实例属性的 Python NotImplementedError

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

请您参考如下方法:

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`