我正在使用borg模式在对象之间共享状态:
class Borg:
__shared_state = {}
def __init__(self):
self.__dict__ = self.__shared_state
现在让我们假设我想根据创建对象时传递的参数创建Borg类的基于上下文的对象。 这是为多个上下文创建Borg模式(状态共享)的正确方法吗?
import random
import cPickle
class Borg:
__shared_state = { }
def __init__(self,*args,**kwargs):
context_key = hash('{0}{1}'.format(cPickle.dumps(args),cPickle.dumps(kwargs)))
self.__shared_state.setdefault(context_key, {})
self.__dict__ = self.__shared_state[context_key]
print(self.__shared_state)
def set_random_property(self):
self.num = str(random.randint(1,100000))
a = Borg(x='ONE')
a.set_random_property()
b = Borg(x = 'TWO')
b.set_random_property()
c = Borg(x = 'TWO')
print('a with ONE has num:{0}'.format(a.num))
print('b with TWO has num:{0}'.format(b.num))
print('c with TWO has num:{0}'.format(c.num))
输出
{7373348246660160089: {}}
{7373348246660160089: {'num': '18322'}, 3334843421982509183: {}}
{7373348246660160089: {'num': '18322'}, 3334843421982509183: {'num': '33084'}}
a with ONE has num:18322
b with TWO has num:33084
c with TWO has num:33084
正常工作。有没有办法即兴发挥这种模式?或者为python 2.7提供更好的替代方案?
答案 0 :(得分:2)
不,你用的是我用的东西;使用字典表示共享状态。
您可以使用dict.setdefault()
的返回值而不是忽略它来简化稍微:
def __init__(self, *args, **kwargs):
context_key = hash('{0}{1}'.format(cPickle.dumps(args),cPickle.dumps(kwargs)))
self.__dict__ = self.__shared_state.setdefault(context_key, {})
所有这些都可以封装在元数据类型中:
class PerArgsBorgMeta(type):
def __new__(mcls, name, bases, attrs):
cls = super(PerArgsBorgMeta, mcls).__new__(mcls, name, bases, attrs)
setattr(cls, '_{}__shared_state'.format(name), {})
return cls
def __call__(cls, *args, **kwargs):
instance = super(PerArgsBorgMeta, cls).__call__(*args, **kwargs)
context_key = hash('{0}{1}'.format(cPickle.dumps(args),cPickle.dumps(kwargs)))
state = getattr(cls, '_{}__shared_state'.format(cls.__name__))
instance.__dict__ = state.setdefault(context_key, {})
return instance
然后将其用作类的__metaclass__
属性:
class SomeBorgClass:
__metaclass__ = PerArgsBorgMeta
请注意,使用hash(cPickle.dumps(kwargs))
仍会为带有冲突的词典创建不同的哈希值:
>>> import cPickle
>>> hash(cPickle.dumps({'a': 42, 'i': 81}))
-7392919546006502834
>>> hash(cPickle.dumps({'i': 81, 'a': 42}))
2932616521978949826
同样适用于集。排序(递归,如果你必须是详尽的)可以在这里提供帮助,但要小心你不会产生假阳性,例如,作为值传入的集合,以及使用相同值的元组。每种方法都有越来越复杂的解决方法,但在某些时候你只需要接受限制而不是使散列代码更复杂。