如何为接受基类任意子类的方法添加类型注解？

我在为基类的子类添加类型注解时遇到了问题：定义了基类Table（所有表子类共享其行为）和管理表集合的Database类，代码运行正常，但为Database的方法使用Table做类型注解时，传入子类（如Ducks）会导致MyPy类型检查失败。尝试过有界泛型但未成功，希望让类型检查器理解该方法可接受Table的任意子类，同时支持特定操作的更具体类型。

原代码 #

#!/usr/bin/env python3
from collections.abc import Callable, Iterable, Mapping, Sequence
from dataclasses import dataclass
from typing import Any, Self, TypeVar

# _Table = TypeVar("_Table", bound="Table") <- Tried this but did not work

@dataclass
class Table:
    @classmethod
    def get_name(cls) -> str:
        return cls.__name__.lower()

@dataclass
class Database:
    records: Mapping[str, Sequence[Table]]

    def filter(
        self, table: type[Table], _filter: Callable[[Table], bool]
    ) -> Iterable[Table]:
        yield from (i for i in self.records[table.get_name()] if _filter(i))
        # USING cast(T) here kind of works but i think is kind a cheat.

@dataclass
class Ducks(Table):
    name: str
    age: int

if __name__ == "__main__":
    records = {
        "ducks": [
            Ducks(name="Patinhas", age=104),
            Ducks(name="Donald", age=30),
            Ducks(name="Huguinho", age=12),
            Ducks(name="Zezinho", age=12),
            Ducks(name="Luizinho", age=12),
        ]
    }

    db = Database(records)
    f: Callable[[Ducks], bool] = lambda t: t.age < 100 # <- BONUS QUESTION: is there a way to not use this long typing lambda and just use the untyped notation ?
    print(*db.filter(Ducks, _filter=f), sep="\n") # <- PROBLEM HERE

运行结果 #

$ python3 ex.py
Ducks(name='Donald', age=30)
Ducks(name='Huguinho', age=12)
Ducks(name='Zezinho', age=12)
Ducks(name='Luizinho', age=12)

MyPy报错信息 #

Mypy lint: ex.py 46 37 error arg-type Argument "_filter" to "filter" of "Database" has incompatible type "Callable[[Ducks], bool]"; expected "Callable[[Table], bool]" (lsp)

一、修复主问题：实现泛型兼容 #

你之前尝试的有界泛型思路是正确的，只是未在filter方法中正确应用。通过绑定到Table的TypeVar，可以让方法的table参数、过滤函数和返回值保持类型一致，让类型检查器认可子类的兼容性：

修改后的完整代码 #

#!/usr/bin/env python3
from collections.abc import Callable, Iterable, Mapping, Sequence
from dataclasses import dataclass
from typing import Self, TypeVar, cast

# 定义绑定到Table的TypeVar，约束为任意Table子类
_Table = TypeVar("_Table", bound="Table")

@dataclass
class Table:
    @classmethod
    def get_name(cls) -> str:
        return cls.__name__.lower()

@dataclass
class Database:
    records: Mapping[str, Sequence[Table]]

    def filter(
        self, table: type[_Table], _filter: Callable[[_Table], bool]
    ) -> Iterable[_Table]:
        # 用cast告诉类型检查器：当前表名对应的记录是_Table类型
        table_records = cast(Sequence[_Table], self.records[table.get_name()])
        yield from (item for item in table_records if _filter(item))

@dataclass
class Ducks(Table):
    name: str
    age: int

if __name__ == "__main__":
    records = {
        "ducks": [
            Ducks(name="Patinhas", age=104),
            Ducks(name="Donald", age=30),
            Ducks(name="Huguinho", age=12),
            Ducks(name="Zezinho", age=12),
            Ducks(name="Luizinho", age=12),
        ]
    }

    db = Database(records)
    # 直接传入lambda，无需额外类型注解
    print(*db.filter(Ducks, lambda t: t.age < 100), sep="\n")

关键修改点 #

1. 定义有界TypeVar：_Table = TypeVar("_Table", bound="Table")，指定该泛型变量只能是Table或其子类
2. 泛型化filter方法：将table参数标注为type[_Table]，_filter标注为Callable[[_Table], bool]，返回值标注为Iterable[_Table]，确保三者类型统一
3. 类型转换：用cast(Sequence[_Table], self.records[table.get_name()])明确告知类型检查器，当前表对应的记录是目标子类类型，避免误报

修改后，MyPy会自动识别：当传入Ducks作为table参数时，过滤函数必须接受Ducks类型，返回的结果也是Iterable[Ducks]，完全兼容子类操作。

二、解决额外问题：简化Lambda类型注解 #

不需要给lambda单独添加冗长的Callable[[Ducks], bool]类型注解。因为泛化后的filter方法已经通过_Table约束了过滤函数的参数类型，当你传入Ducks作为table参数时，类型检查器会自动推断lambda的参数t是Ducks类型，直接写lambda t: t.age < 100即可，MyPy会正确识别且不会报错。

内容的提问来源于stack exchange，提问作者hugo.avlia