[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)
Hope this is helpful,
Dan
Dan Nordlund
Bothell, WA USA
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