如果你不想使用组合库，这里是解决方案:

nums = [1,2,3]
p = [[]]
fnl = [[],nums]

for i in range(len(nums)):
    for j in range(i+1,len(nums)):
        p[-1].append([i,j])

for i in range(len(nums)-3):
    p.append([])
    for m in p[-2]:
        p[-1].append(m+[m[-1]+1])

for i in p:
    for j in i:
        n = []
        for m in j:
            if m < len(nums):
                n.append(nums[m])
        if n not in fnl:
            fnl.append(n)

for i in nums:
    if [i] not in fnl:
        fnl.append([i])

print(fnl)

输出:

[[], [1, 2, 3], [1, 2], [1, 3], [2, 3], [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

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!

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)

这里是另一个解决方案(一行程序)，涉及到使用itertools.combination函数，但这里我们使用了双链表理解式(而不是for循环或sum):

def combs(x):
    return [c for i in range(len(x)+1) for c in combinations(x,i)]

演示:

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