列表列表中的元组列表

Question

Input: List of lists [[1, 2, 3], [5, 6]] something like that.

Required Output : list of combinations in tuples [(1), (1, 2), (1, 2, 3), (5), (5, 6)] .

I can imagine how to solve this, but I guess Python has some handy built-in functions for that

Answer 1

AFAIK没有内置函数，但通过列表理解很容易实现这个结果：

[tuple(seq[:i]) for seq in list_of_lists for i in range(1,len(seq)+1)]

Answer 2

If you actually want all combinations of each sublist, you can use itertools to help:

from itertools import chain, combinations

lst = [[1, 2, 3], [4, 5]]

def powerset(seq, empty=True):
    for i in range(0 if empty else 1, len(seq)+1):
        for comb in combinations(seq, i):
            yield comb

out = list(chain.from_iterable(powerset(l, False) for l in lst))

This gives:

out == [(1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3),   
        (4,), (5,), (4, 5)]

You could modify this to filter just the tuples matching the start of each sublist, but if that's all you want Bakuriu's solution is more efficient.