虽然我从来都不需要这样做,但我突然意识到用Python创建一个不可变对象可能有点棘手。你不能只是覆盖__setattr__,因为这样你甚至不能在__init__中设置属性。子类化一个元组是一个有效的技巧:

class Immutable(tuple):
    
    def __new__(cls, a, b):
        return tuple.__new__(cls, (a, b))

    @property
    def a(self):
        return self[0]
        
    @property
    def b(self):
        return self[1]

    def __str__(self):
        return "<Immutable {0}, {1}>".format(self.a, self.b)
    
    def __setattr__(self, *ignored):
        raise NotImplementedError

    def __delattr__(self, *ignored):
        raise NotImplementedError

但是你可以通过self[0]和self[1]访问a和b变量,这很烦人。

这在Pure Python中可行吗?如果不是,我该如何用C扩展来做呢?

(只能在python3中工作的答案是可以接受的)。

更新:

从Python 3.7开始,要使用的方法是使用@dataclass装饰器,参见最新接受的答案。


当前回答

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/

其他回答

这里有一个优雅的解决方案:

class Immutable(object):
    def __setattr__(self, key, value):
        if not hasattr(self, key):
            super().__setattr__(key, value)
        else:
            raise RuntimeError("Can't modify immutable object's attribute: {}".format(key))

从这个类继承,在构造函数中初始化字段,就完成了所有设置。

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)

另一个想法是完全不允许__setattr__而使用object。构造函数中的__setattr__:

class Point(object):
    def __init__(self, x, y):
        object.__setattr__(self, "x", x)
        object.__setattr__(self, "y", y)
    def __setattr__(self, *args):
        raise TypeError
    def __delattr__(self, *args):
        raise TypeError

当然你可以用object。__setattr__(p, "x", 3)来修改一个Point实例p,但您的原始实现遭受同样的问题(尝试tuple。__setattr__(i, "x", 42)在一个不可变实例)。

您可以在原始实现中应用相同的技巧:去掉__getitem__(),并在属性函数中使用tuple.__getitem__()。

就像字典一样

我有一个开源库,在那里我以函数的方式做事情,所以在不可变对象中移动数据是有帮助的。但是,我不希望必须转换我的数据对象以便客户机与它们交互。所以,我想到了这个-它给你一个字典一样的对象,这是不可变的+一些帮助方法。

这要归功于Sven Marnach对限制属性更新和删除的基本执行的回答。

import json 
# ^^ optional - If you don't care if it prints like a dict
# then rip this and __str__ and __repr__ out

class Immutable(object):

    def __init__(self, **kwargs):
        """Sets all values once given
        whatever is passed in kwargs
        """
        for k,v in kwargs.items():
            object.__setattr__(self, k, v)

    def __setattr__(self, *args):
        """Disables setting attributes via
        item.prop = val or item['prop'] = val
        """
        raise TypeError('Immutable objects cannot have properties set after init')

    def __delattr__(self, *args):
        """Disables deleting properties"""
        raise TypeError('Immutable objects cannot have properties deleted')

    def __getitem__(self, item):
        """Allows for dict like access of properties
        val = item['prop']
        """
        return self.__dict__[item]

    def __repr__(self):
        """Print to repl in a dict like fashion"""
        return self.pprint()

    def __str__(self):
        """Convert to a str in a dict like fashion"""
        return self.pprint()

    def __eq__(self, other):
        """Supports equality operator
        immutable({'a': 2}) == immutable({'a': 2})"""
        if other is None:
            return False
        return self.dict() == other.dict()

    def keys(self):
        """Paired with __getitem__ supports **unpacking
        new = { **item, **other }
        """
        return self.__dict__.keys()

    def get(self, *args, **kwargs):
        """Allows for dict like property access
        item.get('prop')
        """
        return self.__dict__.get(*args, **kwargs)

    def pprint(self):
        """Helper method used for printing that
        formats in a dict like way
        """
        return json.dumps(self,
            default=lambda o: o.__dict__,
            sort_keys=True,
            indent=4)

    def dict(self):
        """Helper method for getting the raw dict value
        of the immutable object"""
        return self.__dict__

辅助方法

def update(obj, **kwargs):
    """Returns a new instance of the given object with
    all key/val in kwargs set on it
    """
    return immutable({
        **obj,
        **kwargs
    })

def immutable(obj):
    return Immutable(**obj)

例子

obj = immutable({
    'alpha': 1,
    'beta': 2,
    'dalet': 4
})

obj.alpha # 1
obj['alpha'] # 1
obj.get('beta') # 2

del obj['alpha'] # TypeError
obj.alpha = 2 # TypeError

new_obj = update(obj, alpha=10)

new_obj is not obj # True
new_obj.get('alpha') == 10 # True

使用冻结的数据类

对于Python 3.7+,你可以使用带frozen=True选项的数据类,这是一种非常Python化和可维护的方式来做你想做的事情。

它看起来是这样的:

from dataclasses import dataclass

@dataclass(frozen=True)
class Immutable:
    a: Any
    b: Any

由于数据类的字段需要类型提示,所以我使用了typing模块中的Any。

不使用命名元组的原因

在Python 3.7之前,经常可以看到命名元组被用作不可变对象。它在很多方面都很棘手,其中之一是命名元组之间的__eq__方法不考虑对象的类。例如:

from collections import namedtuple

ImmutableTuple = namedtuple("ImmutableTuple", ["a", "b"])
ImmutableTuple2 = namedtuple("ImmutableTuple2", ["a", "c"])

obj1 = ImmutableTuple(a=1, b=2)
obj2 = ImmutableTuple2(a=1, c=2)

obj1 == obj2  # will be True

如你所见,即使obj1和obj2的类型不同,即使它们的字段名称不同,obj1 == obj2仍然给出True。这是因为使用的__eq__方法是元组的方法,它只比较给定位置的字段的值。这可能是一个巨大的错误来源,特别是如果您是子类化这些类。