Get all combinations of any length without sub-combinations

Question

I have a list of tuples (item, value) and I want to get all its possible combinations up to a max total value except the “subcombinations” of those already found. Suppose I have list:

items = [("a", 1), ("b", 3), ("b", 3), ("c", 10), ("d", 15)]
combinations = [combo for i in range(len(items), 0, -1) for combo in list(itertools.combinations(items, i)) if sum([item[1] for item in combo]) < 16]

The result is:

[(('a', 1), ('b', 3), ('b', 3)), (('a', 1), ('b', 3), ('c', 10)), (('a', 1), ('b', 3), ('c', 10)), (('a', 1), ('b', 3)), (('a', 1), ('b', 3)), (('a', 1), ('c', 10)), (('a', 1), ('d', 15)), (('b', 3), ('b', 3)), (('b', 3), ('c', 10)), (('b', 3), ('c', 10)), (('a', 1),), (('b', 3),), (('b', 3),), (('c', 10),), (('d', 15),)]

but I want to get:

[(('a', 1), ('b', 3), ('b', 3)), (('a', 1), ('b', 3), ('c', 10)), (('d', 15))]

Is it possible to eliminate the “subcombinations” except doing it by many long ifs (the actual list is much larger)?

Asked By: Kazek Ciaś

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Source

Answer 1

Ok, this should work.

from itertools import combinations

items = [("a", 1), ("b", 3), ("b", 3), ("c", 10), ("d", 15)]

res = []
for l in range(len(items), 0, -1):
    cb = combinations(items, l)
    cbl = [i for i in cb if sum([j[1] for j in i]) < 16]
    for cc in cbl:
        if not any([set(cc).issubset(el) for el in res]):
            res.append(cc)

print(res)

This prints:

[(('a', 1), ('b', 3), ('b', 3)), (('a', 1), ('b', 3), ('c', 10)), (('d', 15),)]

Please note that this code may be slow if you have a lot of items, because it iterates many times to check if a given combination is not a subset of any already accepted combination. Maybe there are more efficient solutions.

Answered By: Valentino

Get all combinations of any length without sub-combinations

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