I believe that Thomas got "mu" wrong. If I understand correctly, the line:
x3 <- rpois(50 * 100, rep(mu, each=100))
should read:
x3 <- rpois(50 * 100, rep(mu, 50))
or just
x3 <- rpois(50 * 100, mu)
Patrick Burns
Burns Statistics [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and "A Guide for the Unwilling S User")
Thomas Lumley wrote:
On Wed, 18 Feb 2004, Haiyan Chen wrote:
Hello,
1. After I generate a 100x50 matrix by x3<-matrix(0,100,50);for (i in
1:100) {x1<-rpois(50, mu[i]);x2<-x1; x2[runif(50)<.01]<-0; x3[i,]<-x2},
YOu can do this without the loop, eg
x3<-rpois(50*100, rep(mu,each=100)) x3<-ifelse(runif(50*100)<0.01, 0, x3) x3<-matrix(x3, ncol=50)
2. I want to calculate means and sample variances of each row and create a
new matrix 100x2;
means<-apply(x3,2,mean) vars<-apply(x3,2,var) cbind(means,vars)
3. I then want to repeat above procedure 500 times so that eventually I
will have 500 100x2 matrices.
make.a.matrix<-function(...){ x3<-rpois(50*100, rep(mu,each=100)) x3<-ifelse(runif(50*100)<0.01, 0, x3) x3<-matrix(x3, ncol=50) cbind(apply(x3,2,mean), apply(x3,2, var)) }
many.matrices<-lapply(1:500, make.a.matrix)
gives a list of 500 matrices. This isn't quite the most efficient solution, but it's not bad.
-thomas
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