William Bernard - 1 year ago 126
Python Question

# Making Combination from lists using itertools

``````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]]]
``````

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 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]]]
``````
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