如何在Python中使用__slots__获取实例属性名称值

Question

定义两个类时，一个带有__dict__实现（A），另一个带有__slots__实现（B）。

是否有一种聪明的方法来获取__slots__类的实例属性名称和值，就像在__dict__类上使用vars（）函数一样？

class A(object):
    def __init__(self, x, y, z):
        self.x = x
        self.y = y
        self.z = z


class B(object):
    __slots__ = ('x', 'y', 'z')
    def __init__(self, x, y, z):
        self.x = x
        self.y = y
        self.z = z


vars(A(1, 2, 3)) # {'y': 2, 'x': 1, 'z': 3}
vars(B(1, 2, 3)) # TypeError: vars() argument must have __dict__ attribute

使用带有检查的.__ slots__或dir（）只返回属性名称，不带值

B(1, 2, 3).__slots__ # ('x', 'y', 'z')

Answer 1

这是我之前使用过的一个函数：

def vars2(obj):
    try:
        return vars(obj)
    except TypeError:
       return {k: getattr(obj, k) for k in obj.__slots__}

Answer 2

In [2]: x = B(1,2,3)

In [3]: {a: x.__getattribute__(a) for a in dir(x)}
Out[3]: 
{'__class__': __main__.B,
 '__delattr__': <method-wrapper '__delattr__' of B object at 0x7f3bb2b48e18>,
 '__doc__': None,
 '__format__': <function __format__>,
 '__getattribute__': <method-wrapper '__getattribute__' of B object at 0x7f3bb2b48e18>,
 '__hash__': <method-wrapper '__hash__' of B object at 0x7f3bb2b48e18>,
 '__init__': <bound method B.__init__ of <__main__.B object at 0x7f3bb2b48e18>>,
 '__module__': '__main__',
 '__new__': <function __new__>,
 '__reduce__': <function __reduce__>,
 '__reduce_ex__': <function __reduce_ex__>,
 '__repr__': <method-wrapper '__repr__' of B object at 0x7f3bb2b48e18>,
 '__setattr__': <method-wrapper '__setattr__' of B object at 0x7f3bb2b48e18>,
 '__sizeof__': <function __sizeof__>,
 '__slots__': ('x', 'y', 'z'),
 '__str__': <method-wrapper '__str__' of B object at 0x7f3bb2b48e18>,
 '__subclasshook__': <function __subclasshook__>,
 'x': 1,
 'y': 2,
 'z': 3}

或者，如果你不想看到魔术方法：

In [4]: {a: x.__getattribute__(a) for a in dir(x) if not a.startswith('__')}
Out[4]: {'x': 1, 'y': 2, 'z': 3}