[R] Replacing NAs in long format
William Dunlap
wdunlap at tibco.com
Sun Nov 4 01:37:37 CET 2012
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.
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