(newbie) N-uples from list of lists

[Dan Bishop]

vd12005 at yahoo.fr wrote:
> Hello,
>
> i think it could be done by using itertools functions even if i can not
> see the trick. i would like to have all available "n-uples" from each
> list of lists.
> example for a list of 3 lists, but i should also be able to handle any
> numbers of items (any len(lol))
>
> lol = (['a0', 'a1', 'a2'], ['b0', 'b1'], ['c0', 'c1', 'c2', 'c3'])
>
> =>
>
> [('a0', 'b0', 'c0'), ('a0', 'b0', 'c1'), ('a0', 'b0', 'c2'), ('a0',
> 'b0', 'c3'), ('a0', 'b1', 'c0'), ('a0', 'b1', 'c1'), ('a0', 'b1',
> 'c2'), ('a0', 'b1', 'c3'), ('a1', 'b0', 'c0'), ('a1', 'b0', 'c1'),
> ('a1', 'b0', 'c2'), ('a1', 'b0', 'c3'), ('a1', 'b1', 'c0'), ('a1',
> 'b1', 'c1'), ('a1', 'b1', 'c2'), ('a1', 'b1', 'c3'), ('a2', 'b0',
> 'c0'), ('a2', 'b0', 'c1'), ('a2', 'b0', 'c2'), ('a2', 'b0', 'c3'),
> ('a2', 'b1', 'c0'), ('a2', 'b1', 'c1'), ('a2', 'b1', 'c2'), ('a2',
> 'b1', 'c3')]
>
> maybe tee(lol, len(lol)) can help ?
>
> it could be done by a recursive call, but i am interested in using and
> understanding generators.

def cross(*args):
   """Iterates over the set cross product of args."""
   if not args:
      return
   elif len(args) == 1:
      for x in args[0]:
         yield (x,)
   else:
      for x in args[0]:
         for y in cross(*args[1:]):
            yield (x,) + y

>>> cross(['a0', 'a1', 'a2'], ['b0', 'b1'], ['c0', 'c1', 'c2', 'c3'])
<generator object at 0xb7df072c>
>>> list(_)
[('a0', 'b0', 'c0'), ('a0', 'b0', 'c1'), ('a0', 'b0', 'c2'), ('a0',
'b0', 'c3'), ('a0', 'b1', 'c0'), ('a0', 'b1', 'c1'), ('a0', 'b1',
'c2'), ('a0', 'b1', 'c3'), ('a1', 'b0', 'c0'), ('a1', 'b0', 'c1'),
('a1', 'b0', 'c2'), ('a1', 'b0', 'c3'), ('a1', 'b1', 'c0'), ('a1',
'b1', 'c1'), ('a1', 'b1', 'c2'), ('a1', 'b1', 'c3'), ('a2', 'b0',
'c0'), ('a2', 'b0', 'c1'), ('a2', 'b0', 'c2'), ('a2', 'b0', 'c3'),
('a2', 'b1', 'c0'), ('a2', 'b1', 'c1'), ('a2', 'b1', 'c2'), ('a2',
'b1', 'c3')]