Pop only one item from lru_cache in Python

Question

I'm using the lru_cache decorator from functools, and I need to invalidate only one item in it, and leave the rest unchanged. The documentation states that the __wrapped__ attribute is there for rewrapping the function with a different cache. I checked the source too, but I can't figure out how to use that. It says that

Users should only access the lru_cache through its public API: cache_info, cache_clear, and f.__wrapped__ The internals of the lru_cache are encapsulated for thread safety and to allow the implementation to change

Is is possible to remove one item from lru_cache (safely), or should I write my own cache functionality?

Answer 1

lru_cache in Python 3.9 does not even expose in Python anything but the public API - which include a cache_clear call that will invalidate the whole cache, not just a key.

Thus, instead of finding a workaround for that, you are clearly better-off writing your own cache where you can have full control.

The __wrapped__ attribute itself is just the original function, with no lru_cache functionality at all. Actually, you can make use of it to bypass the cache for one call completely - and the function will just run with the repeated parameters as an ordinay Python function - but its response won't be cached at all.

That is, for

from functools import lru_cache

@lru_cache
def test(a):
   print("side effect")

test(23)
test.__wrapped__(23)

the function body is run for both calls.