Hello, Alexander,
does
utest <- unlist(test)
utest[ names( utest) == "a"]
come close to what you need?
Hth,
Gerrit
On Tue, 7 Dec 2010, Alexander Senger 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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