Data Manipulation - make diagonal matrix of each element of a matrix

Dear R list,
I have the following data:

set.seed(1)
n ?<- 5 ? ? # number of subjects
tt <- 3 ? ? # number of repeated observation per subject
numco <- 2 ?# number of covariates
x <- matrix(round(rnorm(n*numco),2), ncol=numco) ? # the actual covariates
x

x

? ? ?[,1] ?[,2]
[1,] -0.63 -0.82
[2,] ?0.18 ?0.49
[3,] -0.84 ?0.74
[4,] ?1.60 ?0.58
[5,] ?0.33 -0.31

I need to form a matrix X such that
X = ? ? ?x11 ? ? ?0 ? ? ?0 ? ? x21 ? ? ?0 ? ? ?0
? ? ? ? ? ? ?0 ? x11 ? ? ?0 ? ? ? ?0 ? x21 ? ? ?0
? ? ? ? ? ? ?0 ? ? ?0 ? x11 ? ? ? ?0 ? ? ?0 ? x21
? ? ? ? ? x12 ? ? ?0 ? ? ?0 ? ? x22 ? ? ?0 ? ? ?0
? ? ? ? ? ? ?0 ? x12 ? ? ?0 ? ? ? ?0 ? x22 ? ? ?0
? ? ? ? ? ? ?0 ? ? ?0 ? x12 ? ? ? ?0 ? ? ?0 ? x22
? ? ? ? ? ? ? ? ? ? ? ...
? ? ? ? ? x15 ? ? ?0 ? ? ?0 ? ? x25 ? ? ?0 ? ? ?0
? ? ? ? ? ? ?0 ? x15 ? ? ?0 ? ? ? ?0 ? x25 ? ? ?0
? ? ? ? ? ? ?0 ? ? ?0 ? x15 ? ? ? ?0 ? ? ?0 ? x25
where both tt and numco can change. (So if tt=5 and numco=4, then X
needs to have 20 columns and n*tt rows. Each diagonal matrix should be
5x5 and there will be 4 of them for the 4 covariates.) I wrote this
funky for loop:

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

It works fine, but is there an easier way when n, tt, and numco get
larger and larger?
Thanks,
Tina


--
Clemontina Alexander
Ph.D Student
Department of Statistics
NC State University
Email: ckalexa2 at ncsu.com

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


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