python - 类方法返回实例的类型注解

Question

我应该如何注释返回 cls 实例的 @classmethod？这是一个不好的例子:

class Foo(object):
    def __init__(self, bar: str):
        self.bar = bar

    @classmethod
    def with_stuff_appended(cls, bar: str) -> ???:
        return cls(bar + "stuff")

这会返回一个 Foo，但更准确地说，会返回调用它的 Foo 的子类，因此使用 -> "Foo" 进行注释不会足够好。

Answer 1

诀窍是显式地向cls 参数添加注释，结合TypeVar，对于generics。 , 和 Type, 到 represent a class rather than the instance itself ，像这样:

from typing import TypeVar, Type

# Create a generic variable that can be 'Parent', or any subclass.
T = TypeVar('T', bound='Parent')

class Parent:
    def __init__(self, bar: str) -> None:
        self.bar = bar

    @classmethod
    def with_stuff_appended(cls: Type[T], bar: str) -> T:
        # We annotate 'cls' with a typevar so that we can
        # type our return type more precisely
        return cls(bar + "stuff")

class Child(Parent):
    # If you're going to redefine __init__, make sure it
    # has a signature that's compatible with the Parent's __init__,
    # since mypy currently doesn't check for that.

    def child_only(self) -> int:
        return 3

# Mypy correctly infers that p is of type 'Parent',
# and c is of type 'Child'.
p = Parent.with_stuff_appended("10")
c = Child.with_stuff_appended("20")

# We can verify this ourself by using the special 'reveal_type'
# function. Be sure to delete these lines before running your
# code -- this function is something only mypy understands
# (it's meant to help with debugging your types).
reveal_type(p)  # Revealed type is 'test.Parent*'
reveal_type(c)  # Revealed type is 'test.Child*'

# So, these all typecheck
print(p.bar)
print(c.bar)
print(c.child_only())

通常，您可以不注释cls(和self)，但是如果您需要引用特定的子类，您可以添加一个explicit annotation。 .请注意，此功能仍处于试验阶段，在某些情况下可能存在错误。您可能还需要使用从 Github 克隆的最新版本的 mypy，而不是 pypi 上可用的版本——我不记得该版本是否支持类方法的此功能。