Hello, I believe I can help, or at least, my code is simpler. First, look at your first line:
idd <- length(diag(1,tt)) # length of intercept matrix # not needed: diag(tt) would do the job but it's not needed, why call 2 functions, and one of them, 'diag', uses memory(*), if the result is tt squared? It's much simpler! (*)like you say, "larger and larger" amounts of it My solution to your problem is as follows (as a function, and yours). fun2 <- function(n, tt, numco){ M.Unit <- matrix(rep(diag(1,tt),n), ncol=tt, byrow=TRUE) M <- NULL for(i in 1:numco) M <- cbind(M, M.Unit*rep(x[,i], each=tt)) M } fun1 <- function(n, tt, numco){ idd <- length(diag(1,tt)) # length of intercept matrix X <- matrix(numeric(n*numco*idd),ncol=tt*numco) for(i in 1:numco){ X[,((i-1)*tt+1):(i*tt)] <- matrix( c(matrix(rep(diag(1,tt),n),ncol=tt, byrow=TRUE))* rep(rep(x[,i],each=tt),tt) , ncol=tt) } X } I' ve tested the two with larger values of 'n', 'tt' and 'numco' using the following timing instructions n <- 1000 tt <- 50 numco <- 15 set.seed(1) x <- matrix(round(rnorm(n*numco),2), ncol=numco) # the actual covariates Runs <- 10^1 t1 <- system.time(for(i in 1:Runs) a1 <- fun1(n, tt, numco))[c(1,3)] t2 <- system.time(for(i in 1:Runs) a2 <- fun2(n, tt, numco))[c(1,3)] rbind(t1, t2, t1/t2) user.self elapsed t1 23.210000 31.060000 t2 14.970000 22.540000 1.550434 1.377995 As you can see, it's not a great speed improvement. I hope it's at least usefull. Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Data-Manipulation-make-diagonal-matrix-of-each-element-of-a-matrix-tp4200321p4201305.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.