英文:
how to type a custom callable type in Python
问题
我有一个名为Foo的类：
class Foo:def __init__(self, callable):self.my_attr = "hi"self.callable = callabledef __call__(self, *args, **kwargs):# 调用包装的函数return self.callable(*args, **kwargs)
我想为它的实例（__call__方法和my_attr属性）进行类型注释。
谢谢你的帮助。
英文:
I have a class called Foo:
class Foo:def __init__(self, callable):self.my_attr = "hi"self.callable = callabledef __call__(self, *args, **kwargs):# call the wrapped in functionreturn self.callable(*args, **kwargs)
I would like to type its instances (the __call__ method and the my_attr attribute).
Thank you for your help,
答案1
得分: 1
我使用泛型解决了这个问题：
from typing import ParamSpec, TypeVar, Generic, CallableP = ParamSpec("P")RV = TypeVar("RV")class Foo(Generic[P, RV]):def __init__(self, callable: Callable[P, RV]):self.my_attr = "hi"self.callable = callabledef __call__(self, *args: P.args, **kwargs: P.kwargs) -> RV:# 调用包装的函数return self.callable(*args, **kwargs)def my_decorator(func: Callable[P, RV]) -> Foo[P, RV]:return Foo(func)
现在这些类型标注是有效的：
@my_decoratordef func(a: int, b: str) -> str:raise NotImplementedErrors: str = func(1, "2") # 对于对象调用的有效类型标注ss: str = func.my_attr # 对于属性访问的有效类型标注
英文:
I used Generics to solve the problem:
from typing import ParamSpec, TypeVar, Generic, CallableP = ParamSpec("P")RV = TypeVar("RV")class Foo(Generic[P, RV]):def __init__(self, callable: Callable[P, RV]):self.my_attr = "hi"self.callable = callabledef __call__(self, *args: P.args, **kwargs: P.kwargs) -> RV:# call the wrapped in functionreturn self.callable(*args, **kwargs)def my_decorator(func: Callable[P, RV]) -> Foo[P, RV]:return Foo(func)
Now these typing is valid:
@my_decoratordef func(a: int, b: str) -> str:raise NotImplementedErrors: str = func(1, "2") # valid typing for object calss: str = func.my_attr # valid typing for attribute access
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