All combinations of a list of lists

you need itertools.product:

>>> import itertools
>>> a = [[1,2,3],[4,5,6],[7,8,9,10]]
>>> list(itertools.product(*a))
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 4, 10), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 5, 10), (2, 6, 7), (2, 6, 8), (2, 6, 9), (2, 6, 10), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 4, 10), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 5, 10), (3, 6, 7), (3, 6, 8), (3, 6, 9), (3, 6, 10)]

The most elegant solution is to use itertools.product in python 2.6.

If you aren't using Python 2.6, the docs for itertools.product actually show an equivalent function to do the product the "manual" way:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

listOLists = [[1,2,3],[4,5,6],[7,8,9,10]]
for list in itertools.product(*listOLists):
  print list;

I hope you find that as elegant as I did when I first encountered it.

Nothing wrong with straight up recursion for this task, and if you need a version that works with strings, this might fit your needs:

combinations = []

def combine(terms, accum):
    last = (len(terms) == 1)
    n = len(terms[0])
    for i in range(n):
        item = accum + terms[0][i]
        if last:
            combinations.append(item)
        else:
            combine(terms[1:], item)


>>> a = [['ab','cd','ef'],['12','34','56']]
>>> combine(a, '')
>>> print(combinations)
['ab12', 'ab34', 'ab56', 'cd12', 'cd34', 'cd56', 'ef12', 'ef34', 'ef56']