[英]Test if set is a subset, considering the number (multiplicity) of each element in the set
我知道我可以測試set1是否是set2的子集:
{'a','b','c'} <= {'a','b','c','d','e'} # True
但以下也是正確的:
{'a','a','b','c'} <= {'a','b','c','d','e'} # True
我如何考慮集合中元素的出現次數,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
我知道我可以這樣做:
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
但我想知道是否有更簡潔的東西,比如set(A).issubset(B,count=True)
或者是一種保持列表理解的方式。 謝謝!
如評論中所述,使用Counter
的可能解決方案:
from collections import Counter
def issubset(X, Y):
return len(Counter(X)-Counter(Y)) == 0
結合以前的答案提供了盡可能干凈和快速的解決方案:
def issubset(X, Y):
return all(v <= Y[k] for k, v in X.items())
由於@DSM刪除了他的解決方案,我將借此機會提供一個可以擴展的原型
>>> 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
>>>
對於那些對多包含包含的通常概念感興趣的人來說,測試多包含包含的最簡單方法是使用多重集合的交集:
from collections import Counter
def issubset(X, Y):
return X & Y == X
issubset(Counter("ab"), Counter("aab")) # returns True
issubset(Counter("abc"), Counter("aab")) # returns False
這是冪等半環中使用的標准思想。
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