在Python中__slots__的目的是什么——特别是当我想要使用它时,什么时候不使用它?


当前回答

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

其他回答

如果你要实例化很多(成百上千)同一个类的对象,你会想要使用__slots__。__slots__仅作为内存优化工具存在。

强烈建议使用__slots__来约束属性创建。

使用__slots__ pickle对象将无法使用默认的(最古老的)pickle协议;有必要指定一个更高的版本。

python的其他一些自省特性也可能受到不利影响。

引用雅各布·海伦的话:

The proper use of __slots__ is to save space in objects. Instead of having a dynamic dict that allows adding attributes to objects at anytime, there is a static structure which does not allow additions after creation. [This use of __slots__ eliminates the overhead of one dict for every object.] While this is sometimes a useful optimization, it would be completely unnecessary if the Python interpreter was dynamic enough so that it would only require the dict when there actually were additions to the object. Unfortunately there is a side effect to slots. They change the behavior of the objects that have slots in a way that can be abused by control freaks and static typing weenies. This is bad, because the control freaks should be abusing the metaclasses and the static typing weenies should be abusing decorators, since in Python, there should be only one obvious way of doing something. Making CPython smart enough to handle saving space without __slots__ is a major undertaking, which is probably why it is not on the list of changes for P3k (yet).

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

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
"""