Question

如何在python 3.x中创建给定列表的所有子集的列表？给出的列表如[1,2,3]，我想要一个像

这样的输出

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

Answer 1

您可以使用itertools.combinations获取组合：

>>> 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，你会得到你想要的东西：

>>> [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]]

Answer 2

问题实际上是要询问功率集。这是生成它的另一种方法：（没有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]]

或者您也可以使用more_itertools.powerset（）生成它：

>>> import more_itertools
>>> more_itertools.powerset(L)
<itertools.chain object at 0x00186F28>
>>> list(more_itertools.powerset(L))
[(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)]
>>>

Answer 3

我尝试过了。没有递归。顺利完成。

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

在python 3.x中创建给定列表的所有子集的列表

3 个答案: