 TRUE
> 5 2 9 0 FALSE
> 6 2 10 1 FALSE
> 7 2 11 2 FALSE
> or
> > ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,], any(schyear<=5 &
> year==0)))
> [1] 1 1 1 1 0 0 0
> > flag <- ave(seq_len(nrow(dat2)), dat2$idr, FUN=function(i)with(dat2[i,],
> any(schyear<=5 & year==0)))
> > data.frame(dat2, flag)
> idr schyear year flag
> 1 1 4 -1 1
> 2 1 5 0 1
> 3 1 6 1 1
> 4 1 7 2 1
> 5 2 9 0 0
> 6 2 10 1 0
> 7 2 11 2 0
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
> > -----Original Message-----
> > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> > Of arun
> > Sent: Saturday, November 03, 2012 5:01 PM
> > To: Christopher Desjardins
> > Cc: R help
> > Subject: Re: [R] Replacing NAs in long format
> >
> > Hi,
> > May be this helps:
> > dat2<-read.table(text="
> > idr? schyear? year
> > 1??????? 4????????? -1
> > 1??????? 5??????????? 0
> > 1??????? 6??????????? 1
> > 1??????? 7??????????? 2
> > 2??????? 9??????????? 0
> > 2??????? 10??????????? 1
> > 2??????? 11????????? 2
> > ",sep="",header=TRUE)
> >
> > ?dat2$flag<-unlist(lapply(split(dat2,dat2$idr),function(x)
> > rep(ifelse(any(apply(x,1,function(x) x[2]<=5 &
> x[3]==0)),1,0),nrow(x))),use.names=FALSE)
> > ?dat2
> > #? idr schyear year flag
> > #1?? 1?????? 4?? -1??? 1
> > #2?? 1?????? 5??? 0??? 1
> > #3?? 1?????? 6??? 1??? 1
> > #4?? 1?????? 7??? 2??? 1
> > #5?? 2?????? 9??? 0??? 0
> > #6?? 2????? 10??? 1??? 0
> > #7?? 2????? 11??? 2??? 0
> > A.K.
> >
> >
> >
> >
> > ----- Original Message -----
> > From: Christopher Desjardins <cddesjardins at gmail.com>
> > To: jim holtman <jholtman at gmail.com>
> > Cc: r-help at r-project.org
> > Sent: Saturday, November 3, 2012 7:09 PM
> > Subject: Re: [R] Replacing NAs in long format
> >
> > I have a similar sort of follow up and I bet I could reuse some of this
> > code but I'm not sure how.
> >
> > Let's say I want to create a flag that will be equal to 1 if schyear? < = 5
> > and year = 0 for a given idr. For example
> >
> > > dat
> >
> > idr? schyear? year
> > 1? ? ? ? 4? ? ? ? ? -1
> > 1? ? ? ? 5? ? ? ? ? ? 0
> > 1? ? ? ? 6? ? ? ? ? ? 1
> > 1? ? ? ? 7? ? ? ? ? ? 2
> > 2? ? ? ? 9? ? ? ? ? ? 0
> > 2? ? ? ? 10? ? ? ? ? ? 1
> > 2? ? ? ? 11? ? ? ? ? 2
> >
> > How could I make the data look like this?
> >
> > idr? schyear? year? flag
> > 1? ? ? ? 4? ? ? ? ? -1? ? 1
> > 1? ? ? ? 5? ? ? ? ? ? 0? ? 1
> > 1? ? ? ? 6? ? ? ? ? ? 1? ? 1
> > 1? ? ? ? 7? ? ? ? ? ? 2? ? 1
> > 2? ? ? ? 9? ? ? ? ? ? 0? ? 0
> > 2? ? ? ? 10? ? ? ? ? ? 1? ? 0
> > 2? ? ? ? 11? ? ? ? ? 2? ? 0
> >
> >
> > I am not sure how to end up not getting both 0s and 1s for the 'flag'
> > variable for an idr. For example,
> >
> > dat$flag = ifelse(schyear <= 5 & year ==0, 1, 0)
> >
> > Does not work because it will create:
> >
> > idr? schyear? year? flag
> > 1? ? ? ? 4? ? ? ? ? -1? ? 0
> > 1? ? ? ? 5? ? ? ? ? ? 0? ? 1
> > 1? ? ? ? 6? ? ? ? ? ? 1? ? 0
> > 1? ? ? ? 7? ? ? ? ? ? 2? ? 0
> > 2? ? ? ? 9? ? ? ? ? ? 0? ? 0
> > 2? ? ? ? 10? ? ? ? ? ? 1? ? 0
> > 2? ? ? ? 11? ? ? ? ? 2? ? 0
> >
> > And thus flag changes for an idr. Which it shouldn't.
> >
> > Thanks,
> > Chris
> >
> >
> > On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins <
> > cddesjardins at gmail.com> wrote:
> >
> > > Hi Jim,
> > > Thank you so much. That does exactly what I want.
> > > Chris
> > >
> > >
> > > On Sat, Nov 3, 2012 at 1:30 PM, jim holtman <jholtman at gmail.com> wrote:
> > >
> > >> > x <- read.table(text = "idr? schyear year
> > >> +? 1? ? ? 8? ? 0
> > >> +? 1? ? ? 9? ? 1
> > >> +? 1? ? ? 10? NA
> > >> +? 2? ? ? 4? NA
> > >> +? 2? ? ? 5? -1
> > >> +? 2? ? ? 6? ? 0
> > >> +? 2? ? ? 7? ? 1
> > >> +? 2? ? ? 8? ? 2
> > >> +? 2? ? ? 9? ? 3
> > >> +? 2? ? ? 10? ? 4
> > >> +? 2? ? ? 11? NA
> > >> +? 2? ? ? 12? ? 6
> > >> +? 3? ? ? 4? NA
> > >> +? 3? ? ? 5? -2
> > >> +? 3? ? ? 6? -1
> > >> +? 3? ? ? 7? ? 0
> > >> +? 3? ? ? 8? ? 1
> > >> +? 3? ? ? 9? ? 2
> > >> +? 3? ? ? 10? ? 3
> > >> +? 3? ? ? 11? NA", header = TRUE)
> > >> >? # you did not specify if there might be multiple contiguous NAs,
> > >> >? # so there are a lot of checks to be made
> > >> >? x.l <- lapply(split(x, x$idr), function(.idr){
> > >> +? ? # check for all NAs -- just return indeterminate state
> > >> +? ? if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
> > >> +? ? # repeat until all NAs have been fixed; takes care of contiguous
> > >> ones
> > >> +? ? while (any(is.na(.idr$year))){
> > >> +? ? ? ? # find all the NAs
> > >> +? ? ? ? for (i in which(is.na(.idr$year))){
> > >> +? ? ? ? ? ? if ((i == 1L) && (!is.na(.idr$year[i + 1L]))){
> > >> +? ? ? ? ? ? ? ? .idr$year[i] <- .idr$year[i + 1L] - 1
> > >> +? ? ? ? ? ? } else if ((i > 1L) && (!is.na(.idr$year[i - 1L]))){
> > >> +? ? ? ? ? ? ? ? .idr$year[i] <- .idr$year[i - 1L] + 1
> > >> +? ? ? ? ? ? } else if ((i < nrow(.idr)) && (!is.na(.idr$year[i +
> > >> 1L]))){
> > >> +? ? ? ? ? ? ? ? .idr$year[i] <- .idr$year[i + 1L] -1
> > >> +? ? ? ? ? ? }
> > >> +? ? ? ? }
> > >> +? ? }
> > >> +? ? return(.idr)
> > >> + })
> > >> > do.call(rbind, x.l)
> > >>? ? ? idr schyear year
> > >> 1.1? ? 1? ? ? 8? ? 0
> > >> 1.2? ? 1? ? ? 9? ? 1
> > >> 1.3? ? 1? ? ? 10? ? 2
> > >> 2.4? ? 2? ? ? 4? -2
> > >> 2.5? ? 2? ? ? 5? -1
> > >> 2.6? ? 2? ? ? 6? ? 0
> > >> 2.7? ? 2? ? ? 7? ? 1
> > >> 2.8? ? 2? ? ? 8? ? 2
> > >> 2.9? ? 2? ? ? 9? ? 3
> > >> 2.10? 2? ? ? 10? ? 4
> > >> 2.11? 2? ? ? 11? ? 5
> > >> 2.12? 2? ? ? 12? ? 6
> > >> 3.13? 3? ? ? 4? -3
> > >> 3.14? 3? ? ? 5? -2
> > >> 3.15? 3? ? ? 6? -1
> > >> 3.16? 3? ? ? 7? ? 0
> > >> 3.17? 3? ? ? 8? ? 1
> > >> 3.18? 3? ? ? 9? ? 2
> > >> 3.19? 3? ? ? 10? ? 3
> > >> 3.20? 3? ? ? 11? ? 4
> > >> >
> > >> >
> > >>
> > >>
> > >> On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
> > >> <cddesjardins at gmail.com> wrote:
> > >> > Hi,
> > >> > I have the following data:
> > >> >
> > >> >> data[1:20,c(1,2,20)]
> > >> > idr? schyear year
> > >> > 1? ? ? 8? ? 0
> > >> > 1? ? ? 9? ? 1
> > >> > 1? ? ? 10? NA
> > >> > 2? ? ? 4? NA
> > >> > 2? ? ? 5? -1
> > >> > 2? ? ? 6? ? 0
> > >> > 2? ? ? 7? ? 1
> > >> > 2? ? ? 8? ? 2
> > >> > 2? ? ? 9? ? 3
> > >> > 2? ? ? 10? ? 4
> > >> > 2? ? ? 11? NA
> > >> > 2? ? ? 12? ? 6
> > >> > 3? ? ? 4? NA
> > >> > 3? ? ? 5? -2
> > >> > 3? ? ? 6? -1
> > >> > 3? ? ? 7? ? 0
> > >> > 3? ? ? 8? ? 1
> > >> > 3? ? ? 9? ? 2
> > >> > 3? ? ? 10? ? 3
> > >> > 3? ? ? 11? NA
> > >> >
> > >> > What I want to do is replace the NAs in the year variable with the
> > >> > following:
> > >> >
> > >> > idr? schyear year
> > >> > 1? ? ? 8? ? 0
> > >> > 1? ? ? 9? ? 1
> > >> > 1? ? ? 10? 2
> > >> > 2? ? ? 4? -2
> > >> > 2? ? ? 5? -1
> > >> > 2? ? ? 6? ? 0
> > >> > 2? ? ? 7? ? 1
> > >> > 2? ? ? 8? ? 2
> > >> > 2? ? ? 9? ? 3
> > >> > 2? ? ? 10? ? 4
> > >> > 2? ? ? 11? 5
> > >> > 2? ? ? 12? ? 6
> > >> > 3? ? ? 4? -3
> > >> > 3? ? ? 5? -2
> > >> > 3? ? ? 6? -1
> > >> > 3? ? ? 7? ? 0
> > >> > 3? ? ? 8? ? 1
> > >> > 3? ? ? 9? ? 2
> > >> > 3? ? ? 10? ? 3
> > >> > 3? ? ? 11? 4
> > >> >
> > >> > I have no idea how to do this. What it needs to do is make sure that for
> > >> > each subject (idr) that it either adds a 1 if it is preceded by a value
> > >> in
> > >> > year or subtracts a 1 if it comes before a year value.
> > >> >
> > >> > Does that make sense? I could do this in Excel but I am at a loss for
> > >> how
> > >> > to do this in R. Please reply to me as well as the list if you respond.
> > >> >
> > >> > Thanks!
> > >> > Chris
> > >> >
> > >> >? ? ? ? [[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
> > >> Data Munger Guru
> > >>
> > >> What is the problem that you are trying to solve?
> > >> Tell me what you want to do, not how you want to do it.
> > >>
> > >
> > >
> >
> > ??? [[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.
> >
> >
> > ______________________________________________
> > 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.