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__)。

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)

我找到了一种方法，不用子类化tuple, namedtuple等。你所需要做的就是在初始化后禁用setattr和delattr(如果你想让一个集合成为不可变的，也要禁用setitem和delitem):

def __init__(self, *args, **kwargs):
    # something here

    self.lock()

其中lock可以是这样的:

@classmethod
def lock(cls):
    def raiser(*a):
        raise TypeError('this instance is immutable')

    cls.__setattr__ = raiser
    cls.__delattr__ = raiser
    if hasattr(cls, '__setitem__'):
        cls.__setitem__ = raiser
        cls.__delitem__ = raiser

你可以用这个方法创建类Immutable，并像我展示的那样使用它。

如果你不想在每个init中都写self.lock()，你可以用元类自动实现:

class ImmutableType(type):
    @classmethod
    def change_init(mcs, original_init_method):
        def __new_init__(self, *args, **kwargs):
            if callable(original_init_method):
                original_init_method(self, *args, **kwargs)

            cls = self.__class__

            def raiser(*a):
                raise TypeError('this instance is immutable')

            cls.__setattr__ = raiser
            cls.__delattr__ = raiser
            if hasattr(cls, '__setitem__'):
                cls.__setitem__ = raiser
                cls.__delitem__ = raiser

        return __new_init__

    def __new__(mcs, name, parents, kwargs):
        kwargs['__init__'] = mcs.change_init(kwargs.get('__init__'))
        return type.__new__(mcs, name, parents, kwargs)


class Immutable(metaclass=ImmutableType):
    pass

Test

class SomeImmutableClass(Immutable):
    def __init__(self, some_value: int):
        self.important_attr = some_value

    def some_method(self):
        return 2 * self.important_attr


ins = SomeImmutableClass(3)
print(ins.some_method())  # 6
ins.important_attr += 1  # TypeError
ins.another_attr = 2  # TypeError

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")

另一种方法是创建一个使实例不可变的包装器。

class Immutable(object):

    def __init__(self, wrapped):
        super(Immutable, self).__init__()
        object.__setattr__(self, '_wrapped', wrapped)

    def __getattribute__(self, item):
        return object.__getattribute__(self, '_wrapped').__getattribute__(item)

    def __setattr__(self, key, value):
        raise ImmutableError('Object {0} is immutable.'.format(self._wrapped))

    __delattr__ = __setattr__

    def __iter__(self):
        return object.__getattribute__(self, '_wrapped').__iter__()

    def next(self):
        return object.__getattribute__(self, '_wrapped').next()

    def __getitem__(self, item):
        return object.__getattribute__(self, '_wrapped').__getitem__(item)

immutable_instance = Immutable(my_instance)

这在只有一些实例必须是不可变的情况下很有用(比如函数调用的默认参数)。

也可以用于不可变工厂，如:

@classmethod
def immutable_factory(cls, *args, **kwargs):
    return Immutable(cls.__init__(*args, **kwargs))

也保护对象。__setattr__，但由于Python的动态特性，可能会被其他技巧所绊倒。

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/