# [R] A combinatorial task. How to get rid of loops

Uwe Ligges ligges at statistik.uni-dortmund.de
Thu Aug 2 11:31:03 CEST 2007

Serguei Kaniovski wrote:
> Dear List,
>
> I am looking for a faster way of accomplishing this:
>
> # n is an integer larger than two
> n <- 3
>
> # matrix of 2^n binary outcomes
> bmat <- as.matrix(expand.grid( rep( list(1:0), n))[, n:1])
>
> # I would like to know which rows of "bmat" have "1" in the "i" and "j"
> coordinates, so for n=3 in coordinates 1-2, 1- 3, and 2-3
> # I would like then to construct "choose( n, 2)" binary vectors of lengths
> 2^n to indicate those rows with a "1"
> # The loop below accomplishes this task. Is there a faster way of doing
> the same?
>
>         library(combinat)
>
>         # matrix of all pairwise combinations
>         cmat<-combn( n, 2)

One example is:

n <- 3
bmat <- as.matrix(expand.grid( rep( list(1:0), n))[, n:1])
library(combinat)
cmat<-combn( n, 2)
apply(cmat, 2, function(x) as.numeric(bmat[,x[1]] & bmat[,x[2]]))

Uwe Ligges

>         temp<-matrix(NA, nrow(bmat), ncol(cmat))
>
>         for (i in 1:nrow(bmat))
>         {
>                 for (j in 1:ncol(cmat))
>                 {
>                         temp[ i, j]<-as.numeric( bmat[i,][cmat[1,j]] &
> bmat[i,][cmat[2,j]] == 1)
>                 }
>         }
>
>
> Thank you very much for your help!
> Serguei Kaniovski
>
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