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。
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
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")
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
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
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。