[EMAIL PROTECTED] 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')]
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