考虑集合中每个元素的数量(多重性),测试set是否为子集

Question

我知道我可以测试set1是否是set2的子集:

{'a','b','c'} <= {'a','b','c','d','e'} # True

但以下也是正确的:

{'a','a','b','c'} <= {'a','b','c','d','e'} # True

我如何考虑集合中元素的出现次数,以便:

{'a','b','c'}     <= {'a','b','c','d','e'}      # True
{'a','a','b','c'} <= {'a','b','c','d','e'}      # False since 'a' is in set1 twice but set2 only once
{'a','a','b','c'} <= {'a','a','b','c','d','e'}  # True because both sets have two 'a' elements

我知道我可以这样做:

A, B, C = ['a','a','b','c'], ['a','b','c','d','e'], ['a','a','b','c','d','e']
all([A.count(i) == B.count(i) for i in A]) # False
all([A.count(i) == C.count(i) for i in A]) # True

但我想知道是否有一些更简洁的东西set(A).issubset(B,count=True)或一种方法来保持列表理解.谢谢!

Answer 1

如评论中所述,可能的解决方案使用Counter:

from collections import Counter

def issubset(X, Y):
    return len(Counter(X)-Counter(Y)) == 0

Answer 2

由于@DSM删除了他的解决方案，我将借此机会提供一个原型，您可以在此基础上进行扩展

>>> class Multi_set(Counter):
    def __le__(self, rhs):
        return all(v == rhs[k] for k,v in self.items())


>>> Multi_set(['a','b','c']) <= Multi_set(['a','b','c','d','e'])
True
>>> Multi_set(['a','a','b','c']) <= Multi_set(['a','b','c','d','e'])
False
>>> Multi_set(['a','a','b','c']) <= Multi_set(['a','a','b','c','d','e'])
True
>>>