Re: [R] Permutations with replacement

From: <davidr_at_rhotrading.com>
Date: Sat 19 Aug 2006 - 06:33:20 EST


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
Rho Trading Securities, LLC
Chicago IL 60605

-----Original Message-----
From: r-help-bounces@stat.math.ethz.ch
[mailto:r-help-bounces@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.

Thanks for your help in advance,
Jesse Canchola



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