All combinations of N elements

3 Responses to All combinations of N elements

1. Guyon Moree says:

   2005/04/11 at 05:59

   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. Gustavo Niemeyer says:

   2005/04/11 at 22:06

   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. :-)
3. Anonymous says:

   2005/04/12 at 14:45

   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]