我通过重写__setattr__创建了不可变类，并且如果调用者是__init__，则允许该集合:

import inspect
class Immutable(object):
    def __setattr__(self, name, value):
        if inspect.stack()[2][3] != "__init__":
            raise Exception("Can't mutate an Immutable: self.%s = %r" % (name, value))
        object.__setattr__(self, name, value)

这还不够，因为它允许任何人的___init__来改变对象，但你懂的。

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。

．．如何在C中“正确地”做这件事?

你可以使用Cython为Python创建一个扩展类型:

cdef class Immutable:
    cdef readonly object a, b
    cdef object __weakref__ # enable weak referencing support

    def __init__(self, a, b):
        self.a, self.b = a, b

它既适用于Python 2。X和3。

测试

# compile on-the-fly
import pyximport; pyximport.install() # $ pip install cython
from immutable import Immutable

o = Immutable(1, 2)
assert o.a == 1, str(o.a)
assert o.b == 2

try: o.a = 3
except AttributeError:
    pass
else:
    assert 0, 'attribute must be readonly'

try: o[1]
except TypeError:
    pass
else:
    assert 0, 'indexing must not be supported'

try: o.c = 1
except AttributeError:
    pass
else:
    assert 0, 'no new attributes are allowed'

o = Immutable('a', [])
assert o.a == 'a'
assert o.b == []

o.b.append(3) # attribute may contain mutable object
assert o.b == [3]

try: o.c
except AttributeError:
    pass
else:
    assert 0, 'no c attribute'

o = Immutable(b=3,a=1)
assert o.a == 1 and o.b == 3

try: del o.b
except AttributeError:
    pass
else:
    assert 0, "can't delete attribute"

d = dict(b=3, a=1)
o = Immutable(**d)
assert o.a == d['a'] and o.b == d['b']

o = Immutable(1,b=3)
assert o.a == 1 and o.b == 3

try: object.__setattr__(o, 'a', 1)
except AttributeError:
    pass
else:
    assert 0, 'attributes are readonly'

try: object.__setattr__(o, 'c', 1)
except AttributeError:
    pass
else:
    assert 0, 'no new attributes'

try: Immutable(1,c=3)
except TypeError:
    pass
else:
    assert 0, 'accept only a,b keywords'

for kwd in [dict(a=1), dict(b=2)]:
    try: Immutable(**kwd)
    except TypeError:
        pass
    else:
        assert 0, 'Immutable requires exactly 2 arguments'

如果你不介意索引支持，那么@Sven Marnach建议的collections.namedtuple是更可取的:

Immutable = collections.namedtuple("Immutable", "a b")

从Python 3.7开始，你可以在你的类中使用@dataclass装饰器，它将像结构体一样是不可变的!不过，它可能会也可能不会将__hash__()方法添加到类中。引用:

hash() is used by built-in hash(), and when objects are added to hashed collections such as dictionaries and sets. Having a hash() implies that instances of the class are immutable. Mutability is a complicated property that depends on the programmer’s intent, the existence and behavior of eq(), and the values of the eq and frozen flags in the dataclass() decorator. By default, dataclass() will not implicitly add a hash() method unless it is safe to do so. Neither will it add or change an existing explicitly defined hash() method. Setting the class attribute hash = None has a specific meaning to Python, as described in the hash() documentation. If hash() is not explicit defined, or if it is set to None, then dataclass() may add an implicit hash() method. Although not recommended, you can force dataclass() to create a hash() method with unsafe_hash=True. This might be the case if your class is logically immutable but can nonetheless be mutated. This is a specialized use case and should be considered carefully.

下面是上面链接的文档中的例子:

@dataclass
class InventoryItem:
    '''Class for keeping track of an item in inventory.'''
    name: str
    unit_price: float
    quantity_on_hand: int = 0

    def total_cost(self) -> float:
        return self.unit_price * self.quantity_on_hand

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()

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