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)

使用列表推导式:

def selfCombine( list2Combine, length ):
    listCombined = str( ['list2Combine[i' + str( i ) + ']' for i in range( length )] ).replace( "'", '' ) \
                     + 'for i0 in range(len( list2Combine ) )'
    if length > 1:
        listCombined += str( [' for i' + str( i ) + ' in range( i' + str( i - 1 ) + ', len( list2Combine ) )' for i in range( 1, length )] )\
            .replace( "', '", ' ' )\
            .replace( "['", '' )\
            .replace( "']", '' )

    listCombined = '[' + listCombined + ']'
    listCombined = eval( listCombined )

    return listCombined

list2Combine = ['A', 'B', 'C']
listCombined = selfCombine( list2Combine, 2 )

输出将是:

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

如文档中所述

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)


x = [2, 3, 4, 5, 1, 6, 4, 7, 8, 3, 9]
for i in combinations(x, 2):
    print i

下面是一个惰性一行代码，同样使用itertools:

from itertools import compress, product

def combinations(items):
    return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) )
    # alternative:                      ...in product([0,1], repeat=len(items)) )

这个答案背后的主要思想是:有2^N种组合——与长度为N的二进制字符串的数量相同。对于每个二进制字符串，您选择与“1”对应的所有元素。

items=abc * mask=###
 |
 V
000 -> 
001 ->   c
010 ->  b
011 ->  bc
100 -> a
101 -> a c
110 -> ab
111 -> abc

需要考虑的事情:

This requires that you can call len(...) on items (workaround: if items is something like an iterable like a generator, turn it into a list first with items=list(_itemsArg)) This requires that the order of iteration on items is not random (workaround: don't be insane) This requires that the items are unique, or else {2,2,1} and {2,1,1} will both collapse to {2,1} (workaround: use collections.Counter as a drop-in replacement for set; it's basically a multiset... though you may need to later use tuple(sorted(Counter(...).elements())) if you need it to be hashable)

Demo

>>> list(combinations(range(4)))
[set(), {3}, {2}, {2, 3}, {1}, {1, 3}, {1, 2}, {1, 2, 3}, {0}, {0, 3}, {0, 2}, {0, 2, 3}, {0, 1}, {0, 1, 3}, {0, 1, 2}, {0, 1, 2, 3}]

>>> list(combinations('abcd'))
[set(), {'d'}, {'c'}, {'c', 'd'}, {'b'}, {'b', 'd'}, {'c', 'b'}, {'c', 'b', 'd'}, {'a'}, {'a', 'd'}, {'a', 'c'}, {'a', 'c', 'd'}, {'a', 'b'}, {'a', 'b', 'd'}, {'a', 'c', 'b'}, {'a', 'c', 'b', 'd'}]

我喜欢这个问题，因为有很多方法来实现它。我决定为未来创造一个参考答案。

在生产中使用什么?

intertools的文档有一个独立的例子，为什么不在你的代码中使用它呢?一些人建议使用more_itertools。Powerset，但它具有完全相同的实现!如果我是你，我不会为一个小东西安装整个软件包。也许这是最好的方法:

import itertools

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

其他可能的方法

方法0:使用组合

import itertools

def subsets(nums):
    result = []
    for i in range(len(nums) + 1):
        result += itertools.combinations(nums, i)
    return result

方法1:简单的递归

def subsets(nums):
    result = []

    def powerset(alist, index, curr):
        if index == len(alist):
            result.append(curr)
            return

        powerset(alist, index + 1, curr + [alist[index]])
        powerset(alist, index + 1, curr)

    powerset(nums, 0, [])
    return result

方法2:回溯

def subsets(nums):
    result = []

    def backtrack(index, curr, k):
        if len(curr) == k:
            result.append(list(curr))
            return
        for i in range(index, len(nums)):
            curr.append(nums[i])
            backtrack(i + 1, curr, k)
            curr.pop()

    for k in range(len(nums) + 1):
        backtrack(0, [], k)
    return result

or

def subsets(nums):
    result = []

    def dfs(nums, index, path, result):
        result.append(path)
        for i in range(index, len(nums)):
            dfs(nums, i + 1, path + [nums[i]], result)

    dfs(nums, 0, [], result)
    return result

方法3:位掩码

def subsets(nums):
    res = []
    n = len(nums)
    for i in range(1 << n):
        aset = []
        for j in range(n):
            value = (1 << j) & i  # value = (i >> j) & 1
            if value:
                aset.append(nums[j])
        res.append(aset)
    return res

或者(不是位掩码，直觉上是2^n个子集)

def subsets(nums):
    subsets = []
    expected_subsets = 2 ** len(nums)

    def generate_subset(subset, nums):
        if len(subsets) >= expected_subsets:
            return
        if len(subsets) < expected_subsets:
            subsets.append(subset)
        for i in range(len(nums)):
            generate_subset(subset + [nums[i]], nums[i + 1:])

    generate_subset([], nums)
    return subsets

方法4:级联

def subsets(nums):
    result = [[]]
    for i in range(len(nums)):
        for j in range(len(result)):
            subset = list(result[j])
            subset.append(nums[i])
            result.append(subset)
    return result

下面是一个使用递归的例子:

>>> import copy
>>> def combinations(target,data):
...     for i in range(len(data)):
...         new_target = copy.copy(target)
...         new_data = copy.copy(data)
...         new_target.append(data[i])
...         new_data = data[i+1:]
...         print new_target
...         combinations(new_target,
...                      new_data)
...                      
... 
>>> target = []
>>> data = ['a','b','c','d']
>>> 
>>> combinations(target,data)
['a']
['a', 'b']
['a', 'b', 'c']
['a', 'b', 'c', 'd']
['a', 'b', 'd']
['a', 'c']
['a', 'c', 'd']
['a', 'd']
['b']
['b', 'c']
['b', 'c', 'd']
['b', 'd']
['c']
['c', 'd']
['d']