不能用kwargs挑选新款式吗？

Question

在我的类实例化期间,我初始化一些不可选择的字段.因此,为了能够(un)正确地挑选我的类,我希望我的init方法可以在unpickling上调用.这似乎是它与旧式课程一起使用的方式.

对于新的样式类,我需要使用__new__和__getnewargs__.这是我做的:

import cPickle


class Blubb(object):

  def __init__(self, value):
    self.value = value


class Bla(Blubb):

  def __new__(cls, value):
    instance = super(Bla, cls).__new__(cls)
    instance.__init__(value)
    return instance

  def __getnewargs__(self):
    return self.value,

  def __getstate__(self):
    return {}

  def __setstate__(self, dct):
    pass

x = Bla(2)
print x.value
pickled = cPickle.dumps(x, 2)
x_ = cPickle.loads(pickled)
assert x_.value == 2

如果不是这样的话,那就没问题了obj = C.__new__(C, *args).现在有**kwargs.所以我在我__new__和__init__方法中仅限于非关键字参数.

有谁知道解决这个问题的方法？这真的很不方便.

Answer 1

pickle协议2想要cls.__new__(cls, *args)默认调用,但有一种方法可以解决这个问题.如果您使用,__reduce__您可以返回一个将您的参数映射到的函数__new__.我能够修改你的例子来开始**kwargs工作:

import cPickle

class Blubb(object):

    def __init__(self, value, foo=None, bar=None):
        self.value = value
        self.foo = foo
        self.bar = bar

def _new_Bla(cls, value, kw):
    "A function to map kwargs into cls.__new__"
    return cls.__new__(cls, value, **kw)

class Bla(Blubb):

    def __new__(cls, value, **kw):
        instance = super(Bla, cls).__new__(cls)
        instance.__init__(value, **kw)
        return instance

    def __reduce__(self):
        kwargs = {'foo': self.foo, 'bar': self.bar}
        return _new_Bla, (self.__class__, self.value, kwargs), None

x = Bla(2, bar=[1, 2, 3])
pickled = cPickle.dumps(x, 2)
y = cPickle.loads(pickled)
assert y.value == 2
assert y.bar == [1, 2, 3]