Marc,
Thanks,
sapply(ls(pat = "^name"),get)
was exactly what I was after. The default behavior for vectors of
equal length is nice too, but I was most interested in the ragged
case, which produces a list.
url: www.econ.uiuc.edu/~roger Roger Koenker
email rkoen...@uiuc.edu Department of Economics
vox: 217-333-4558 University of Illinois
fax: 217-244-6678 Urbana, IL 61801
On Jul 23, 2009, at 2:41 PM, Marc Schwartz wrote:
On Jul 23, 2009, at 9:19 AM, roger koenker wrote:
This is an attempt to rescue an old R-help question that apparently
received
no response from the oblivion of collective silence, and besides
I'm also
curious about the answer
From: Griffith Feeney (gfee...@hawaii.edu)
Date: Fri 28 Jan 2000 - 07:48:45 EST wrote (to R-help)
Constructing lists with
list(name1=name1, name2=name2, ...)
is tedious when there are many objects and names are long. Is
there an R
function that takes a character vector of object names as an
argument and
returns a list with each objected tagged by its name?
The idiom
lapply(ls(pat = "^name"), function(x) eval(as.name(x)))
makes the list, but (ironically) doesn't assign the names to the
components.
Roger,
How about something like this:
name1 <- 1:3
name2 <- 1:5
name3 <- 1:9
name4 <- 1:7
> ls(pat = "^name")
[1] "name1" "name2" "name3" "name4"
> sapply(ls(pat = "^name"), get, simplify = FALSE)
$name1
[1] 1 2 3
$name2
[1] 1 2 3 4 5
$name3
[1] 1 2 3 4 5 6 7 8 9
$name4
[1] 1 2 3 4 5 6 7
Is that what you are after? With sapply(), you can take advantage of
the USE.NAMES argument, which defaults to TRUE and then set simplify
to FALSE to force the result to be a list rather than a matrix. Of
course, in the case I have above, when there are uneven length
elements, the result will be a list anyway.
HTH,
Marc Schwartz
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