# [R] Permutations with replacement (final final final)

Jesse Albert Canchola jesse.canchola.b at bayer.com
Mon Aug 21 19:48:54 CEST 2006

```Hi Daniel,

Turns out, your code, however simple, is quite elegant for my needs
(sometimes I overanalyze  :O).  Here is my last code to do what I need to
do:

####### begin R code ########
# generate a matrix of ten thousand rows of 1-8
z <- t(matrix(rep(1:8,10000),8,10000))
library(Matrix)
# use the R sample function in a loop to sample each line with replacement
zcomb=Matrix()
for (i in 1:dim(z)[1]) {
z1 <- t(matrix(sample(z,8,replace=TRUE)))
zcomb = rbind(zcomb,z1)
}
zcomb
####### end R code #########

Regards,
Jesse A. Canchola

"Daniel Nordlund" <res90sx5 at verizon.net>
Sent by: r-help-bounces at stat.math.ethz.ch
08/18/2006 05:16 PM

To
"'Jesse Albert Canchola'" <jesse.canchola.b at bayer.com>, "'r-help'"
<r-help at stat.math.ethz.ch>
cc

Subject
Re: [R] Permutations with replacement

> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch]
> On Behalf Of Jesse Albert Canchola
> Sent: Friday, August 18, 2006 1:02 PM
> To: r-help
> Subject: [R] Permutations with replacement
>
> Is there a simple function or process that will create permutations with
> replacement?
>
> I know that using the combinat package
>
> ###### begin R code ######
> > library(combinat)
> > m <- t(array(unlist(permn(3)), dim = c(3, 6)))
>
> # we can get the permutations, for example 3!=6
> # gives us
>
> > m
>      [,1] [,2] [,3]
> [1,]    1    2    3
> [2,]    1    3    2
> [3,]    3    1    2
> [4,]    3    2    1
> [5,]    2    3    1
> [6,]    2    1    3
> ###### end R code ##########
>
> I'd like to include the "with replacement possibilities" such as
>
> 1,1,3
> 1,1,2
> 2,3,3
>
Isn't what you want just sampling with replacement?

x <- c(1,2,3)
sample(x,3,replace=TRUE)

Dan

Dan Nordlund
Bothell, WA  USA

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and provide commented, minimal, self-contained, reproducible code.

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