如何获取一个集合的所有子集？（幂集）

def power_set(lst):
    pw_set = [[]]

    for i in range(0,len(lst)):
        for j in range(0,len(pw_set)):
            ele = pw_set[j].copy()
            ele = ele + [lst[i]]
            pw_set = pw_set + [ele]

    return pw_set

使用空集，它是所有子集的一部分，可以使用以下内容：

def subsets(iterable):
    for n in range(len(iterable) + 1):
        yield from combinations(iterable, n)

我只想提供最易懂的解决方案，即反代码高尔夫版本。

from itertools import combinations

l = ["x", "y", "z", ]

def powerset(items):
    combo = []
    for r in range(len(items) + 1):
        #use a list to coerce a actual list from the combinations generator
        combo.append(list(combinations(items,r)))
    return combo

l_powerset = powerset(l)

for i, item in enumerate(l_powerset):
    print "All sets of length ", i
    print item

结果

所有长度为0的集合

[()]

所有长度为1的集合

[('x',), ('y',), ('z',)]

所有长度为2的集合

[('x', 'y'), ('x', 'z'), ('y', 'z')]

所有长度为3的集合

[('x', 'y', 'z')]

有关更多信息，请参见itertools文档，以及维基百科词条幂集

from itertools import combinations
def subsets(arr: set) -> list:
   subsets = []
   [subsets.extend(list(combinations(arr,n))) for n in range(len(arr))]
   return subsets 

a = {1,2,3}
print(subsets(a))

输出：

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

# sorted subsets
print(sorted(subsets(a)))

输出：

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

def powerSet(s):
    sets = [[]]
    for i in s:
        newsets = []
        for k in sets:
            newsets.append(k+[i])
        sets += newsets
    return sets

对于那些想要简单回答的人，首先编写代码。我在这里有一个好的解释：https://leetcode.com/problems/subsets/solutions/3138042/simple-python-solution/

但简短的答案是，你从空集合开始，即“sets = [[]]”。我建议在“for i in s”下面放置一个打印语句，即“print(sets)”，看到它在每个元素i中都会增加一倍。

快速复习一下幂集！

一个集合X的幂集，就是包含X所有子集（包括空集）的集合。

例如，令X = (a,b,c)

幂集= { { a , b , c } , { a , b } , { a , c } , { b , c } , { a } , { b } , { c } , { } }

以下是另一种找到幂集的方法：

def power_set(input):
    # returns a list of all subsets of the list a
    if (len(input) == 0):
        return [[]]
    else:
        main_subset = [ ]
        for small_subset in power_set(input[1:]):
            main_subset += [small_subset]
            main_subset += [[input[0]] + small_subset]
        return main_subset

print(power_set([0,1,2,3]))

感谢原作者的贡献

from itertools import combinations
someSet = {0, 1, 2, 3}
([x for i in range(len(someSet)+1) for x in combinations(someSet,i)])

def powerset(x):
    m=[]
    if not x:
        m.append(x)
    else:
        A = x[0]
        B = x[1:]
        for z in powerset(B):
            m.append(z)
            r = [A] + z
            m.append(r)
    return m

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

输出：

[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3], [4], [1, 4], [2, 4], [1, 2, 4], [3, 4], [1, 3, 4], [2, 3, 4], [1, 2, 3, 4]]

使用递归获取所有子集。疯狂的一行代码。

from typing import List

def subsets(xs: list) -> List[list]:
    return subsets(xs[1:]) + [x + [xs[0]] for x in subsets(xs[1:])] if xs else [[]]

subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = map (x:) (subsets xs) ++ subsets xs

def findsubsets(s, n): 
    return list(itertools.combinations(s, n)) 

def allsubsets(s) :
    a = []
    for x in range(1,len(s)+1):
        a.append(map(set,findsubsets(s,x)))      
    return a