Python 在子 class 中覆盖方法 return 的类型提示，而不重新定义方法签名

Question

我有一个基础 class，在方法的 return.

上有类型提示 float

在子 class 中，在不重新定义签名的情况下，我能否以某种方式将方法 return 上的类型提示更新为 int？

示例代码

#!/usr/bin/env python3.6


class SomeClass:
    """This class's some_method will return float."""

    RET_TYPE = float

    def some_method(self, some_input: str) -> float:
        return self.RET_TYPE(some_input)


class SomeChildClass(SomeClass):
    """This class's some_method will return int."""

    RET_TYPE = int


if __name__ == "__main__":
    ret: int = SomeChildClass().some_method("42"). # 
    ret2: float = SomeChildClass().some_method("42")

我的 IDE 抱怨类型不匹配：

这是因为我的 IDE 仍在使用来自 SomeClass.some_method 的类型提示。

研究

我认为解决方案可能是使用泛型，但我不确定是否有更简单的方法。

建议可以使用 instance variable annotations，但我不确定如何为 return 类型执行此操作。

Answer 1

你可以使用类似的东西：

from typing import TypeVar, Generic


T = TypeVar('T', float, int) # types you support


class SomeClass(Generic[T]):
    """This class's some_method will return float."""

    RET_TYPE = float

    def some_method(self, some_input: str) -> T:
        return self.RET_TYPE(some_input)


class SomeChildClass(SomeClass[int]):
    """This class's some_method will return int."""

    RET_TYPE = int


if __name__ == "__main__":
    ret: int = SomeChildClass().some_method("42")
    ret2: float = SomeChildClass().some_method("42")

但是有一个问题。那个我不知道怎么解决。对于 SomeChildClass 方法 some_method IDE 将显示通用提示。至少 pycharm（我猜你是这个）不会显示为错误。

Answer 2

以下代码在 PyCharm 上运行良好。我添加了 complex 案例以使其更清楚。

我基本上将方法提取为通用 class，然后将其用作每个子 class 的混合。请格外小心使用，因为它似乎很不标准。

from typing import ClassVar, Generic, TypeVar, Callable


S = TypeVar('S', bound=complex)


class SomeMethodImplementor(Generic[S]):
    RET_TYPE: ClassVar[Callable]

    def some_method(self, some_input: str) -> S:
        return self.__class__.RET_TYPE(some_input)


class SomeClass(SomeMethodImplementor[complex]):
    RET_TYPE = complex


class SomeChildClass(SomeClass, SomeMethodImplementor[float]):
    RET_TYPE = float


class OtherChildClass(SomeChildClass, SomeMethodImplementor[int]):
    RET_TYPE = int


if __name__ == "__main__":
    ret: complex = SomeClass().some_method("42")
    ret2: float = SomeChildClass().some_method("42")
    ret3: int = OtherChildClass().some_method("42")
    print(ret, type(ret), ret2, type(ret2), ret3, type(ret3))

如果您将 ret2: float 更改为 ret2: int，它将正确显示类型错误。

遗憾的是，mypy 在这种情况下是否显示错误（版本 0.770），

otherhint.py:20: error: Incompatible types in assignment (expression has type "Type[float]", base class "SomeClass" defined the type as "Type[complex]")
otherhint.py:24: error: Incompatible types in assignment (expression has type "Type[int]", base class "SomeClass" defined the type as "Type[complex]")
otherhint.py:29: error: Incompatible types in assignment (expression has type "complex", variable has type "float")
otherhint.py:30: error: Incompatible types in assignment (expression has type "complex", variable has type "int")

第一个错误可以"fixed"写

    RET_TYPE: ClassVar[Callable] = int

每个子class。现在，错误减少到

otherhint.py:29: error: Incompatible types in assignment (expression has type "complex", variable has type "float")
otherhint.py:30: error: Incompatible types in assignment (expression has type "complex", variable has type "int")

这与我们想要的恰恰相反，但如果你只关心PyCharm，那其实并不重要。

Answer 3

好吧，我可以尝试并结合 and 的答案得出一个解决方案，其中：

我的IDE（PyCharm）可以正确推断出return类型
mypy 报告类型是否不匹配
即使类型提示指示不匹配，运行时也不会出错

这正是我想要的行为。非常感谢大家:)

#!/usr/bin/env python3

from typing import TypeVar, Generic, ClassVar, Callable


T = TypeVar("T", float, int)  # types supported


class SomeBaseClass(Generic[T]):
    """This base class's some_method will return a supported type."""

    RET_TYPE: ClassVar[Callable]

    def some_method(self, some_input: str) -> T:
        return self.RET_TYPE(some_input)


class SomeChildClass1(SomeBaseClass[float]):
    """This child class's some_method will return a float."""

    RET_TYPE = float


class SomeChildClass2(SomeBaseClass[int]):
    """This child class's some_method will return an int."""

    RET_TYPE = int


class SomeChildClass3(SomeBaseClass[complex]):
    """This child class's some_method will return a complex."""

    RET_TYPE = complex


if __name__ == "__main__":
    some_class_1_ret: float = SomeChildClass1().some_method("42")
    some_class_2_ret: int = SomeChildClass2().some_method("42")

    # PyCharm can infer this return is a complex.  However, running mypy on
    # this will report (this is desirable to me):
    # error: Value of type variable "T" of "SomeBaseClass" cannot be "complex"
    some_class_3_ret = SomeChildClass3().some_method("42")

    print(
        f"some_class_1_ret = {some_class_1_ret} of type {type(some_class_1_ret)}\n"
        f"some_class_2_ret = {some_class_2_ret} of type {type(some_class_2_ret)}\n"
        f"some_class_3_ret = {some_class_3_ret} of type {type(some_class_3_ret)}\n"
    )

Python 在子 class 中覆盖方法 return 的类型提示，而不重新定义方法签名

Python overriding type hint on a method's return in child class, without redefining method signature

python

oop

type-hinting

method-signature