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

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

不使用itertools:

def combine(inp):
    return combine_helper(inp, [], [])


def combine_helper(inp, temp, ans):
    for i in range(len(inp)):
        current = inp[i]
        remaining = inp[i + 1:]
        temp.append(current)
        ans.append(tuple(temp))
        combine_helper(remaining, temp, ans)
        temp.pop()
    return ans


print(combine(['a', 'b', 'c', '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')

看看itertools.combination:

itertools.combinations (iterable, r) 返回元素的r长度子序列 输入迭代对象。 组合是按字典排序顺序发出的。那么，如果 Input iterable已排序，则 组合元组将在 排序顺序。

从2.6开始，电池包括在内!

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))