您可以使用itertools.combinations来获得以下组合：

>>> import itertools
>>> xs = [1, 2, 3]

>>> itertools.combinations(xs, 2)  # returns an iterator
<itertools.combinations object at 0x7f88f838ff48>
>>> list(itertools.combinations(xs, 2))  # yields 2-length subsequences
[(1, 2), (1, 3), (2, 3)]

>>> for i in range(0, len(xs) + 1):  # to get all lengths: 0 to 3
...     for subset in itertools.combinations(xs, i):
...         print(list(subset))
... 
[]
[1]
[2]
[3]
[1, 2]
[1, 3]
[2, 3]
[1, 2, 3]

结合列表理解，你会得到你想要的：

>>> [list(subset) for i in range(0, len(xs) + 1)
                  for subset in itertools.combinations(xs, i)]
[[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]

这个问题实际上是问幂集，这里有另一种生成它的方法：（不使用itertools.combinations（））

>>> L = [1, 2, 3]
>>> res = [[]]
>>> for e in L:
    res += [sub + [e] for sub in res]

>>> res
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

或者，您可以使用此第三方库more_itertools：powerset（这个名称恰当地反映了它的用途）。

>>> import more_itertools

>>> list(more_itertools.powerset(L))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
>>>

**如果只使用more_intertools中的一个函数，则无需安装more_intertools。**您可以查看itertools文档中的Itertools Recipes，它包含以下配方：

from itertools import chain, combinations

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

实际上，more_itertools.powerset具有完全相同的实现。

我试过了，没有递归，结果很成功。
A是列表，K是每个子集的长度

def subsets(A,K):
       sets = []
       for j in range(len(A)-K+1):
           mySet = []
           for i in range(j,K+j):  
               mySet.append(A[i])
           sets.append(mySet)
    return sets