Cartesian Product of two lists (itertools)

[Mensanator]

On Jan 25, 3:12�pm, Thorsten Kampe <thors... at thorstenkampe.de> wrote:
> Hi,
>
> is there a way to make itertools.product generate triples instead of
> pairs from two lists?
>
> For example:>>> list1 = [1, 2]; list2 = [4, 5]; list3 = [7, 8]
> >>> from itertools import product
> >>> list(product(list1, list2, list3))
>
> [(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
> 5, 7), (2, 5, 8)]
>
> so far so good... Now...>>> list(product(product(list1, list2), list3))
>
> [((1, 4), 7), ((1, 4), 8), ((1, 5), 7), ((1, 5), 8), ((2, 4), 7), ((2,
> 4), 8), ((2, 5), 7), ((2, 5), 8)]
>
> Oops, pairs of pairs instead triples. Not what I wanted.
>
> What's the best way to pre-process the arguments to "itertools.product"
> or to post-process the result of "itertools.product" to get what I
> want?!
>
> I have an older utility which I would like to replace with
> itertools.product. The old one uses a rather clumsy way to indicate that
> a triple was wanted:
>
> def cartes(seq0, seq1, modus = 'pair'):
> � � """ return the Cartesian Product of two sequences """
> � � if � modus == 'pair':
> � � � � return [[item0, item1] for item0 in seq0 for item1 in seq1]
> � � elif modus == 'triple':
> � � � � return [item0 + [item1] for item0 in seq0 for item1 in seq1]
>
> Thorsten

Will this work for you?

>>> list4 = [(i,) for i in list3]

>>> list4
[(7,), (8,)]

>>> a = list(itertools.product(itertools.product(list1, list2), list4))
>>> a
[((1, 4), (7,)), ((1, 4), (8,)), ((1, 5), (7,)), ((1, 5), (8,)), ((2,
4), (7,)), ((2, 4), (8,)), ((2, 5), (7,)), ((2, 5), (8,))]

>>> def flatten(listOfLists):
    return tuple(itertools.chain.from_iterable(listOfLists))

>>> list5 = [flatten(i) for i in a]
>>> list5
[(1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 4, 7), (2, 4, 8), (2,
5, 7), (2, 5, 8)]
>>>