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 <david@davidkatzconsulting.com>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.
>
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