fast subsetting of lists in lists
First, subset 'test' once, e.g.
testT <- test[1:3];
and then use sapply() on that, e.g.
val <- sapply(testT, FUN=function (x) { x$a })
Then you can avoid one level of function calls, by
val <- sapply(testT, FUN="[[", "a")
Second, there is some overhead in "[[", "$" etc. You can use
.subset2() to avoid this, e.g.
val <- sapply(testT, FUN=.subset2, "a")
Third, it may be that using sapply() to structure you results is a bit
overkill. If you know that the 'a' element is always of the same
dimension, you can do it yourself, e.g.
val <- lapply(testT, FUN=.subset2, "a")
val <- unlist(val, use.names=FALSE) # use.names=FALSE is much faster than TRUE
See what that does
/Henrik
On Tue, Dec 7, 2010 at 6:47 AM, Alexander Senger
<senger at physik.hu-berlin.de> wrote:
Hello,
my data is contained in nested lists (which seems not necessarily to be
the best approach). What I need is a fast way to get subsets from the data.
An example:
test <- list(list(a = 1, b = 2, c = 3), list(a = 4, b = 5, c = 6),
list(a = 7, b = 8, c = 9))
Now I would like to have all values in the named variables "a", that is
the vector c(1, 4, 7). The best I could come up with is:
val <- sapply(1:3, function (i) {test[[i]]$a})
which is unfortunately not very fast. According to R-inferno this is due
to the fact that apply and its derivates do looping in R rather than
rely on C-subroutines as the common [-operator.
Does someone now a trick to do the same as above with the faster
built-in subsetting? Something like:
test[<somesubsettingmagic>]
Thank you for your advice
Alex
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