Generic protocols: mypy error: Argument 1 has incompatible type ...; expected

Question

我正在尝试实施通用协议。我的意图是拥有一个 Widget[key_type、value_type] 协议和一个简单的 getter。 Mypy 抱怨 Protocol[K, T] 所以变成了 Protocol[K_co, T_co]。我已经去掉了所有其他约束，但我什至无法让最基本的情况 widg0: Widget[Any, Any] = ActualWidget() 工作。 ActualWidget.get 应该与 get(self, key: K) -> Any 完全兼容，这让我觉得我在某些方面使用了 generics/protocol 错误，或者 mypy 无法处理这个问题。

command/error 来自 mypy:

$ mypy cat_example.py
cat_example.py:34: error: Argument 1 to "takes_widget" has incompatible type "ActualWidget"; expected "Widget[Any, Any]"
cat_example.py:34: note: Following member(s) of "ActualWidget" have conflicts:
cat_example.py:34: note:     Expected:
cat_example.py:34: note:         def [K] get(self, key: K) -> Any
cat_example.py:34: note:     Got:
cat_example.py:34: note:         def get(self, key: str) -> Cat
Found 1 error in 1 file (checked 1 source file)

或者，如果我尝试使用 widg0: Widget[Any, Any] = ActualWidget():

强制分配

error: Incompatible types in assignment (expression has type "ActualWidget", variable has type "Widget[Any, Any]")

完整代码：

from typing import Any, TypeVar
from typing_extensions import Protocol, runtime_checkable

K = TypeVar("K")  # ID/Key Type
T = TypeVar("T")  # General type
K_co = TypeVar("K_co", covariant=True)  # ID/Key Type or subclass
T_co = TypeVar("T_co", covariant=True)  # General type or subclass
K_contra = TypeVar("K_contra", contravariant=True)  # ID/Key Type or supertype
T_contra = TypeVar("T_contra", contravariant=True)  # General type or supertype

class Animal(object): ...

class Cat(Animal): ...


@runtime_checkable
class Widget(Protocol[K_co, T_co]):
    def get(self, key: K) -> T_co: ...

class ActualWidget(object):
    def get(self, key: str) -> Cat:
        return Cat()

def takes_widget(widg: Widget):
    return widg

if __name__ == '__main__':
    widg0 = ActualWidget()
    #widg0: Widget[str, Cat] = ActualWidget()
    #widg0: Widget[Any, Any] = ActualWidget()

    print(isinstance(widg0, Widget))
    print(isinstance({}, Widget))
    takes_widget(widg0)

Answer 1

将我在评论中的内容放在这里。

为了使你的问题的示例工作，你需要使输入参数 contravariant 和输出参数 covariant 像这样：

from typing import TypeVar
from typing_extensions import Protocol, runtime_checkable

T_co = TypeVar("T_co", covariant=True)  # General type or subclass
K_contra = TypeVar("K_contra", contravariant=True)  # ID/Key Type or supertype

class Animal: ...

class Cat(Animal): ...

@runtime_checkable
class Widget(Protocol[K_contra, T_co]):
    def get(self, key: K_contra) -> T_co: ...

class ActualWidget:
    def get(self, key: str) -> Cat:
        return Cat()

def takes_widget(widg: Widget):
    return widg

class StrSub(str):
    pass

if __name__ == '__main__':
    widget_0: Widget[str, Cat] = ActualWidget()
    widget_1: Widget[StrSub, Cat] = ActualWidget()
    widget_2: Widget[str, object] = ActualWidget()
    widget_3: Widget[StrSub, object] = ActualWidget()

    takes_widget(widget_0)
    takes_widget(widget_1)
    takes_widget(widget_2)
    takes_widget(widget_3)

ActualWidget()，即 Widget[str, Cat]，然后可分配给 Widget[SubStr, object] for widget_3，这意味着 Widget[str, Cat] 是 [=13] 的子类=].

Widget[str, Cat] 可以采用所有 SubStr 加上其他 str 子类型（子类关系中的输入类型可以不那么具体，因此是逆变的）并且可以有一个输出是至少一个 object，加上 str 属性（子类关系中的输出类型可以更具体，因此协变）。另见 Wikipedia - Function Types，它正式说明了这一观察：

For example, functions of type Animal -> Cat, Cat -> Cat, and Animal -> Animal can be used wherever a Cat -> Animal was expected.

In other words, the → type constructor is contravariant in the parameter (input) type and covariant in the return (output) type.

Generic protocols: mypy error: Argument 1 has incompatible type ...; expected

Generic protocols: mypy error: Argument 1 has incompatible type ...; expected

python

generics

types

structural-typing

mypy