实例属性的 Python NotImplementedError

Question

如何将 instance 属性标记为未在基类中实现？ (与讨论将 class 属性标记为未实现的 this 问题不同，但也许我没有正确理解基类...)

例如我想要类似的东西

class Base():
    def __init__(self):
        self.x = NotImplemented

class GoodSub(Base):
    def __init__(self, x):
        super().__init__()
        self.x = x #good

class BadSub(Base):
    def __init__(self):
       super().__init__()
       #forgot to set self.x

good = GoodSub(5)
bad = BadSub(-1)    
good.x #returns 5
bad.x #throws error because x not implemented

或者，是否有更好的方法强制 Base 的所有子类在初始化时设置 self.x 属性？

编辑:link to related question

Answer 1

一种带有类装饰器和描述符的解决方案(__get__ 方法):

def abstract_variables(*args):
    class av:
        def __init__(self, error_message):
            self.error_message = error_message

        def __get__(self, *args, **kwargs):
            raise NotImplementedError(self.error_message)

    def f(klass):
        for arg in args:
            setattr(klass, arg, av('Descendants must set variable `{}`'.format(arg)))
        return klass

    return f


@abstract_variables('x', 'y')
class Base:
    def __init__(self):
        pass

class Derived(Base):
    x = 10


b = Base()
d = Derived()
print(d.x)    # prints 10
print(d.y)    # raises NotImplementedError

打印:

10
Traceback (most recent call last):
  File "main.py", line 28, in <module>
    print(d.y)
  File "main.py", line 7, in __get__
    raise NotImplementedError(self.error_message)
NotImplementedError: Descendants must set variable `y`