How do I properly decorate a `classmethod` with `functools.lru_cache`?

Question

I tried to decorate a classmethod with functools.lru_cache. My attempt failed:

import functools
class K:
    @functools.lru_cache(maxsize=32)
    @classmethod
    def mthd(i, stryng: str): \
        return stryng

obj = K()

The error message comes from functools.lru_cache:

TypeError: the first argument must be callable

`classmethod` has to be used last.. which means it has to be at the top of your decorator chain — SyntaxVoid, Oct 31 '19 at 18:48

score 27 · Accepted Answer · answered Oct 31 '19 at 18:49

A class method is, itself, not callable. (What is callable is the object return by the class method's __get__ method.)

As such, you want the function decorated by lru_cache to be turned into a class method instead.

@classmethod
@functools.lru_cache(maxsize=32)
def mthd(cls, stryng: str):
    return stryng

score 3 · Answer 2 · answered Nov 09 '19 at 13:18

The selected answer is totally correct but adding another post. If you want to bind the cache storages to each classes, instead of sharing the single storage to all its subclasses, there is another option methodtools

import functools
import methodtools


class K:
    @classmethod
    @functools.lru_cache(maxsize=1)
    def mthd(cls, s: str):
        print('functools', s)
        return s

    @methodtools.lru_cache(maxsize=1)  # note that methodtools wraps classmethod
    @classmethod
    def mthd2(cls, s: str):
        print('methodtools', s)
        return s


class L(K):
    pass


K.mthd('1')
L.mthd('2')
K.mthd2('1')
L.mthd2('2')

K.mthd('1')  # functools share the storage
L.mthd('2')
K.mthd2('1')  # methodtools doesn't share the storage
L.mthd2('2')

Then the result is

$ python example.py
functools 1
functools 2
methodtools 1
methodtools 2
functools 1
functools 2

How do I properly decorate a `classmethod` with `functools.lru_cache`?

2 Answers2