是否可以修饰其实例将被酸洗的Python类？

Question

下面的代码运行正常：

import pickle 

class MyClass(): 
    def __init__(self, arg): 
     self.arg = arg 

a = MyClass('my arg') 
with open('/home/mahikeulbody/mypickle', 'wb') as file: 
    pickle.dump(a, file) 

，但增加了装饰以获得多例类：

import pickle 

def multiton(cls): 
    instances = {} 
    def getinstance(arg): 
     if arg not in instances: 
      instances[arg] = cls(arg) 
     return instances[arg] 
    return getinstance 

@multiton 
class MyClass(): 
    def __init__(self, arg): 
     self.arg = arg 

a = MyClass('my arg') 
with open('/home/michel/mypickle', 'wb') as file: 
    pickle.dump(a, file) 

产生以下错误：

pickle.dump(a, file) 
_pickle.PicklingError: Can't pickle <class '__main__.MyClass'>: it's not the same object as __main__.MyClass 

有什么不对？

Answer 1

Pickle必须能够直接加载类。你的装饰器用工厂函数替换了这个类，这使得pickle不可能导入类本身。

使用一个单独的工厂函数，而不是装饰，返回一个“私人”类（但仍然可导入直接）：

class _MyClass(): 
    def __init__(self, arg): 
     self.arg = arg 

def MyClass(arg, instances={}): 
    if arg not in instances: 
     instances[arg] = _MyClass(arg) 
    return instances[arg] 

Answer 2

要做到这一点，我会用dill，可序列化几乎所有的东西蟒蛇。

>>> def multiton(cls): 
...  instances = {} 
...  def getinstance(arg): 
...   if arg not in instances: 
...    instances[arg] = cls(arg) 
...   return instances[arg] 
...  return getinstance 
... 
>>> @multiton 
... class MyClass(): 
...  def __init__(self, arg): 
...   self.arg = arg 
... 
>>> import dill 
>>>      
>>> a = MyClass('my arg') 
>>> b = dill.loads(dill.dumps(a)) 
>>> a 
<__main__.MyClass instance at 0x4d64558> 
>>> b 
<__main__.MyClass instance at 0x4d64800> 

萝也有some good tools帮助您了解是什么原因导致你的酸洗当你的代码没有失败。