Re: [R] problem with lapply(x, subset, ...) and variable select argument

From: joerg van den hoff <>
Date: Tue 11 Oct 2005 - 18:18:24 EST

Gabor Grothendieck wrote:

> 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 <> 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.
>>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)
>>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
>> mailing list
>>PLEASE do read the posting guide!

many thanks to thomas and gabor for their help. both solutions solve my problem perfectly.

but just as an attempt to improve my understanding of the inner workings of R (similar problems are sure to come up ...) two more question:

why does the call of the "[" function (thomas' solution) behave different from "subset" in that the look up of the variable "n" works without providing lapply with the current environment (which is nice)?

using 'subset' in this context becomes more cumbersome, if sapply is used. it seems that than I need
environment(sapply) <- environment(lapply) <- environment() sapply(x, subset, select = n))
to get it working (and that means you must know, that sapply uses lapply). or can I somehow avoid the additional explicit definition of the lapply-environment?

again: many thanks

joerg mailing list PLEASE do read the posting guide! Received on Tue Oct 11 18:23:42 2005

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