problem with lapply(x, subset, ...) and variable select argument

As Gabor said, the issue here is that subset.data.frame() evaluates 
the value of the `select' argument in the parent.frame(); Thus, if you 
create a local function within lapply() (or sapply()) it works:

tt <- function (n) {
    x <- list(data.frame(a = 1, b = 2), data.frame(a = 3, b = 4))
    print(lapply(x, function(y, n) subset(y, select = n), n = n))
    print(sapply(x, function(y, n) subset(y, select = n), n = n))
}

tt("a")


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm



----- Original Message ----- 
From: "joerg van den hoff" <j.van_den_hoff at fz-rossendorf.de>
To: "Gabor Grothendieck" <ggrothendieck at gmail.com>; "Thomas Lumley" 
<tlumley at u.washington.edu>
Cc: "r-help" <r-help at stat.math.ethz.ch>
Sent: Tuesday, October 11, 2005 10:18 AM
Subject: Re: [R] problem with lapply(x, subset,...) and variable 
select argument

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 <j.van_den_hoff at fz-rossendorf.de> 
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

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:

1.
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)?

2.
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

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