Question

解决组合问题，并尝试输出列表列表作为输入的列表。我找到的最接近的解决方案是在这里：All combinations of a list of lists

但是，我不需要列表之间的所有组合，而希望在每个列表中。例如

[[1],[2,3],[4,5,6]] -> [[1],[2],[3],[2,3],[4],[5],[6],[4,5],[4,6],            
[5,6],[4,5,6]]

Answer 1

贷记How can I find all the subsets of a set, with exactly n elements?

首先找到一种方法来查找一个列表的所有子集（也称为幂集）：

from itertools import chain, combinations

def powerset(iterable):
    """
    powerset([1,2,3]) --> [[1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]
    """
    xs = list(iterable)
    return [list(x) for x in chain.from_iterable(combinations(xs,n) for n in range(len(xs)+1)) if x]

然后迭代每个列表：

list_of_list = [[1],[2,3],[4,5,6]]
result = []
for x in list_of_list:
    result += powerset(x)
print(result)

输出：

[[1], [2], [3], [2, 3], [4], [5], [6], [4, 5], [4, 6], [5, 6], [4, 5, 6]]

Answer 2

下面，我定义了一个辅助函数，以从序列中获取所有组合，然后将其应用于输入列表的每个子列表，并将结果chain一起使用。

from itertools import chain, combinations

l=[[1],[2,3],[4,5,6]]

def all_comb(seq):
    return chain.from_iterable(combinations(seq, i) for i in range(1, len(seq)+1))

print(list(chain.from_iterable(map(all_comb, l))))
# [(1,), (2,), (3,), (2, 3), (4,), (5,), (6,), (4, 5), (4, 6), (5, 6), (4, 5, 6)]

Answer 3

与itertools.combinations：

from itertools import combinations

l = [[1],[2,3],[4,5,6]]

combos = sum([[list(c) for c in combinations(x, i)] for x in l for i in range(1, len(x)+1)], [])

combos
>>> [[1], [2], [3], [2, 3], [4], [5], [6], [4, 5], [4, 6], [5, 6], [4, 5, 6]]

列表内列表的组合

3 个答案: