from itertools import combinations


features = ['A', 'B', 'C']
tmp = []
for i in range(len(features)):
    oc = combinations(features, i + 1)
    for c in oc:
        tmp.append(list(c))

输出

[
 ['A'],
 ['B'],
 ['C'],
 ['A', 'B'],
 ['A', 'C'],
 ['B', 'C'],
 ['A', 'B', 'C']
]

这一行代码给出了所有的组合(如果原始列表/set包含n个不同的元素，则在0到n个元素之间)，并使用本机方法itertools.combination:

Python 2

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([map(list, combinations(input, i)) for i in range(len(input) + 1)], [])

Python 3

from itertools import combinations

input = ['a', 'b', 'c', 'd']

output = sum([list(map(list, combinations(input, i))) for i in range(len(input) + 1)], [])

输出将是:

[[],
 ['a'],
 ['b'],
 ['c'],
 ['d'],
 ['a', 'b'],
 ['a', 'c'],
 ['a', 'd'],
 ['b', 'c'],
 ['b', 'd'],
 ['c', 'd'],
 ['a', 'b', 'c'],
 ['a', 'b', 'd'],
 ['a', 'c', 'd'],
 ['b', 'c', 'd'],
 ['a', 'b', 'c', 'd']]

在网上试试吧:

http://ideone.com/COghfX

这个答案漏掉了一个方面:OP要求所有的组合……不仅仅是长度为r的组合。

所以你要么要遍历所有长度为L的循环:

import itertools

stuff = [1, 2, 3]
for L in range(len(stuff) + 1):
    for subset in itertools.combinations(stuff, L):
        print(subset)

或者——如果你想变得时髦(或者让那些在你之后阅读你的代码的人动脑筋)——你可以生成“组合()”生成器链，并遍历它:

from itertools import chain, combinations
def all_subsets(ss):
    return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))

for subset in all_subsets(stuff):
    print(subset)

In comments under the highly upvoted answer by @Dan H, mention is made of the powerset() recipe in the itertools documentation—including one by Dan himself. However, so far no one has posted it as an answer. Since it's probably one of the better if not the best approach to the problem—and given a little encouragement from another commenter, it's shown below. The function produces all unique combinations of the list elements of every length possible (including those containing zero and all the elements).

注意:如果略有不同，目标是只获得唯一元素的组合，将s = list(iterable)一行更改为s = list(set(iterable))以消除任何重复的元素。无论如何，iterable最终被转换为列表这一事实意味着它将与生成器一起工作(与其他几个答案不同)。

from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)  # allows duplicate elements
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

stuff = [1, 2, 3]
for i, combo in enumerate(powerset(stuff), 1):
    print('combo #{}: {}'.format(i, combo))

输出:

combo #1: ()
combo #2: (1,)
combo #3: (2,)
combo #4: (3,)
combo #5: (1, 2)
combo #6: (1, 3)
combo #7: (2, 3)
combo #8: (1, 2, 3)

下面是itertools.combination的两个实现

返回一个列表的函数

def combinations(lst, depth, start=0, items=[]):
    if depth <= 0:
        return [items]
    out = []
    for i in range(start, len(lst)):
        out += combinations(lst, depth - 1, i + 1, items + [lst[i]])
    return out

一个返回一个生成器

def combinations(lst, depth, start=0, prepend=[]):
    if depth <= 0:
        yield prepend
    else:
        for i in range(start, len(lst)):
            for c in combinations(lst, depth - 1, i + 1, prepend + [lst[i]]):
                yield c

请注意，建议为它们提供一个helper函数，因为prepend参数是静态的，不会随着每次调用而改变

print([c for c in combinations([1, 2, 3, 4], 3)])
# [[1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]

# get a hold of prepend
prepend = [c for c in combinations([], -1)][0]
prepend.append(None)

print([c for c in combinations([1, 2, 3, 4], 3)])
# [[None, 1, 2, 3], [None, 1, 2, 4], [None, 1, 3, 4], [None, 2, 3, 4]]

这是一个很肤浅的例子，但小心为妙

这是我的实现

def get_combinations(list_of_things):
"""gets every combination of things in a list returned as a list of lists

Should be read : add all combinations of a certain size to the end of a list for every possible size in the
the list_of_things.

"""
list_of_combinations = [list(combinations_of_a_certain_size)
                        for possible_size_of_combinations in range(1,  len(list_of_things))
                        for combinations_of_a_certain_size in itertools.combinations(list_of_things,
                                                                                     possible_size_of_combinations)]
return list_of_combinations