# [R] How to repeat a procedure

Haiyan Chen hxc05 at health.state.ny.us
Wed Feb 18 20:14:08 CET 2004

```Thanks, Patrick. I was aware of that and was able to fix it.

Thanks Tom, Wayne and Peter for all your help.

Heyen

Patrick Burns
<pburns at pburns.se        To:       Thomas Lumley <tlumley at u.washington.edu>
anet.com>                cc:       Haiyan Chen <hxc05 at health.state.ny.us>, R-help at stat.math.ethz.ch
Subject:  Re: [R] How to repeat a procedure
02/18/2004 01:09
PM

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
patrick at burns-stat.com
+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<-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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>
>
>

```

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