抽象类子类的函数注解

问题描述

我正在尝试使用函数注释，希望我的编辑器能够更好地重构.但是，我遇到了以下问题:

I'm trying to use function annotations in the hope that my editor will be better at refactoring. I however am stumbling over the following problem:

我有一个抽象基类算法.

I have an abstract base class Algorithm.

class Algorithm(metaclass=ABCMeta):
     def __init__(self):   
          self.foo = 'bar'

我还有一个使用算法子类实例的函数

I also have a function that uses instances of subclasses of Algorithm

def get_foo(foo_algorithm):
    return foo_algoritm.foo

输入的 foo_algorithm 可以是 Algorithm 的任何子类的实例.我如何明智地注释此输入?我正在寻找类似的东西:

the input foo_algorithm can be an instance of any of the subclasses of Algorithm. How do I sensibly annotate this input? I'm looking for something along the lines off:

def get_foo(foo_algorithm: subclassof(Algorithm)):
    return foo_algoritm.foo

但我找不到正确的方法来做到这一点.

But I couldn't find the right way to do this.

推荐答案

直接使用Algorithm:

def get_foo(foo_algorithm: Algorithm):
    return foo_algoritm.foo

并且自动子类的任何实例都是可以接受的(isinstance(foo_algorithm, Algorithm) 必须为真，这适用于基类的所有子类).

and automatically any instance of a subclass will be acceptable (isinstance(foo_algorithm, Algorithm) must be true, which applies to all subclasses of a baseclass).

如果你只能接受类，那么使用Type[Algorithm]作为类型提示:

If you can only accept classes, then use Type[Algorithm] as the type hint:

def get_foo(foo_algorithm: Type[Algorithm]):
    return foo_algoritm().foo

参见类型PEP 484 的类对象部分 -- 类型提示:

有时您想谈论类对象，特别是从给定类继承的类对象.这可以拼写为 Type[C]，其中 C 是一个类.澄清:C(当用作注解时)指的是 C 类的实例，Type[C] 指的是 的子类>C.

这里我调用类对象，因为根据您的代码示例，.foo 是一个实例属性；从Algorithm 派生的类本身不会有这样的属性.

Here I called the class object, since .foo is an instance attribute according to your code example; a class derived from Algorithm would not have such an attribute itself.

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