Comments on: All combinations of N elements

By: Anonymous

Anonymous — Tue, 12 Apr 2005 14:45:00 +0000

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

import probstat # probstat.sf.net
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]

By: Gustavo Niemeyer

Gustavo Niemeyer — Mon, 11 Apr 2005 22:06:00 +0000

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.pop()
        else:
            queue[-1] = i
            yield [chr(97+j) for j in queue]
            if len(queue) < n:
                queue.append(-1)

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.

By: Guyon Moree

Guyon Moree — Mon, 11 Apr 2005 05:59:00 +0000

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!