如何在 Python 中从现有的超类类型对象实例化子类类型变量

Question

我有一种情况，我用几个属性扩展了一个类：

class SuperClass:
    def __init__(self, tediously, many, attributes):
        # assign the attributes like "self.attr = attr"

class SubClass:
    def __init__(self, id, **kwargs):
        self.id = id
        super().__init__(**kwargs)

然后我想创建实例，但我知道这会导致子类只能像这样实例化：

super_instance = SuperClass(tediously, many, attributes)

sub_instance = SubClass(id, tediously=super_instance.tediously, many=super_instance.many, attributes=super_instance.attributes)

我的问题是，是否可以通过复制超类实例的属性来实例化子类更漂亮/更简洁，而无需编写一段香肠代码来手动执行它（在构造函数调用或构造函数的主体中） )... 类似：

utopic_sub_instance = SubClass(id, **super_instance)

Answer 1

也许你想要一些关于如何不写这么多代码的具体想法？ 所以一种方法是这样的：

class A:
    def __init___(self, a, b, c):
        self.a = a
        self.b = b
        self.c = c


class B:
    def __init__(self, x, a, b, c):
        self.x = x
        super().__init__(a, b, c)


a = A(1, 2, 3)
b = B('x', 1, 2, 3)


# so your problem is that you want to avoid passing 1,2,3 manually, right?
# So as a comment suggests, you should use alternative constructors here.
# Alternative constructors are good because people not very familiar with
#  Python could also understand them.
# Alternatively, you could use this syntax, but it is a little dangerous and prone to producing 
# bugs in the future that are hard to spot


class BDangerous:
    def __init__(self, x, a, b, c):
        self.x = x
        kwargs = dict(locals())
        kwargs.pop('x')
        kwargs.pop('self')

        # This is dangerous because if in the future someone adds a variable in this 
        # scope, you need to remember to pop that also
        # Also, if in the future, the super constructor acquires the same parameter that
        # someone else adds as a variable here... maybe you will end up passing an argument
        # unwillingly. That might cause a bug
        # kwargs.pop(...pop all variable names you don't want to pass)
        super().__init__(**kwargs)


class BSafe:
    def __init__(self, x, a, b, c):
        self.x = x
        bad_kwargs = dict(locals())

        # This is safer: you are explicit about which arguments you're passing
        good_kwargs = {}
        for name in 'a,b,c'.split(','):
            good_kwargs[name] = bad_kwargs[name]

        # but really, this solution is not that much better compared to simply passing all 
        # parameters explicitly
        super().__init__(**good_kwargs)
  

或者，让我们更疯狂一点。我们将使用自省来动态构建要作为参数传递的 dict。我的示例中没有包含仅关键字参数、默认值、*args 或 **kwargs 的情况

class A:
    def __init__(self, a,b,c):
        self.a = a
        self.b = b
        self.c = c


class B(A):
    def __init__(self, x,y,z, super_instance):
        import inspect
        spec = inspect.getfullargspec(A.__init__)

        positional_args = []
        super_vars = vars(super_instance)

        for arg_name in spec.args[1:]:  # to exclude 'self'
            positional_args.append(super_vars[arg_name])

        # ...but of course, you must have the guarantee that constructor
        # arguments will be set as instance attributes with the same names
        super().__init__(*positional_args)

Answer 2

我最终成功地使用了 alt 构造函数和 super_instance 的 __dict__ 属性的组合。

class SuperClass:
    def __init__(self, tediously, many, attributes):
        self.tediously = tediously 
        self.many = many 
        self.attributes = attributes

class SubClass(SuperClass):
    def __init__(self, additional_attribute, tediously, many, attributes):
        self.additional_attribute = additional_attribute
        super().__init__(tediously, many, attributes)

    @classmethod
    def from_super_instance(cls, additional_attribute, super_instance):
        return cls(additional_attribute=additional_attribute, **super_instance.__dict__)

super_instance = SuperClass("tediously", "many", "attributes")

sub_instance = SubClass.from_super_instance("additional_attribute", super_instance)

注意：请记住，python 是按顺序执行语句的，因此如果您想覆盖继承属性的值，请将super().__init__() 放在SubClass.__init__ 中的其他赋值语句之前。

注意 2：pydantic 有这个非常好的特性，他们的 BaseModel 类自动生成 .__init__() 方法，有助于属性类型验证并为此类模型提供 .dict() 方法（它与 @ 基本相同987654329@ 虽然）。