如何将带槽的python类转换为字典？

Question

我正在使用带槽的类来减少实例占用的内存.现在,我如何将插槽实例转换为字典？

插槽类看起来像这样:

class Foo(object):
       __slots__ = ['x','y','z']
       def __init__(self):
           self.x = 1
           self.y = 2 
           self.z = 3

我期待这样的事情:

y = Foo()
y.__dict__
{'x': 1, 'y': 2, 'z': 3}

Answer 1

在字典理解中使用__slots__属性plus getattr():

{s: getattr(obj, s) for s in obj.__slots__ if hasattr(obj, s)}

跳过未设置的任何属性.

另外,将缺少的属性设置为None:

{s: getattr(obj, s, None) for s in obj.__slots__}

演示:

>>> class Foo(object):
...     __slots__ = ('bar', 'spam')
... 
>>> obj = Foo()
>>> obj.bar = 42
>>> {s: getattr(obj, s) for s in obj.__slots__ if hasattr(obj, s)}
{'bar': 42}
>>> {s: getattr(obj, s, None) for s in obj.__slots__}
{'spam': None, 'bar': 42}

您甚至可以将其作为该类的属性并vars()使用它:

>>> class Foo(object):
...     __slots__ = ('bar', 'spam')
...     @property
...     def __dict__(self):
...         return {s: getattr(self, s) for s in self.__slots__ if hasattr(self, s)}
... 
>>> f = Foo()
>>> f.bar = 42
>>> f.__dict__
{'bar': 42}
>>> f.spam = 'eggs'
>>> f.__dict__
{'spam': 'eggs', 'bar': 42}
>>> vars(f)
{'spam': 'eggs', 'bar': 42}
>>> f.hello = 'world'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'Foo' object has no attribute 'hello'