Hello,
I don't think he advantage here is speed but simplicity, lapply does it
in one line of code.
Rui Barradas
Em 11-11-2012 18:02, Bert Gunter escreveu:
Clemontina:
As you have seen, the answer is yes, but my question is why bother?
What's wrong with a for() loop?
lapply() should offer no advantage in speed over a loop. Vectorization
could, but lapply() is not vectorization.
-- Bert
On Sun, Nov 11, 2012 at 8:33 AM, Clemontina Davenport <ckale...@ncsu.edu> wrote:
Hi all,
I have the following code:
set.seed(1)
x1 <- matrix(sample(1:12), ncol=3)
x2 <- matrix(sample(1:12), ncol=3)
x3 <- matrix(sample(1:12), ncol=3)
X <- list(x1,x2,x3)
tt <- matrix(round(runif(5*4),2), ncol=4)
Is there a way I can construct a new list where
newlist[[i]] = tt[i,] %*% X[[i]]
without using a for loop? Each element of newlist will be 3 x 1 vector.
Thanks
--
Tina Alexander
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