返回实例的类方法的MyPy注释

Question

如何注释返回cls的实例的@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但更准确地返回取的Foo子类是呼吁，与-> "Foo"这样标注是不够好。

Answer 1

诀窍是注释明确添加到cls参数，结合TypeVar，为generics，并且Type，以represent a class rather then 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()) 

通常情况下，你可以离开cls（和self ）未加注释，但如果您需要参考特定的子类，则可以添加一个explicit annotation。请注意，此功能仍处于试验阶段，在某些情况下可能会出现问题。您可能还需要使用从Github克隆的mypy的最新版本，而不是pypi上可用的 - 我不记得该版本是否支持classmethods的此功能。