是否可以装饰其实例将被腌制的 Python 类？

Question

以下代码运行正确：

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)

但是添加一个装饰器来获得一个 multiton 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

怎么了？

Answer 1

Pickle 必须能够直接加载类。您的装饰器替换类与工厂函数，使泡菜无法导入类本身。

使用单独的工厂函数，而不是装饰器，返回一个“私有”类（但仍可直接导入）：

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

为此，我会使用dill，它可以在 python 中序列化几乎任何东西。

>>> 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>

Dill 还拥有some good tools，可帮助您了解在代码失败时导致酸洗失败的原因。