Я хотел бы определить протокол, который может быть удовлетворен объектом, который может быть модулем:
from typing import Protocol
class MyType(Protocol):
def foo(self) -> int:
...
class MyClass:
def foo(self) -> int:
return 1
# this is OK
a = MyClass() # type: MyType
import mymodule
# mymodule.py just contains
# def foo() -> int:
# return 42
# but this is not.
b = mymodule # type: MyType
# mypy complains saying
# Incompatible types in assignment (expression has type Module, variable has type "MyType")
import inspect
print(inspect.signature(a.foo))
print(inspect.signature(b.foo))
# both print: () -> int
Суть здесь: https://gist.github.com/hjwp/e322c86d14ce0b11f08b27d7b17f7791