Create x lists in python dynamically

Question

(First, I chose to do this in Python because I never programmed in it and it would be good practice.)

Someone asked me to implement a little "combination" program that basically outputs all possible combinations of a set of group of numbers. Example, if you have:
(1,2,3) as the first set,
(4,5,6) as the second, and
(7,8,9) as the third, then one combination would be (1,4,7) and so on, with a total of 27 possible combinations. This person just wants to do a 6rows x 6cols matrix or a 5rows x 6cols matrix. However, I want to make my little program as flexible as possible.
The next requirement is to only output combinations with X even numbers. If he wants 0 even numbers, then a possible combination would be (1,5,7). You get the idea. For the permutation part, I used itertools.product(), which works perfectly. It would be easy if I just assume that the number of numbers in each set (cols) is fixed as 6. In that case, I could manually create 6 lists and append each combination to the right list. However and again, I want this to work with N number of cols.

I'm thinking of 2 ways I might be able to do this, but tried with no luck. So my question is: How can I create?

li_1 = [] 
li_2 = [] 
...
li_x = [] 

The one way I tried using "lists of lists":

for combination in itertools.product(*li):
    total_combinations = total_combinations + 1
    #Counts number of even numbers in a single combination
    for x in range(numberInRows):
        if combination[x] % 2 == 0:
            even_counter = even_counter + 1
    print "Even counter:",even_counter
    num_evens[even_counter].append(combination)
    print "Single set:",num_evens
    even_counter = 0

    print combination
print "Num_evens:",num_evens

print '\nTotal combinations:', total_combinations

Answer 1

Ls = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
import collections
import itertools

def products_by_even_count(seq):
    ret = collections.defaultdict(set)
    for p in itertools.product(*seq):
        n_even = sum(1 for n in p if n % 2 == 0)
        ret[n_even].add(p)
    return ret

import pprint
# Calling dict() is only necessary for pretty pprint output.
pprint.pprint(dict(products_by_even_count(Ls)))

Output:

{0: set([(1, 5, 7), (1, 5, 9), (3, 5, 7), (3, 5, 9)]),
 1: set([(1, 4, 7),
         (1, 4, 9),
         (1, 5, 8),
         (1, 6, 7),
         (1, 6, 9),
         (2, 5, 7),
         (2, 5, 9),
         (3, 4, 7),
         (3, 4, 9),
         (3, 5, 8),
         (3, 6, 7),
         (3, 6, 9)]),
 2: set([(1, 4, 8),
         (1, 6, 8),
         (2, 4, 7),
         (2, 4, 9),
         (2, 5, 8),
         (2, 6, 7),
         (2, 6, 9),
         (3, 4, 8),
         (3, 6, 8)]),
 3: set([(2, 4, 8), (2, 6, 8)])}

Answer 2

num_evens = {} 
for combination in itertools.product(*li):
    even_counter = len([ y for y in combination if y & 1 == 0 ])
    num_evens.setdefault(even_counter,[]).append(combination)

import pprint
pprint.pprint(num_evens)

Answer 3

from itertools import product
from collections import defaultdict
num_evens = defaultdict(list)
for comb in product(*li):
    num_evens[sum(y%2==0 for y in comb)].append(comb)

import pprint
pprint.pprint(num_evens)