如何将带槽的 python 类转换为字典？

我使用带有插槽的类来减少实例占用的内存。现在，如何将插槽实例转换为字典？

插槽类如下所示：

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

我期望这样的事情：

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

Use the __slots__属性加成getattr()在字典理解中：

{
    s: getattr(obj, s)
    for s in {
        s
        for cls in type(obj).__mro__
        for s in getattr(cls, '__slots__', ())
    }
    if hasattr(obj, s)
}

它从所有基类收集插槽名称并跳过任何未设置的属性。

或者，将缺失的属性设置为None:

{
    s: getattr(obj, s, None)
    for s in {
        s
        for cls in type(obj).__mro__
        for s in getattr(cls, '__slots__', ())
    }
}

Demo:

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

您甚至可以将其作为班级的财产vars()将利用它：

>>> class Foo(object):
...     __slots__ = ('bar', 'spam')
...     @property
...     def __dict__(self):
...         return {
...             s: getattr(self, s)
...             for s in {
...                 s
...                 for cls in type(self).__mro__
...                 for s in getattr(cls, '__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'