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'}]

flag = 0
requiredCals =12
from itertools import chain, combinations

def powerset(iterable):
    s = list(iterable)  # allows duplicate elements
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

stuff = [2,9,5,1,6]
for i, combo in enumerate(powerset(stuff), 1):
    if(len(combo)>0):
        #print(combo , sum(combo))
        if(sum(combo)== requiredCals):
            flag = 1
            break
if(flag==1):
    print('True')
else:
    print('else')

你可以使用以下简单的代码在Python中生成列表的所有组合:

import itertools

a = [1,2,3,4]
for i in xrange(0,len(a)+1):
   print list(itertools.combinations(a,i))

结果将是:

[()]
[(1,), (2,), (3,), (4,)]
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
[(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
[(1, 2, 3, 4)]

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)

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

如果有人正在寻找一个反向列表，就像我一样:

stuff = [1, 2, 3, 4]

def reverse(bla, y):
    for subset in itertools.combinations(bla, len(bla)-y):
        print list(subset)
    if y != len(bla):
        y += 1
        reverse(bla, y)

reverse(stuff, 1)