combinatorics via __future__ generators

[Chris Rebert]

On Wed, Nov 18, 2009 at 4:58 PM, Phlip <phlip2005 at gmail.com> wrote:
> Python:
>
> I have a quaint combinatorics problem. Before I solve it, or find a
> solution among "generators", I thought y'all might like to show off
> any solutions.
>
> Given an array like this...
>
>  [0, 4, 3]
>
> Produce an array like this:
>
>  [
>    [0, 0, 0],
>    [0, 1, 0],
>    [0, 2, 0],
>    [0, 3, 0],
>    [0, 1, 1],
>    [0, 2, 1],
>    [0, 3, 1],
>    [0, 1, 2],
>    [0, 2, 2],
>    [0, 3, 2],
> ]
>
> The first array is the counts of options in 4 slots, and the second is
> all combinations of indexes of each option, such that every option
> associates once with every other option. The leading 0 simply
> represents a slot with no options; the algorithm must preserve those.
>
> This should be child's play for the generator package, right?

from itertools import imap, product

def special_range(n):
    return (xrange(n) if n else [0])

def something(arr):
    return product(*imap(special_range, arr))

print list(something([0,4,3]))

Cheers,
Chris
--
http://blog.rebertia.com