返回實例的類方法的MyPy註釋

Question

如何註釋返回cls的實例的@classmethod？這是一個不好的例子：

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

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

這會返回一個Foo但更準確地返回取的Foo子類是呼籲，與-> "Foo"這樣標註是不夠好。

Answer 1

訣竅是註釋明確添加到cls參數，結合TypeVar，爲generics，並且Type，以represent a class rather then the instance itself，像這樣：

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()) 

通常情況下，你可以離開cls（和self ）未加註釋，但如果您需要參考特定的子類，則可以添加一個explicit annotation。請注意，此功能仍處於試驗階段，在某些情況下可能會出現問題。您可能還需要使用從Github克隆的mypy的最新版本，而不是pypi上可用的 - 我不記得該版本是否支持classmethods的此功能。