Test if set is a subset, considering the number (multiplicity) of each element in the set

Question

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 a

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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
>>> 

Answer 5

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