获取描述符对象的简洁方法

Question

在Python 3中

class A(object):
    attr = SomeDescriptor()
    ...
    def somewhere(self):
        # need to check is type of self.attr is SomeDescriptor()
        desc = self.__class__.__dict__[attr_name]
        return isinstance(desc, SomeDescriptor)

有更好的方法吗？我不喜欢这个self.__class__.__dict__东西

Answer 1

A.attr导致Python来调用SomeDescriptor().__get__(None, A)所以如果你有SomeDescriptor.__get__回报self的时候inst是None,那么A.attr将返回描述:

class SomeDescriptor():
    def __get__(self, inst, instcls):
        if inst is None:
            # instance attribute accessed on class, return self
            return self

然后使用.访问描述符

desc  = type(self).attr

如果属性的名称仅作为字符串知道attr_name,那么您将使用

desc  = getattr(type(self), attr_name)

这个工程即使self是一个实例子类的A,而

desc = self.__class__.__dict__[attr_name]

只有self作为一个实例才会起作用A.

class SomeDescriptor():
    def __get__(self, inst, instcls):
        if inst is None:
            # instance attribute accessed on class, return self
            return self
        return 4

class A():
    attr = SomeDescriptor()
    def somewhere(self):
        attr_name = 'attr'
        desc  = getattr(type(self), attr_name)
        # desc = self.__class__.__dict__[attr_name]  # b.somewhere() would raise KeyError
        return isinstance(desc, SomeDescriptor)

这显示A.attr返回描述符,并按a.somewhere()预期工作:

a = A()
print(A.attr)
# <__main__.SomeDescriptor object at 0xb7395fcc>    
print(a.attr)
# 4    
print(a.somewhere())
# True

这表明它也适用于子类A.如果取消注释 desc = self.__class__.__dict__[attr_name],您将看到 b.somewhere()引发KeyError:

class B(A): pass
b = B()
print(B.attr)
# <__main__.SomeDescriptor object at 0xb7395fcc>   
print(b.attr)
# 4    
print(b.somewhere())
# True

顺便说一句,即使你没有完全控制SomeDescriptor的定义,你仍然可以将它包装在一个描述符中,该描述符返回self时inst为None:

def wrapper(Desc):
    class Wrapper(Desc):
        def __get__(self, inst, instcls):
            if inst is None: return self
            return super().__get__(inst, instcls)
    return Wrapper

class A():
    attr = wrapper(SomeDescriptor)()
    def somewhere(self):
        desc  = type(self).attr
        # desc = self.__class__.__dict__[attr_name]  # b.somewhere() would raise KeyError
        return isinstance(desc, SomeDescriptor)

所以没有必要使用

desc = self.__class__.__dict__[attr_name]

要么

desc = vars(type(self))['attr']

它遇到了同样的问题.

Answer 2

是的,要访问类的描述符,防止descriptor.__get__被调用的唯一方法是通过类__dict__.

您可以使用type(self)访问当前类,并vars()以__dict__更多符合API的方式访问:

desc = vars(type(self))['attr']

如果您的描述是完全自定义的,你可以随时返回self从SomeDescriptor.__get__()当实例参数是None,当你直接访问类上的描述符发生.您可以放弃vars()通话直接进入:

desc = type(self).attr

该property()描述符对象正是这样做的.