Cache implementations and the "in" operator / __contains__ method

[bozhodimitrov]

Consider the following examples:

from cachetools import LRUCache

cache = LRUCache(2)
cache[1] = 1
cache[2] = 2
cache[1]  # get first key
cache[3] = 3
cache

Expected and actual result:

LRUCache([(1, 1), (3, 3)], maxsize=2, currsize=2)

But if you use the in operator or the __contains__ method instead, this doesn't count as usage:

from cachetools import LRUCache

cache = LRUCache(2)
cache[1] = 1
cache[2] = 2
1 in cache  # lookup first key
cache[3] = 3
cache

Expected result:

LRUCache([(1, 1), (3, 3)], maxsize=2, currsize=2)

Actual result:

LRUCache([(2, 2), (3, 3)], maxsize=2, currsize=2)

Hanse the question -- is this by design?
I saw only one other kinda-related issue #26, which is the opposite case of this one -- missing vs contains.

[tkem]

Yes, using the in operator doesn't count as usage and should not affect caching behavior.
It surely can be argued both ways, but frankly it just never occurred to me that __contains__() could be seen as "accessing" the cached item.

[bozhodimitrov]

Thank you for answering. My use case requires this behavior, so I wanted to know if this is by design.
Example with lookup (this use case doesn't care about the actual stored value):

from cachetools import LRUCache


class LookupCache(LRUCache):
    def __contains__(self, key):
        found = super().__contains__(key)
        self[key] = None
        return found


cache = LookupCache(2)
cache[1] = 1
cache[2] = 2
1 in cache  # lookup first key
cache[3] = 3
cache