All combinations of N elements

Here’s a snippet to compute all combinations of N elements over K positions.

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3 Responses to All combinations of N elements

  1. Guyon Moree says:

    hey, that someone was me! thanks for the code.

    I don’t see a ‘border’ though ;)

    What I mean is that you take the border from the length of your list. At least that is what I think I see happening here.

    What I meant was having a list of possible characters like [‘a’,’b’,(..) ‘z’]. And then also give a ‘border’. Let’s say you give a border of 3. This will effectively give you all possible 3-character combinations of *all* given characters.

    I hope I explained this good enough, thanks again!

  2. Hello Guyon! I think I understand what you meant now. The algorithm is pretty much the same:

    def allcombinations(n):
        queue = [-1]
        while queue:
            i = queue[-1]+1
            if i == 26:
                queue[-1] = i
                yield [chr(97+j) for j in queue]
                if len(queue) < n:

    The total combinations is computed by:

    def total(n):
        return sum([26**i for i in range(1,n+1)])

    Which means 12356630 entries for n=5!!! Be careful. :-)

  3. Anonymous says:

    This list is the sum of three cartesian products,
    here it is cheating a bit and not in exactly the same order.

    import probstat #
    def total(mylist):
      for (n) in range(len(mylist)):
        for (item) in probstat.Cartesian([mylist]*(n+1)):
          yield item

    for the list [1,2,3] it does the cartesian product of [1,2,3]
    then [1,2,3] x [1,2,3] then [1,2,3] x [1,2,3] x [1,2,3]

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