Python3 pass lists to function with functools.lru_cache

It should not throw an error, rather convert into hash-able form within decorator without user even knowing it. You can fix this problem by decorating your functions like this:

#Custom Decorator function
def listToTuple(function):
    def wrapper(*args):
        args = [tuple(x) if type(x) == list else x for x in args]
        result = function(*args)
        result = tuple(result) if type(result) == list else result
        return result
    return wrapper

#your cached function
@listToTuple
@lru_cache(maxsize=cacheMaxSize)
def checkIfAdminAcquired(self, adminId) -> list:
    query = "SELECT id FROM public.admins WHERE id IN ({}) and 
    confirmed_at IS NOT NULL"
    response = self.handleQuery(query, "int", adminId)
    return response

You might want to use yet another decorator after lru_cache to make sure that output of the function is not a tuple, but a list, since right now it will return tuple.

This fails because a list is unhashable. This would make it hard for Python to know what values are cached. A way to fix this is by converting lists to tuples before passing them to a cached function: since tuples are immutable and hashable, they can be cached.

TL;DR

Use a tuple instead of a list:

>>> @lru_cache(maxsize=2)
... def my_function(args):
...     pass
...
>>> my_function([1,2,3])
Traceback (most recent call last):
  File "<input>", line 1, in <module>
    my_function([1,2,3])
TypeError: unhashable type: 'list'

>>> # TO FIX: use a tuple 

>>> my_function(tuple([1,2,3]))
>>>