问题
I know I can test if set1 is a subset of set2 with:
{'a','b','c'} <= {'a','b','c','d','e'} # True
But the following is also True:
{'a','a','b','c'} <= {'a','b','c','d','e'} # True
How do I have it consider the number of times an element in the set occurs so that:
{'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
I know I could do something like:
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
But I was wondering if there was something more succinct like set(A).issubset(B,count=True)
or a way to stay from list comprehensions. Thanks!
回答1:
As @DSM deleted his solution , I will take the opportunity to provide a prototype based on which you can expand
>>> 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
>>>
回答2:
As stated in the comments, a possible solution using Counter:
from collections import Counter
def issubset(X, Y):
return len(Counter(X)-Counter(Y)) == 0
回答3:
The short answer to your question is there is no set operation that does this, because the definition of a set does not provide those operations. IE defining the functionality you're looking for would make the data type not a set.
Sets by definition have unique, unordered, members:
>>> print {'a', 'a', 'b', 'c'}
set(['a', 'c', 'b'])
>>> {'a', 'a', 'b', 'c'} == {'a', 'b', 'c'}
True
回答4:
Combining previous answers gives a solution which is as clean and fast as possible:
def issubset(X, Y):
return all(v <= Y[k] for k, v in X.items())
- No instances created instead of 3 in @A.Rodas version (both arguments must already be of type Counter, since this is the Pythonic way to handle multisets).
- Early return (short-circuit) as soon as predicate is falsified.
回答5:
For those that are interested in the usual notion of multiset inclusion, the easiest way to test for multiset inclusion is to use intersection of multisets:
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
This is a standard idea used in idempotent semirings.
来源:https://stackoverflow.com/questions/15208369/test-if-set-is-a-subset-considering-the-number-multiplicity-of-each-element-i