两个列表的组合（不是元素）

Question

I have two lists:

a = ['a', 'b']
b = [1, 2, 3]

I want to get the combinations produced between the elements of list b and the elements of list a but treating elements of a as pairs (or triples etc. etc.) as the example below which gives len(b) ** len(a) number of combinations.

c = ["a_1 b_1", "a_1 b_2", "a_1 b_3", "a_2 b_1", "a_2 b_2", "a_2 b_3", "a_3 b_1", "a_3 b_2" "a_3 b_3"]

I have tried to use itertools.product (as described here ) but this will give only the 6 possible combinations.

Answer 1

You can use itertools.product(..) but specify repeat to be repeat=len(a) . So you can use:

from itertools import product

def mul_product(a,b):
    for tup in product(b,repeat=len(a)):
        yield ' '.join('%s_%s'%t for t in zip(a,tup))

The product(..) will generate tuples like:

>>> list(product(b,repeat=len(a)))
[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]

So here the first element of the tuple is the one that is attached to a_ , the second one to b_ . Now we zip(..) them together with the a list, producing:

>>> list(map(lambda bi:list(zip(a,bi)),product(b,repeat=len(a))))
[[('a', 1), ('b', 1)], [('a', 1), ('b', 2)], [('a', 1), ('b', 3)], [('a', 2), ('b', 1)], [('a', 2), ('b', 2)], [('a', 2), ('b', 3)], [('a', 3), ('b', 1)], [('a', 3), ('b', 2)], [('a', 3), ('b', 3)]]

Now it is only a matter of formatting ( '%s_%s'%t ), and ' '.join(..) ining them together and yield them (or you can use list comprehension to produce a list).

The result for your sample input is:

>>> list(mul_product(a,b))
['a_1 b_1', 'a_1 b_2', 'a_1 b_3', 'a_2 b_1', 'a_2 b_2', 'a_2 b_3', 'a_3 b_1', 'a_3 b_2', 'a_3 b_3']

Note that the elements here are generated lazily. This can be useful if you are for instance only interested in the first k ones, or when you do not want to generate all of them at once.

Answer 2

You could explicitly create your pairwise items using itertools.product , then operate on those pairs again with itertools.product

import itertools
a = ['a', 'b']
b = [1, 2, 3]
pairs = [list(itertools.product([ai], b)) for ai in a]

pairs will contain the two lists that can fed into itertools.product again.

list(itertools.product(*pairs))

The result is:

[(('a', 1), ('b', 1)),
 (('a', 1), ('b', 2)),
 (('a', 1), ('b', 3)),
 (('a', 2), ('b', 1)),
 (('a', 2), ('b', 2)),
 (('a', 2), ('b', 3)),
 (('a', 3), ('b', 1)),
 (('a', 3), ('b', 2)),
 (('a', 3), ('b', 3))]