修饰类将失去对其属性的访问权限

Question

我实现了一个像魅力一样工作的装饰器，直到我给装饰类添加属性。当我实例化类时，它不能访问类属性。以以下最小的工作示例为例:

from module import specialfunction
class NumericalMathFunctionDecorator:
    def __init__(self, enableCache=True):
        self.enableCache = enableCache
    def __call__(self, wrapper):
        def numericalmathfunction(*args, **kwargs):
            func = specialfunction(wrapper(*args, **kwargs))
            """
            Do some setup to func with decorator arguments (e.g. enableCache)
            """
        return numericalmathfunction
@NumericalMathFunctionDecorator(enableCache=True)
class Wrapper:
    places = ['home', 'office']
    configs = {
               'home':
                  {
                   'attr1': 'path/at/home',
                   'attr2': 'jhdlt'
                  },
               'office':
                  {
                   'attr1': 'path/at/office',
                   'attr2': 'sfgqs'
                  }
              }
    def __init__(self, where='home'):
        # Look for setup configuration on 'Wrapper.configs[where]'.
        assert where in Wrapper.places, "Only valid places are {}".format(Wrapper.places)
        self.__dict__.update(Wrapper.configs[where])
    def __call__(self, X):
        """Do stuff with X and return the result
        """
        return X ** 2
model = Wrapper()

当我实例化Wrapper类(#1)时，我得到以下错误:

---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-5-a99bd3d544a3> in <module>()
     15         assert where in Wrapper.places, "Only valid places are {}".format(Wrapper.places)
     16 
---> 17 model = Wrapper()
<ipython-input-5-a99bd3d544a3> in numericalmathfunction(*args, **kwargs)
      5     def __call__(self, wrapper):
      6         def numericalmathfunction(*args, **kwargs):
----> 7             func = wrapper(*args, **kwargs)
      8         return numericalmathfunction
      9 
<ipython-input-5-a99bd3d544a3> in __init__(self, where)
     13     def __init__(self, where='home'):
     14         # Look for setup configuration on 'Wrapper.configs[where]'.
---> 15         assert where in Wrapper.places, "Only valid places are {}".format(Wrapper.places)
     16 
     17 model = Wrapper()
AttributeError: 'function' object has no attribute 'places'

我猜有了这个装饰器，Wrapper就变成了一个失去对其属性访问的函数…

有什么办法可以解决这个问题吗?也许有一个解决方法

Answer 1

您将Wrapper(这是一个类)替换为numericalmathfunction函数对象。该对象没有任何类属性，没有

实际上，装饰器是这样做的:

class Wrapper:
    # ...
Wrapper = NumericalMathFunctionDecorator(enableCache=True)(Wrapper)

所以无论NumericalMathFunctionDecorator.__call__方法返回什么，现在已经取代了类;所有对Wrapper的引用现在都引用该返回值。当您在__init__方法中使用名称Wrapper时，您引用的是该全局变量，而不是原始类。

您仍然可以使用type(self)访问当前类，或者仅通过self引用这些属性(其中名称查找将通过类):

def __init__(self, where='home'):
    # Look for setup configuration on 'Wrapper.configs[where]'.
    assert where in self.places, "Only valid places are {}".format(self.places)
    self.__dict__.update(self.configs[where])

或

def __init__(self, where='home'):
    # Look for setup configuration on 'Wrapper.configs[where]'.
    cls = type(self)
    assert where in cls.places, "Only valid places are {}".format(cls.places)
    self.__dict__.update(cls.configs[where])

在这两种情况下，如果你曾经做过子类Wrapper，你最终都可以引用子类上的属性(在这种情况下你无论如何都不能这样做，因为你必须从decorator闭包中钓出类)。

或者，您可以将原始类存储为返回函数的属性:

def __call__(self, wrapper):
    def numericalmathfunction(*args, **kwargs):
        func = specialfunction(wrapper(*args, **kwargs))
        """
        Do some setup to func with decorator arguments (e.g. enableCache)
        """
    numericalmathfunction.__wrapped__ = wrapper
    return numericalmathfunction

然后在你的__init__:

中使用该引用

def __init__(self, where='home'):
    # Look for setup configuration on 'Wrapper.configs[where]'.
    cls = Wrapper
    while hasattr(cls, '__wrapped__'):
        # remove any decorator layers to get to the original
        cls = cls.__wrapped__
    assert where in cls.places, "Only valid places are {}".format(cls.places)
    self.__dict__.update(cls.configs[where])