The problem is that subset looks into its parent frame but in this
case the parent frame is not the environment in tt but the environment
in lapply since tt does not call subset directly but rather lapply does.
Try this which is similar except we have added the line beginning
with environment before the print statement.
tt <- function (n) {
x <- list(data.frame(a=1,b=2), data.frame(a=3,b=4))
environment(lapply) <- environment()
print(lapply(x, subset, select = n))
}
n <- "b"
tt("a")
What this does is create a new version of lapply whose
parent is the environment in tt.
On 10/10/05, joerg van den hoff <[EMAIL PROTECTED]> wrote:
> I need to extract identically named columns from several data frames in
> a list. the column name is a variable (i.e. not known in advance). the
> whole thing occurs within a function body. I'd like to use lapply with a
> variable 'select' argument.
>
>
> example:
>
> tt <- function (n) {
> x <- list(data.frame(a=1,b=2), data.frame(a=3,b=4))
> for (xx in x) print(subset(xx, select = n)) ### works
> print (lapply(x, subset, select = a)) ### works
> print (lapply(x, subset, select = "a")) ### works
> print (lapply(x, subset, select = n)) ### does not work as intended
> }
> n = "b"
> tt("a") #works (but selects not the intended column)
> rm(n)
> tt("a") #no longer works in the lapply call including variable 'n'
>
>
> question: how can I enforce evaluation of the variable n such that
> the lapply call works? I suspect it has something to do with eval and
> specifying the correct evaluation frame, but how? ....
>
>
> many thanks
>
> joerg
>
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