Thanks these different examples work perfectly.
Chris

On Sat, Nov 3, 2012 at 8:32 PM, arun <smartpink...@yahoo.com> wrote:

> 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.
>
>
> ______________________________________________
> 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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