您可以覆盖setattr，仍然使用init来设置变量。你可以使用超类setattr。这是代码。

class Immutable:
    __slots__ = ('a','b')
    def __init__(self, a , b):
        super().__setattr__('a',a)
        super().__setattr__('b',b)

    def __str__(self):
        return "".format(self.a, self.b)

    def __setattr__(self, *ignored):
        raise NotImplementedError

    def __delattr__(self, *ignored):
        raise NotImplementedError

我使用了与Alex相同的想法:一个元类和一个“init marker”，但结合重写__setattr__:

>>> from abc import ABCMeta
>>> _INIT_MARKER = '_@_in_init_@_'
>>> class _ImmutableMeta(ABCMeta):
... 
...     """Meta class to construct Immutable."""
... 
...     def __call__(cls, *args, **kwds):
...         obj = cls.__new__(cls, *args, **kwds)
...         object.__setattr__(obj, _INIT_MARKER, True)
...         cls.__init__(obj, *args, **kwds)
...         object.__delattr__(obj, _INIT_MARKER)
...         return obj
...
>>> def _setattr(self, name, value):
...     if hasattr(self, _INIT_MARKER):
...         object.__setattr__(self, name, value)
...     else:
...         raise AttributeError("Instance of '%s' is immutable."
...                              % self.__class__.__name__)
...
>>> def _delattr(self, name):
...     raise AttributeError("Instance of '%s' is immutable."
...                          % self.__class__.__name__)
...
>>> _im_dict = {
...     '__doc__': "Mix-in class for immutable objects.",
...     '__copy__': lambda self: self,   # self is immutable, so just return it
...     '__setattr__': _setattr,
...     '__delattr__': _delattr}
...
>>> Immutable = _ImmutableMeta('Immutable', (), _im_dict)

注意:我直接调用元类，以使它在Python 2中都能工作。X和3.x。

>>> class T1(Immutable):
... 
...     def __init__(self, x=1, y=2):
...         self.x = x
...         self.y = y
...
>>> t1 = T1(y=8)
>>> t1.x, t1.y
(1, 8)
>>> t1.x = 7
AttributeError: Instance of 'T1' is immutable.

它也适用于插槽…:

>>> class T2(Immutable):
... 
...     __slots__ = 's1', 's2'
... 
...     def __init__(self, s1, s2):
...         self.s1 = s1
...         self.s2 = s2
...
>>> t2 = T2('abc', 'xyz')
>>> t2.s1, t2.s2
('abc', 'xyz')
>>> t2.s1 += 'd'
AttributeError: Instance of 'T2' is immutable.

．.． 和多重继承:

>>> class T3(T1, T2):
... 
...     def __init__(self, x, y, s1, s2):
...         T1.__init__(self, x, y)
...         T2.__init__(self, s1, s2)
...
>>> t3 = T3(12, 4, 'a', 'b')
>>> t3.x, t3.y, t3.s1, t3.s2
(12, 4, 'a', 'b')
>>> t3.y -= 3
AttributeError: Instance of 'T3' is immutable.

但是请注意，可变属性仍然是可变的:

>>> t3 = T3(12, [4, 7], 'a', 'b')
>>> t3.y.append(5)
>>> t3.y
[4, 7, 5]

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)

最简单的方法是使用__slots__:

class A(object):
    __slots__ = []

A的实例现在是不可变的，因为您不能在它们上设置任何属性。

如果你想让类实例包含数据，你可以将this和derived from tuple结合起来:

from operator import itemgetter
class Point(tuple):
    __slots__ = []
    def __new__(cls, x, y):
        return tuple.__new__(cls, (x, y))
    x = property(itemgetter(0))
    y = property(itemgetter(1))

p = Point(2, 3)
p.x
# 2
p.y
# 3

编辑:如果你想摆脱索引，你可以重写__getitem__():

class Point(tuple):
    __slots__ = []
    def __new__(cls, x, y):
        return tuple.__new__(cls, (x, y))
    @property
    def x(self):
        return tuple.__getitem__(self, 0)
    @property
    def y(self):
        return tuple.__getitem__(self, 1)
    def __getitem__(self, item):
        raise TypeError

注意，不能使用operator。在这种情况下，属性的itemgetter，因为这将依赖于Point.__getitem__()而不是tuple.__getitem__()。此外，这不会阻止使用元组。__getitem__(p, 0)，但我很难想象这应该如何构成一个问题。

我不认为创建不可变对象的“正确”方法是编写C扩展。Python通常依赖于库实现者和库用户是成年人，而不是真正强制执行接口，接口应该在文档中清楚地说明。这就是为什么我不认为通过调用object.__setattr__()来规避被重写的__setattr__()是一个问题的可能性。如果有人这么做，风险自负。

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

你可以在init的最后一条语句中重写setAttr。那么你可以构建，但不能改变。显然，你仍然可以用usint对象重写。但在实践中，大多数语言都有某种形式的反射，因此不可变始终是一个有漏洞的抽象。不可变性更多的是防止客户端意外地违反对象的契约。我使用:

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

最初提供的解决方案是不正确的，这是基于使用这里的解决方案的评论而更新的

原来的解决方案是错误的，这是一种有趣的方式，所以它被包括在底部。

===============================

class ImmutablePair(object):

    __initialised = False # a class level variable that should always stay false.
    def __init__(self, a, b):
        try :
            self.a = a
            self.b = b
        finally:
            self.__initialised = True #an instance level variable

    def __setattr__(self, key, value):
        if self.__initialised:
            self._raise_error()
        else :
            super(ImmutablePair, self).__setattr__(key, value)

    def _raise_error(self, *args, **kw):
        raise NotImplementedError("Attempted To Modify Immutable Object")

if __name__ == "__main__":

    immutable_object = ImmutablePair(1,2)

    print immutable_object.a
    print immutable_object.b

    try :
        immutable_object.a = 3
    except Exception as e:
        print e

    print immutable_object.a
    print immutable_object.b

输出:

1
2
Attempted To Modify Immutable Object
1
2

======================================

最初的实现:

评论中指出，这实际上是行不通的，因为它阻止了在重写类setattr方法时创建多个对象，这意味着不能作为self创建第二个对象。A =将在第二次初始化时失败。

class ImmutablePair(object):

    def __init__(self, a, b):
        self.a = a
        self.b = b
        ImmutablePair.__setattr__ = self._raise_error

    def _raise_error(self, *args, **kw):
        raise NotImplementedError("Attempted To Modify Immutable Object")