HI Bill,
It is much simpler.
# with aggregate() and merge()
res1<-with(dat2,aggregate(seq_len(nrow(dat2)),by=list(idr=idr),FUN=function(i)
with(dat2[i,], any(schyear<=5 & year ==0))))
res2<-merge(dat2,res1,by="idr")
colnames(res2)[4]<-"flag"
within(res2,{flag<-as.integer(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
A.K.
----- Original Message -----
From: William Dunlap <wdun...@tibco.com>
To: arun <smartpink...@yahoo.com>; Christopher Desjardins
<cddesjard...@gmail.com>
Cc: R help <r-help@r-project.org>
Sent: Saturday, November 3, 2012 9:21 PM
Subject: RE: [R] Replacing NAs in long format
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-boun...@r-project.org [mailto:r-help-boun...@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-boun...@r-project.org [mailto:r-help-boun...@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 <cddesjard...@gmail.com>
> > To: jim holtman <jholt...@gmail.com>
> > Cc: r-help@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$flag = ifelse(schyear <= 5 & year ==0, 1, 0)
> >
> > Does not work because it will create:
> >
> > idr schyear year flag
> > 1 4 -1 0
> > 1 5 0 1
> > 1 6 1 0
> > 1 7 2 0
> > 2 9 0 0
> > 2 10 1 0
> > 2 11 2 0
> >
> > And thus flag changes for an idr. Which it shouldn't.
> >
> > Thanks,
> > Chris
> >
> >
> > On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins <
> > cddesjard...@gmail.com> wrote:
> >
> > > Hi Jim,
> > > Thank you so much. That does exactly what I want.
> > > Chris
> > >
> > >
> > > On Sat, Nov 3, 2012 at 1:30 PM, jim holtman <jholt...@gmail.com> wrote:
> > >
> > >> > x <- read.table(text = "idr schyear year
> > >> + 1 8 0
> > >> + 1 9 1
> > >> + 1 10 NA
> > >> + 2 4 NA
> > >> + 2 5 -1
> > >> + 2 6 0
> > >> + 2 7 1
> > >> + 2 8 2
> > >> + 2 9 3
> > >> + 2 10 4
> > >> + 2 11 NA
> > >> + 2 12 6
> > >> + 3 4 NA
> > >> + 3 5 -2
> > >> + 3 6 -1
> > >> + 3 7 0
> > >> + 3 8 1
> > >> + 3 9 2
> > >> + 3 10 3
> > >> + 3 11 NA", header = TRUE)
> > >> > # you did not specify if there might be multiple contiguous NAs,
> > >> > # so there are a lot of checks to be made
> > >> > x.l <- lapply(split(x, x$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
> > >> <cddesjard...@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@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@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@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@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.
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