这种方式不停止对象。__setattr__从工作，但我仍然发现它有用:

class A(object):

    def __new__(cls, children, *args, **kwargs):
        self = super(A, cls).__new__(cls)
        self._frozen = False  # allow mutation from here to end of  __init__
        # other stuff you need to do in __new__ goes here
        return self

    def __init__(self, *args, **kwargs):
        super(A, self).__init__()
        self._frozen = True  # prevent future mutation

    def __setattr__(self, name, value):
        # need to special case setting _frozen.
        if name != '_frozen' and self._frozen:
            raise TypeError('Instances are immutable.')
        else:
            super(A, self).__setattr__(name, value)

    def __delattr__(self, name):
        if self._frozen:
            raise TypeError('Instances are immutable.')
        else:
            super(A, self).__delattr__(name)

你可能需要根据用例重写更多的东西(比如__setitem__)。

所以，我在写python 3的相关内容:

I)借助数据类装饰器并设置frozen=True。 我们可以在python中创建不可变对象。

为此需要从data classes lib导入data class，并需要设置frozen=True

ex.

从数据类导入数据类

@dataclass(frozen=True)
class Location:
    name: str
    longitude: float = 0.0
    latitude: float = 0.0

o/p:

>>> l = Location("Delhi", 112.345, 234.788)
>>> l.name
'Delhi'
>>> l.longitude
112.345
>>> l.latitude
234.788
>>> l.name = "Kolkata"
dataclasses.FrozenInstanceError: cannot assign to field 'name'
>>> 

来源:https://realpython.com/python-data-classes/

如果您对具有行为的对象感兴趣，那么namedtuple几乎是您的解决方案。

正如namedtuple文档底部所描述的，您可以从namedtuple派生自己的类;然后，你可以添加你想要的行为。

例如(代码直接取自文档):

class Point(namedtuple('Point', 'x y')):
    __slots__ = ()
    @property
    def hypot(self):
        return (self.x ** 2 + self.y ** 2) ** 0.5
    def __str__(self):
        return 'Point: x=%6.3f  y=%6.3f  hypot=%6.3f' % (self.x, self.y, self.hypot)

for p in Point(3, 4), Point(14, 5/7):
    print(p)

这将导致:

Point: x= 3.000  y= 4.000  hypot= 5.000
Point: x=14.000  y= 0.714  hypot=14.018

这种方法适用于Python 3和Python 2.7(在IronPython上也进行了测试)。 唯一的缺点是继承树有点奇怪;但这不是你经常玩的东西。

这里没有包括的是完全不可变性……不仅仅是父对象，还有所有的子对象。例如，元组/frozensets可能是不可变的，但它所属的对象可能不是。下面是一个小的(不完整的)版本，它在执行不变性方面做得很好:

# Initialize lists
a = [1,2,3]
b = [4,5,6]
c = [7,8,9]

l = [a,b]

# We can reassign in a list 
l[0] = c

# But not a tuple
t = (a,b)
#t[0] = c -> Throws exception
# But elements can be modified
t[0][1] = 4
t
([1, 4, 3], [4, 5, 6])
# Fix it back
t[0][1] = 2

li = ImmutableObject(l)
li
[[1, 2, 3], [4, 5, 6]]
# Can't assign
#li[0] = c will fail
# Can reference
li[0]
[1, 2, 3]
# But immutability conferred on returned object too
#li[0][1] = 4 will throw an exception

# Full solution should wrap all the comparison e.g. decorators.
# Also, you'd usually want to add a hash function, i didn't put
# an interface for that.

class ImmutableObject(object):
    def __init__(self, inobj):
        self._inited = False
        self._inobj = inobj
        self._inited = True

    def __repr__(self):
        return self._inobj.__repr__()

    def __str__(self):
        return self._inobj.__str__()

    def __getitem__(self, key):
        return ImmutableObject(self._inobj.__getitem__(key))

    def __iter__(self):
        return self._inobj.__iter__()

    def __setitem__(self, key, value):
        raise AttributeError, 'Object is read-only'

    def __getattr__(self, key):
        x = getattr(self._inobj, key)
        if callable(x):
              return x
        else:
              return ImmutableObject(x)

    def __hash__(self):
        return self._inobj.__hash__()

    def __eq__(self, second):
        return self._inobj.__eq__(second)

    def __setattr__(self, attr, value):
        if attr not in  ['_inobj', '_inited'] and self._inited == True:
            raise AttributeError, 'Object is read-only'
        object.__setattr__(self, attr, value)

你可以创建一个@immutable装饰器，它覆盖__setattr__并将__slots__更改为一个空列表，然后用它装饰__init__方法。

编辑:正如OP所指出的，改变__slots__属性只会阻止新属性的创建，而不会阻止修改。

Edit2:下面是一个实现:

Edit3:使用__slots__会破坏这段代码，因为if会停止对象__dict__的创建。我正在寻找替代方案。

Edit4:嗯，就是这样。这是一个很粗鄙的问题，但可以作为练习:-)

class immutable(object):
    def __init__(self, immutable_params):
        self.immutable_params = immutable_params

    def __call__(self, new):
        params = self.immutable_params

        def __set_if_unset__(self, name, value):
            if name in self.__dict__:
                raise Exception("Attribute %s has already been set" % name)

            if not name in params:
                raise Exception("Cannot create atribute %s" % name)

            self.__dict__[name] = value;

        def __new__(cls, *args, **kws):
            cls.__setattr__ = __set_if_unset__

            return super(cls.__class__, cls).__new__(cls, *args, **kws)

        return __new__

class Point(object):
    @immutable(['x', 'y'])
    def __new__(): pass

    def __init__(self, x, y):
        self.x = x
        self.y = y

p = Point(1, 2) 
p.x = 3 # Exception: Attribute x has already been set
p.z = 4 # Exception: Cannot create atribute z

下面的基本解决方案针对以下场景:

__init__()可以像往常一样访问属性。 在此之后，对象仅冻结属性更改:

其思想是覆盖__setattr__方法，并在每次对象冻结状态改变时替换其实现。

因此，我们需要一些方法(_freeze)来存储这两个实现，并在请求时在它们之间切换。

这个机制可以在用户类内部实现，也可以从一个特殊的freeze类继承，如下所示:

class Freezer:
    def _freeze(self, do_freeze=True):
        def raise_sa(*args):            
            raise AttributeError("Attributes are frozen and can not be changed!")
        super().__setattr__('_active_setattr', (super().__setattr__, raise_sa)[do_freeze])

    def __setattr__(self, key, value):        
        return self._active_setattr(key, value)

class A(Freezer):    
    def __init__(self):
        self._freeze(False)
        self.x = 10
        self._freeze()