disable `functools.lru_cache` from inside function

Question

I want to have a function that can use functools.lru_cache, but not by default. I am looking for a way to use a function parameter that can be used to disable the lru_cache. Currently, I have a two versions of the function, one with lru_cache and one without. Then I have another function that wraps these with a parameter that can be used to control which function is used

def _no_cache(foo):
    print('_no_cache')
    return 1


@lru_cache()
def _with_cache(foo):
    print('_with_cache')
    return 0


def cache(foo, use_cache=False):
    if use_cache:
        return _with_cache(foo)
    return _no_cache(foo)

Is there a simpler way to do this?

Answer 1

You can't disable the cache from inside the decorated function. However, you can simplify your code a bit by accessing the function directly via the __wrapped__ attribute.

From the documentation:

The original underlying function is accessible through the __wrapped__ attribute. This is useful for introspection, for bypassing the cache, or for rewrapping the function with a different cache.

Demo:

from functools import lru_cache

@lru_cache()
def f(arg):
    print(f"called with {arg}")
    return arg    

def call(arg, use_cache=False):
    if use_cache:
        return f(arg)
    return f.__wrapped__(arg)

call(1)
call(1, True)
call(2, True)
call(1, True)

Output:

called with 1
called with 1
called with 2