我可以得到“所有者”的参考吗？描述符的 init 方法期间的类？

发布于 2024-11-08 13:08:57 字数 308 浏览 0 评论 0原文

是否可以在描述符的 __init__ 函数期间访问描述符内的“所有者”类，而无需像本示例中那样手动传递它？

class FooDescriptor(object):
    def __init__(self, owner):
        #do things to owner here
        setattr(owner, 'bar_attribute', 'bar_value')


class BarClass(object):
    foo_attribute = FooDescriptor(owner=BarClass)

原文

Is it possible to access the 'owner' class inside a descriptor during the __init__ function of that descriptor, without passing it in manually as in this example?

class FooDescriptor(object):
    def __init__(self, owner):
        #do things to owner here
        setattr(owner, 'bar_attribute', 'bar_value')


class BarClass(object):
    foo_attribute = FooDescriptor(owner=BarClass)

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夜巴黎 2024-11-15 13:08:57

执行此类操作的一种方法是使用元类。只要确保它确实是你想要的，如果你不明白它是如何工作的，不要盲目复制。

class Descriptor(object):
    pass

class Meta(type):
    def __new__(cls, name, bases, attrs):
        obj = type.__new__(cls, name, bases, attrs)
        # obj is now a type instance

        # this loop looks for Descriptor subclasses
        # and instantiates them, passing the type as the first argument
        for name, attr in attrs.iteritems():
            if isinstance(attr, type) and issubclass(attr, Descriptor):
                setattr(obj, name, attr(obj))

        return obj

class FooDescriptor(Descriptor):
    def __init__(self, owner):
        owner.foo = 42

class BarClass(object):
    __metaclass__ = Meta
    foo_attribute = FooDescriptor # will be instantiated by the metaclass

print BarClass.foo

如果您需要传递其他参数，您可以使用例如 (class, args) 的元组代替类，或者使 FooDescriptor 成为一个装饰器，它将返回构造函数中只接受一个参数的类。

One way to do something like that is with a metaclass. Just make sure it's really what you want, and don't just copy blindly if you don't understand how it works.

class Descriptor(object):
    pass

class Meta(type):
    def __new__(cls, name, bases, attrs):
        obj = type.__new__(cls, name, bases, attrs)
        # obj is now a type instance

        # this loop looks for Descriptor subclasses
        # and instantiates them, passing the type as the first argument
        for name, attr in attrs.iteritems():
            if isinstance(attr, type) and issubclass(attr, Descriptor):
                setattr(obj, name, attr(obj))

        return obj

class FooDescriptor(Descriptor):
    def __init__(self, owner):
        owner.foo = 42

class BarClass(object):
    __metaclass__ = Meta
    foo_attribute = FooDescriptor # will be instantiated by the metaclass

print BarClass.foo

If you need to pass additional arguments, you could use e.g. a tuple of (class, args) in the place of the class, or make FooDescriptor a decorator that would return a class that takes only one argument in the ctor.

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黎夕旧梦 2024-11-15 13:08:57

从Python 3.6开始，您可以使用__set_name__特殊方法：

class FooDescriptor(object):
    def __set_name__(self, owner, name):
        owner.foo = 42

class BarClass(object):
    foo_attribute = FooDescriptor()

# foo_attribute.__set_name__(BarClass, "foo_attribute") called after class definition

在创建类后立即在类中的所有描述符上自动调用__set_name__。
有关更多详细信息，请参阅 PEP 487。

Since Python 3.6, you can use the __set_name__ special method:

class FooDescriptor(object):
    def __set_name__(self, owner, name):
        owner.foo = 42

class BarClass(object):
    foo_attribute = FooDescriptor()

# foo_attribute.__set_name__(BarClass, "foo_attribute") called after class definition

__set_name__ is automatically called on all descriptors in a class immediately after the class is created.
See PEP 487 for more details.

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