Test if python Counter is contained in another Cou

问题:

How to test if a python Counter is contained in another one using the following definition:

A Counter a is contained in a Counter b if, and only if, for every key k in a, the value a[k] is less or equal to the value b[k]. The Counter({'a': 1, 'b': 1}) is contained in Counter({'a': 2, 'b': 2}) but it is not contained in Counter({'a': 2, 'c': 2}).

I think it is a poor design choice but in python 2.x the comparison operators (<, <=, >=, >) do not use the previous definition, so the third Counter is considered greater-than the first. In python 3.x, instead, Counter is an unorderable type.

回答1:

While Counter instances are not comparable with the < and > operators, you can find their difference with the - operator. The difference never returns negative counts, so if A - B is empty, you know that B contains all the items in A.

def contains(larger, smaller):
    return not smaller - larger

回答2:

The best I came up with is to convert the definition i gave in code:

def contains(container, contained):
    return all(container[x] >= contained[x] for x in contained)

But if feels strange that python don't have an out-of-the-box solution and I have to write a function for every operator (or make a generic one and pass the comparison function).

回答3:

For all the keys in smaller Counter make sure that no value is greater than its counterpart in the bigger Counter:

def containment(big, small):
    return not any(v > big[k] for (k, v) in small.iteritems())

>>> containment(Counter({'a': 2, 'b': 2}), Counter({'a': 1, 'b': 1}))
True
>>> containment(Counter({'a': 2, 'c': 2, 'b': 3}), Counter({'a': 2, 'b': 2}))
True
>>> print containment(Counter({'a': 2, 'b': 2}), Counter({'a': 2, 'b': 2, 'c':1}))
False
>>> print containment(Counter({'a': 2, 'c': 2}), Counter({'a': 1, 'b': 1})
False