除了在这里的其他答案中描述的无数优点-内存意识的紧凑实例，比更易变的__dict__承载实例更不容易出错等等-我发现使用__slots__提供了更清晰的类声明，因为类的实例变量显式地公开。

为了解决__slots__声明的继承问题，我使用了这个元类:

import abc

class Slotted(abc.ABCMeta):
    
    """ A metaclass that ensures its classes, and all subclasses,
        will be slotted types.
    """
    
    def __new__(metacls, name, bases, attributes, **kwargs):
        """ Override for `abc.ABCMeta.__new__(…)` setting up a
            derived slotted class.
        """
        if '__slots__' not in attributes:
            attributes['__slots__'] = tuple()
        
        return super(Slotted, metacls).__new__(metacls, name, # type: ignore
                                                        bases,
                                                        attributes,
                                                      **kwargs)

…如果在继承塔中声明为基类的元类，则确保从该基类派生的所有内容都将正确继承__slots__属性，即使中间类没有声明任何属性。像这样:

# note no __slots__ declaration necessary with the metaclass:
class Base(metaclass=Slotted):
    pass

# class is properly slotted, no __dict__:
class Derived(Base):
    __slots__ = 'slot', 'another_slot'

# class is also properly slotted:
class FurtherDerived(Derived):
    pass

除了其他答案，这里还有一个使用__slots__的例子:

>>> class Test(object):   #Must be new-style class!
...  __slots__ = ['x', 'y']
... 
>>> pt = Test()
>>> dir(pt)
['__class__', '__delattr__', '__doc__', '__getattribute__', '__hash__', 
 '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__', 
 '__repr__', '__setattr__', '__slots__', '__str__', 'x', 'y']
>>> pt.x
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: x
>>> pt.x = 1
>>> pt.x
1
>>> pt.z = 2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'Test' object has no attribute 'z'
>>> pt.__dict__
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'Test' object has no attribute '__dict__'
>>> pt.__slots__
['x', 'y']

因此，要实现__slots__，它只需要额外的一行(并使您的类成为一个新样式的类，如果它还不是的话)。通过这种方式，您可以将这些类的内存占用减少5倍，代价是必须编写自定义pickle代码(如果需要的话)。

类实例的属性有3个属性:实例、属性名和属性值。

在常规属性访问中，实例充当字典，属性名充当字典查找值中的键。

实例(属性)——>值

在__slots__访问中，属性的名称充当字典，实例充当字典查找值中的键。

属性(实例)——>值

在flyweight模式中，属性的名称充当字典，值充当查找实例的字典中的键。

属性(value)——>实例

从Python 3.9开始，字典可用于通过__slots__向属性添加描述。没有描述的属性可以使用None，即使给出了描述，私有变量也不会出现。

class Person:

    __slots__ = {
        "birthday":
            "A datetime.date object representing the person's birthday.",
        "name":
            "The first and last name.",
        "public_variable":
            None,
        "_private_variable":
            "Description",
    }


help(Person)
"""
Help on class Person in module __main__:

class Person(builtins.object)
 |  Data descriptors defined here:
 |
 |  birthday
 |      A datetime.date object representing the person's birthday.
 |
 |  name
 |      The first and last name.
 |
 |  public_variable
"""

除了在这里的其他答案中描述的无数优点-内存意识的紧凑实例，比更易变的__dict__承载实例更不容易出错等等-我发现使用__slots__提供了更清晰的类声明，因为类的实例变量显式地公开。

为了解决__slots__声明的继承问题，我使用了这个元类:

import abc

class Slotted(abc.ABCMeta):
    
    """ A metaclass that ensures its classes, and all subclasses,
        will be slotted types.
    """
    
    def __new__(metacls, name, bases, attributes, **kwargs):
        """ Override for `abc.ABCMeta.__new__(…)` setting up a
            derived slotted class.
        """
        if '__slots__' not in attributes:
            attributes['__slots__'] = tuple()
        
        return super(Slotted, metacls).__new__(metacls, name, # type: ignore
                                                        bases,
                                                        attributes,
                                                      **kwargs)

…如果在继承塔中声明为基类的元类，则确保从该基类派生的所有内容都将正确继承__slots__属性，即使中间类没有声明任何属性。像这样:

# note no __slots__ declaration necessary with the metaclass:
class Base(metaclass=Slotted):
    pass

# class is properly slotted, no __dict__:
class Derived(Base):
    __slots__ = 'slot', 'another_slot'

# class is also properly slotted:
class FurtherDerived(Derived):
    pass

最初的问题是关于一般用例，而不仅仅是关于内存。 因此，这里应该提到的是，当实例化大量对象时，您也会获得更好的性能——有趣的是，当将大型文档解析为对象或从数据库中解析时。

下面是使用插槽和不使用插槽创建具有一百万个条目的对象树的比较。作为对树使用普通字典时的性能参考(OSX上的Py2.7.10):

********** RUN 1 **********
1.96036410332 <class 'css_tree_select.element.Element'>
3.02922606468 <class 'css_tree_select.element.ElementNoSlots'>
2.90828204155 dict
********** RUN 2 **********
1.77050495148 <class 'css_tree_select.element.Element'>
3.10655999184 <class 'css_tree_select.element.ElementNoSlots'>
2.84120798111 dict
********** RUN 3 **********
1.84069895744 <class 'css_tree_select.element.Element'>
3.21540498734 <class 'css_tree_select.element.ElementNoSlots'>
2.59615707397 dict
********** RUN 4 **********
1.75041103363 <class 'css_tree_select.element.Element'>
3.17366290092 <class 'css_tree_select.element.ElementNoSlots'>
2.70941114426 dict

测试类(标识，除了槽):

class Element(object):
    __slots__ = ['_typ', 'id', 'parent', 'childs']
    def __init__(self, typ, id, parent=None):
        self._typ = typ
        self.id = id
        self.childs = []
        if parent:
            self.parent = parent
            parent.childs.append(self)

class ElementNoSlots(object): (same, w/o slots)

Testcode，详细模式:

na, nb, nc = 100, 100, 100
for i in (1, 2, 3, 4):
    print '*' * 10, 'RUN', i, '*' * 10
    # tree with slot and no slot:
    for cls in Element, ElementNoSlots:
        t1 = time.time()
        root = cls('root', 'root')
        for i in xrange(na):
            ela = cls(typ='a', id=i, parent=root)
            for j in xrange(nb):
                elb = cls(typ='b', id=(i, j), parent=ela)
                for k in xrange(nc):
                    elc = cls(typ='c', id=(i, j, k), parent=elb)
        to =  time.time() - t1
        print to, cls
        del root

    # ref: tree with dicts only:
    t1 = time.time()
    droot = {'childs': []}
    for i in xrange(na):
        ela =  {'typ': 'a', id: i, 'childs': []}
        droot['childs'].append(ela)
        for j in xrange(nb):
            elb =  {'typ': 'b', id: (i, j), 'childs': []}
            ela['childs'].append(elb)
            for k in xrange(nc):
                elc =  {'typ': 'c', id: (i, j, k), 'childs': []}
                elb['childs'].append(elc)
    td = time.time() - t1
    print td, 'dict'
    del droot