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copied!<p>Here is a brute-force-and-dirty solution, which may work for different number of groups, but you really should test it before use. Moreover, as it uses <code>permn</code>, it will be unusable very fast depending on the size of your vector :</p> <pre><code>library(combinat) split.groups <- function(x, nb.groups) { length.groups <- length(x)/nb.groups perm <- permn(x) perm <- lapply(perm, function(v) { m <- as.data.frame(matrix(v, length.groups, nb.groups)) m <- apply(m,2,sort) m <- t(m) m <- m[order(m[,1]),] rownames(m) <- NULL m}) unique(perm) } </code></pre> <p>Which gives, for example :</p> <pre><code>R> split.groups(1:4, 2) [[1]] [,1] [,2] [1,] 1 2 [2,] 3 4 [[2]] [,1] [,2] [1,] 1 4 [2,] 2 3 [[3]] [,1] [,2] [1,] 1 3 [2,] 2 4 </code></pre> <p>Or :</p> <pre><code>R> split.groups(1:6, 3) [[1]] [,1] [,2] [1,] 1 2 [2,] 3 4 [3,] 5 6 [[2]] [,1] [,2] [1,] 1 2 [2,] 3 6 [3,] 4 5 [[3]] [,1] [,2] [1,] 1 6 [2,] 2 3 [3,] 4 5 [[4]] [,1] [,2] [1,] 1 2 [2,] 3 5 [3,] 4 6 [[5]] [,1] [,2] [1,] 1 6 [2,] 2 5 [3,] 3 4 [[6]] [,1] [,2] [1,] 1 5 [2,] 2 6 [3,] 3 4 [[7]] [,1] [,2] [1,] 1 5 [2,] 2 3 [3,] 4 6 [[8]] [,1] [,2] [1,] 1 5 [2,] 2 4 [3,] 3 6 [[9]] [,1] [,2] [1,] 1 6 [2,] 2 4 [3,] 3 5 [[10]] [,1] [,2] [1,] 1 4 [2,] 2 3 [3,] 5 6 [[11]] [,1] [,2] [1,] 1 4 [2,] 2 6 [3,] 3 5 [[12]] [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 [[13]] [,1] [,2] [1,] 1 3 [2,] 2 5 [3,] 4 6 [[14]] [,1] [,2] [1,] 1 3 [2,] 2 6 [3,] 4 5 [[15]] [,1] [,2] [1,] 1 3 [2,] 2 4 [3,] 5 6 </code></pre>

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