[R] Data Manipulation - make diagonal matrix of each element of a matrix

Rui Barradas ruipbarradas at sapo.pt
Thu Dec 15 20:04:47 CET 2011


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))

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))*
           , ncol=tt)

I' ve tested the two with larger values of 'n', 'tt' and 'numco'
using the following timing instructions

n  <- 1000
tt <- 50
numco <- 15
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

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