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    <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 &lt;- function(x, nb.groups) { length.groups &lt;- length(x)/nb.groups perm &lt;- permn(x) perm &lt;- lapply(perm, function(v) { m &lt;- as.data.frame(matrix(v, length.groups, nb.groups)) m &lt;- apply(m,2,sort) m &lt;- t(m) m &lt;- m[order(m[,1]),] rownames(m) &lt;- NULL m}) unique(perm) } </code></pre> <p>Which gives, for example :</p> <pre><code>R&gt; 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&gt; 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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    1. POFind all possible ways to split a list of elements into a a given number of group of the same size
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    1. USjuba
    1. USjuba
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    1. COWhy did you put `as.data.frame` in there?
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    2. CO@Roland In order to be able to sort each column independently. I told you it was dirty :)
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