Making Combination from lists using itertools

Question

import itertools
a = [[2, 3], [3, 4]]
b = [[5, 6], [7, 8], [9, 10]]
c = [[11, 12], [13, 14]]
d = [[15, 16], [17, 18]]
e = [[12,16],[13,17],[14,18],[15,19]]

q=[]
q=list(itertools.combinations((a, b, b,c, c, d,e),7)
print q

How would I go about using the combination function from itertools properly to use list a one time, b 2 times without replacement, c 2 times without replacement, and d and e one time each?

[[[2, 3],[5, 6],[7, 8],[11, 12],[13, 14],[15, 16],[12,16]], 
[[2, 3],[5, 6],[7, 8],[11, 12],[13, 14],[15, 16],[13,17]],
[[2, 3],[5, 6],[7, 8],[11, 12],[13, 14],[15, 16],[14,18]],
[[2, 3],[5, 6],[7, 8],[11, 12],[13, 14],[15, 16],[15,19]], 
[[2, 3],[5, 6],[7, 8],[11, 12],[13, 14],[15, 16],[12,16]],...
[[3, 4],[7, 8],[9, 10],[11, 12], [13, 14],[17, 18],[15,19]]]

Answer 1

Updated given clarification of expected output:

You want itertools.product:

itertools.product(a, b, b, c, c, c, c, d, e)

Which will pick one element from each of its arguments on each iteration, cycling the rightmost element the fastest, leftmost slowest.

You can use extended argument unpacking to express the repetition of certain arguments a little more obviously in Python 3:

itertools.product(a, *[b]*2, *[c]*4, d, e)

Or use tobias_k's solution for more general repetition of sequences (that will also work on Py2).

Answer 2

It seems like you are looking for a combination of combinations and product: Use combinations to get the possible combinations without replacement for the repeated lists, then use product to combine all those combinations. You can put the lists and counts in two lists, zip those lists, and use a generator expression to get all the combinations.

from itertools import product, combinations, chain
lists = [a,b,c,d,e]
counts = [1,2,2,1,1]
combs = product(*(combinations(l, c) for l, c in zip(lists, counts)))

For this example, the combs generator has 48 elements, among others:

[(([2, 3],), ([5, 6], [7, 8]), ([11, 12], [13, 14]), ([15, 16],), ([12, 16],)),
 ...
 (([2, 3],), ([5, 6], [7, 8]), ([11, 12], [13, 14]), ([17, 18],), ([15, 19],)),
 (([2, 3],), ([5, 6], [9, 10]),([11, 12], [13, 14]), ([15, 16],), ([12, 16],)),
 ...
 (([3, 4],), ([5, 6], [7, 8]), ([11, 12], [13, 14]), ([15, 16],), ([12, 16],)),
 ...
 (([3, 4],), ([5, 6], [7, 8]), ([11, 12], [13, 14]), ([17, 18],), ([15, 19],)),
 ...
 (([3, 4],), ([7, 8], [9, 10]),([11, 12], [13, 14]), ([17, 18],), ([15, 19],))]

If you want flattened lists, just chain them:

>>> combs = (list(chain(*p)) for p in product(*(combinations(l, c) for l, c in zip(lists, counts))))
>>> list(combs)
[[[2, 3], [5, 6], [7, 8], [11, 12], [13, 14], [15, 16], [12, 16]],
 ...
 [[3, 4], [7, 8], [9, 10], [11, 12], [13, 14], [17, 18], [15, 19]]]

Answer 3

What your are trying to achieve is the Cartesian product of input iterables and not the combinations of the item present in the list. Hence you have to use itertools.product() instead.

In case repetition is allowed among the lists which used more than once, answer is simple:

>>> import itertools
>>> a = [1,2]
>>> b = [3,4]
>>> [i for i in itertools.product(a, b, b)]
[(1, 3, 3), (1, 3, 4), (1, 4, 3), (1, 4, 4), (2, 3, 3), (2, 3, 4), (2, 4, 3), (2, 4, 4)]

But in case repetition is not allowed within the same lists, it will become little nasty and you need to combine the above answer with combinations() and chain() (same as mentioned by tobias_k). This code will give the list of all combinations:

>>> from itertools import chain, product, combinations
>>> lists, counts = [a, b], [1, 2]  # to track that a is to be used once, and b twice
>>> list(list(chain(*p)) for p in product(*(combinations(l, c) for l, c in zip(lists, counts))))
[[1, 3, 4], [2, 3, 4]]

However, in case you need permutations instead of combinations, you have to update the above code with permutations():

>>> from itertools import chain, product, permutations
>>> list(list(chain(*p)) for p in product(*(permutations(l, c) for l, c in zip(lists, counts))))
[[1, 3, 4], [1, 4, 3], [2, 3, 4], [2, 4, 3]]