python - 如何 pickle 继承自 A 的类 B(具有许多变量)的对象，该对象定义了 __setstate__ 和 __getstate__

Question

我的问题是:

class A(object):
    def __init__(self):
        #init
    def __setstate__(self,state):
        #A __setstate__ code here            
    def __getstate__(self):
        #A __getstate__ code here
        return state

class B(A):
    def __init__(self):
        #creates many object variables here

A 来自外部图书馆。

硬解

我想避免这种情况

当 pickle B 时，pickle 当然使用类 A 的 __setstate__, __getstate__ 方法，所以为了 pickle 工作我应该做这样的事情:

class B(A):
    def __init__(self):
        #creates many object variables here

    def __setstate__(self,state)
        A.__setstate__(self,state)
        #B __setstate__ code here
        #getting various variables from state for example
        self._a0 = state['a0']
        self._a1 = state['a1']
        #...
        self._a100 = state['a100']
        self._a101 = state['a101']

    def __getstate__(self):
        state = A.__getstate__(self)
        #B __setstate__ code here
        #filling state with various variables  
        #getting various variables from state for example
        state['a0'] =  self._a0
        state['a1'] =  self._a1
        #...
        state['a100'] =  self._a100
        state['a101'] =  self._a101           
        return state

我的问题是:

如何避免在 B 中定义 __setstate__ 和 __getstate__ 以便 pickle 自己完成 pickle 变量的工作？B 中的所有变量都是 pickle 可以自行 pickle(handle) 的类型。因此，如果 B 不继承自 A，则可能会产生良好的结果:

b = B()
path = 'path.temp'
fout = open(path,'w')
pickler = pickl.Pickler(fout)

pickler.dump(b)
fout.close()

fin = open(path,'r')
upickler = pickl.Unpickler(fin)
b = unpickler.load()
fin.close()
#b has all variables

显而易见的解决方案

class B(object):
    def __init__(self):
        #creates many object variables here
        a = A()            

但是我希望 B 继承自 A。知道如何解决这个问题，或者至少在 B 中自动化 pickling/unpickling 变量吗？

解决方法:

至于硬解中的自动化 pickle

向 B 添加一个包含变量的字典以 pickle:

class B(A):
    __picklableObjects__ = {'_a0', '_a1', ... ,'_a101'}

    def __init__(self):
        #creates many object variables here
        A.__init__(self)
        self._a0 = ...
        ...
        self._a101 = ...

    @staticmethod
    def getPicklableObjects():
        return B.__picklableObjects__

    def __setstate__(self,state):
        A.__setstate__(self,state)
        for po in B.getPicklableObjects():
           __dict__[po] = state[po]

    def __getstate__(self):
        state = A.__getstate__(self)
        for po in B.getPicklableObjects():
            state[po] = copy.deepcopy(__dict__[po])
        return state

还有其他想法吗？

A的图书馆:

好的，对于任何感兴趣的人来说，A 是 graph_tool.Graph: A src code

line 786: class Graph(object)

...

line 1517: __getstate__

...

line 1533: __setstate__

Answer 1

根据文档，当未定义 __getstate__ 时，实例的 __dict__ 会被 pickle，因此您可以使用它来定义自己的状态方法作为组合A 方法和实例的 __dict__:

import pickle

class A(object):
    def __init__(self):
        self.a = 'A state'

    def __getstate__(self):
        return {'a': self.a}

    def __setstate__(self, state):
        self.a = state['a']

class B(A):
    def __init__(self):
        A.__init__(self)
        self.b = 'B state'

    def __getstate__(self):
        a_state = A.__getstate__(self)
        b_state = self.__dict__
        return (a_state, b_state)

    def __setstate__(self, state):
        a_state, b_state = state
        self.__dict__ = b_state
        A.__setstate__(self, a_state)

b = pickle.loads(pickle.dumps(B()))
print b.a
print b.b