# [R] Permutations with replacement

Jesse Albert Canchola jesse.canchola.b at bayer.com
Mon Aug 21 01:32:12 CEST 2006

```Thanks, David.  That worked fabulously!

Here is the R code for the hypercube test example:

########## begin R code ############
library(combinat)
x <- rep(3,3)   # for partitions of 3 units into the three classes {1,2,3}

hcube(x, scale=1, transl=0)
########### end R code ############

For the larger one I want (i.e., 8^8), I will take a random sample of
10,000 from the 16,777,216 possibilities.

Regards,
Jesse Canchola

Sent by: r-help-bounces at stat.math.ethz.ch
08/18/2006 01:33 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

If you also want 1,1,1 and so on, the number of these is n^n,
(n choices for each of n slots.)
In that case, you could use hcube from combinat.

David L. Reiner
Chicago  IL  60605

-----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 3:26 PM
To: r-help
Subject: [R] Permutations with replacement

Is there a simple function or process that will create a matrix of
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

and so on.  This will eventually be done on 8!=40,320 rather than the
development version using 3! as above.

If no function exists (I've Googled on CRAN with no palpable luck), then

perhaps this is more of a bootstrap type problem.

Jesse Canchola

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