Yes, but note that such comparisons don't necessarily tell you much.
Using your l1 on my computer:
> system.time(r1 <- sapply(l1, `[`, 1))
user system elapsed
1.2 0.0 1.2
> system.time(r2<- rapply(l1,function(x)x[1]))
user system elapsed
0.810.000.81
## But
> syste
Or
rapply(l,function(x) x[1])
#[1] 1 3 7
set.seed(42)
l1 <- replicate(1e6, list(sample(1:5,sample(8),replace=T)))
system.time(r1 <- sapply(l1, `[`, 1))
# user system elapsed
# 1.324 0.000 1.326
system.time(r2 <- rapply(l1, function(x) x[1]))
# user system elapsed
# 0.736 0.004
Hi Carol,
On 07/21/2014 09:10 PM, Richard M. Heiberger wrote:
l = list(c(1,2), c(3,5,6), c(7))
sapply(l, `[`, 1)
Using sapply() works but won't be very efficient if you have a very long
list. If you worry about efficiency, you can do the following (using the
IRanges package from Bioconductor)
l = list(c(1,2), c(3,5,6), c(7))
sapply(l, `[`, 1)
On Mon, Jul 21, 2014 at 3:55 PM, carol white wrote:
> Hi,
> If we have a list of vectors of different lengths, how is it possible to
> retrieve the first element of the vectors of the list?
>
>
> l = list(c(1,2), c(3,5,6), c(7))
>
> 1,3,7 shoul
Hi,
If we have a list of vectors of different lengths, how is it possible to
retrieve the first element of the vectors of the list?
l = list(c(1,2), c(3,5,6), c(7))
1,3,7 should be retrieved
Thanks
Carol
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