Hi:
Try this for your second loop instead:
for(s in school.list){
print(s)
print(subset(input.data, sch == s))
}
[1] 1
sch pop
1 1 100
2 1 200
[1] 2
sch pop
3 2 300
4 2 400
Don't confound the 'sch' variable in your data frame with the
index in your loop :)
HTH,
Dennis
On Mon, Feb 1, 2010 at 8:17 PM, David Katz wrote:
>
> I was surprised to see this unexpected behavior of subset in a for loop. I
> looked in subset.data.frame and it seemed to me that both versions should
> work, since the subset call should be evaluated in the global environment -
> but perhaps I don't understand environments well enough. Can someone
> enlighten me? In any case, this is a bit of a gotcha for naive users of
> subset.
>
> input.data <-
> data.frame(sch=c(1,1,2,2),
> pop=c(100,200,300,400))
>
> school.var <- "sch"
>
> school.list <- 1:2
>
> for(sch in school.list){
> print(sch)
> #do this before subset!:
> right.sch.p <-
> input.data[,school.var] == sch
> print( subset(input.data,right.sch.p)) #this is what I expected
> }
>
> ## [1] 1
> ## sch pop
> ## 1 1 100
> ## 2 1 200
> ## [1] 2
> ## sch pop
> ## 3 2 300
> ## 4 2 400
>
>
> for(sch in school.list){
> print(sch)
> print(subset(input.data,input.data[,school.var] == sch)) #note - compact
> version fails!
> }
>
> ## [1] 1
> ## sch pop
> ## 1 1 100
> ## 2 1 200
> ## 3 2 300
> ## 4 2 400
> ## [1] 2
> ## sch pop
> ## 1 1 100
> ## 2 1 200
> ## 3 2 300
> ## 4 2 400
>
> --
> View this message in context:
> http://n4.nabble.com/subset-function-unexpected-behavior-tp1459535p1459535.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> 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.
>
[[alternative HTML version deleted]]