我应该如何注释返回@classmethod
实例的cls
?这是一个糟糕的例子:
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Foo
,但会更准确地返回调用此Foo
的子类,因此使用-> "Foo"
进行注释并不够好。
答案 0 :(得分:11)
诀窍是明确向s2
参数添加注释,并结合cls
,generics和TypeVar
添加到represent a class rather then the instance itself,像这样:
Type
通常情况下,您可以保留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())
(和cls
)未注释,但如果您需要引用特定的子类,则可以添加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问题。您可能还需要使用从Github克隆的最新版mypy,而不是pypi上可用的内容 - 我不记得该版本是否支持类方法的此功能。
答案 1 :(得分:2)
出于完整性考虑,在Python 3.7中,您可以通过在文件的开头导入postponed evaluation of annotations
来使用PEP563中定义的from __future__ import annotations
。
然后输入您的代码
from __future__ import annotations
class Foo(object):
def __init__(self, bar: str):
self.bar = bar
@classmethod
def with_stuff_appended(cls, bar: str) -> Foo:
return cls(bar + "stuff")