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 > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
