我同意Dan H的观点，Ben确实要求所有的组合。itertools.combination()不会给出所有的组合。

另一个问题是，如果输入iterable很大，返回一个生成器而不是列表中的所有内容可能会更好:

iterable = range(10)
for s in xrange(len(iterable)+1):
  for comb in itertools.combinations(iterable, s):
    yield comb

来自itertools的组合

import itertools
col_names = ["aa","bb", "cc", "dd"]
all_combinations = itertools.chain(*[itertools.combinations(col_names,i+1) for i,_ in enumerate(col_names)])
print(list(all_combinations))

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'}]

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']

这个怎么样?使用字符串而不是列表，但同样的事情..string可以像Python中的列表一样处理:

def comb(s, res):
    if not s: return
    res.add(s)
    for i in range(0, len(s)):
        t = s[0:i] + s[i + 1:]
        comb(t, res)

res = set()
comb('game', res) 

print(res)

还可以使用more_itertools包中的powerset函数。

from more_itertools import powerset

l = [1,2,3]
list(powerset(l))

# [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]

我们也可以验证，它满足OP的要求

from more_itertools import ilen

assert ilen(powerset(range(15))) == 32_768