functools.cache - notify that the result is cached

Question

import functools

@functools.cache
def get_some_results():
    return results

Is there a way to notify the user of the function that the results they are getting are a cached version of the original for any other time they are calling the function?

Answer 1

This isn't a perfect approach, but you could use a custom decorator instead of @functools.cache which would then wrap your function with functools.cache and gather the cache stats before and after the call to determine if the lookup resulted in a cache hit.

This was hastily thrown together but seems to work:

def cache_notify(func):
    func = functools.cache(func)
    def notify_wrapper(*args, **kwargs):
        stats = func.cache_info()
        hits = stats.hits
        results = func(*args, **kwargs)
        stats = func.cache_info()
        if stats.hits > hits:
            print(f"NOTE: {func.__name__}() results were cached")
        return results
    return notify_wrapper

As an example of a simple function:

@cache_notify
def f(x):
    return -x

print("calling f")
print(f"f(1) returned {f(1)}")
print("calling f again")
print(f"f(1) returned {f(1)}")

Results:

calling f
f(1) returned -1
calling f again
NOTE: f() results were cached
f(1) returned -1

How to "notify the user" can be tailored as needed.

Also note that it's possible for the cache stats to be a bit misleading in a multi-threaded environment; see Python lru_cache: how can currsize < misses < maxsize? for details.