我不知道__setstate__
和__getstate__
方法的作用,所以请帮我一个简单的例子。
答案 0 :(得分:52)
这是Python 2的一个非常简单的示例,它应该补充pickle docs。
class Foo(object):
def __init__(self, val=2):
self.val = val
def __getstate__(self):
print "I'm being pickled"
self.val *= 2
return self.__dict__
def __setstate__(self, d):
print "I'm being unpickled with these values:", d
self.__dict__ = d
self.val *= 3
import pickle
f = Foo()
f_string = pickle.dumps(f)
f_new = pickle.loads(f_string)
答案 1 :(得分:22)
最小例子
无论getstate
出现什么,都会进入setstate
。它不需要是一个字典。
getstate
发出的任何内容都必须是可视的,例如由基本的内置插件组成,例如int
,str
,list
。
class C(object):
def __init__(self, i):
self.i = i
def __getstate__(self):
return self.i
def __setstate__(self, i):
self.i = i
assert pickle.loads(pickle.dumps(C(1), -1)).i == 1
默认__setstate__
默认__setstate__
需要dict
。
self.__dict__
是https://stackoverflow.com/a/1939384/895245中的一个不错的选择,但我们可以自己构建一个以更好地了解正在发生的事情:
class C(object):
def __init__(self, i):
self.i = i
def __getstate__(self):
return {'i': self.i}
assert pickle.loads(pickle.dumps(C(1), -1)).i == 1
默认__getstate__
类似于__setstate__
。
class C(object):
def __init__(self, i):
self.i = i
def __setstate__(self, d):
self.i = d['i']
assert pickle.loads(pickle.dumps(C(1), -1)).i == 1
__slots__
个对象没有__dict__
如果对象有__slots__
,则它没有__dict__
如果您要同时实施get
和setstate
,默认方式是:
class C(object):
__slots__ = 'i'
def __init__(self, i):
self.i = i
def __getsate__(self):
return { slot: getattr(self, slot) for slot in self.__slots__ }
def __setsate__(self, d):
for slot in d:
setattr(self, slot, d[slot])
assert pickle.loads(pickle.dumps(C(1), -1)).i == 1
__slots__
默认get和set需要一个元组
如果您想重复使用默认的__getstate__
或__setstate__
,则必须将元组传递为:
class C(object):
__slots__ = 'i'
def __init__(self, i):
self.i = i
def __getsate__(self):
return (None, { slot: getattr(self, slot) for slot in self.__slots__ })
assert pickle.loads(pickle.dumps(C(1), -1)).i == 1
我不确定这是为了什么。
<强>继承强>
首先看到酸洗工作默认情况下:
class C(object):
def __init__(self, i):
self.i = i
class D(C):
def __init__(self, i, j):
super(D, self).__init__(i)
self.j = j
d = pickle.loads(pickle.dumps(D(1, 2), -1))
assert d.i == 1
assert d.j == 2
继承自定义__getstate__
没有__slots__
这很容易,因为__dict__
的{{1}}包含D
的{{1}},所以我们不需要触及{{} 1}}:
__dict__
继承和C
使用C
,我们需要转发到基类,并且可以传递元组:
class C(object):
def __init__(self, i):
self.i = i
class D(C):
def __init__(self, i, j):
super(D, self).__init__(i)
self.j = j
def __getstate__(self):
return self.__dict__
def __setstate__(self, d):
self.__dict__ = d
d = pickle.loads(pickle.dumps(D(1, 2), -1))
assert d.i == 1
assert d.j == 2
不幸的是,无法重用基数的默认__slots__
和__slots__
:https://groups.google.com/forum/#!topic/python-ideas/QkvOwa1-pHQ我们被迫定义它们。
在Python 2.7.12上测试过。 GitHub upstream
答案 2 :(得分:4)
这些方法用于控制pickle模块对对象进行pickle和unpickled的方式。这通常是自动处理的,因此除非您需要覆盖类的pickle或unpickled方式,否则您不必担心它。
答案 3 :(得分:2)
对@BrainCore 的回答的澄清。在实践中,您可能不想修改 self
内的 __getstate__
。而是构造一个将被腌制的新对象,保持原始对象不变以供进一步使用。效果如下:
import pickle
class Foo:
def __init__(self, x:int=2, y:int=3):
self.x = x
self.y = y
self.z = x*y
def __getstate__(self):
# Create a copy of __dict__ to modify values and return;
# you could also construct a new dict (or other object) manually
out = self.__dict__.copy()
out["x"] *= 3
out["y"] *= 10
# You can remove attributes, but note they will not get set with
# some default value in __setstate__ automatically; you would need
# to write a custom __setstate__ method yourself; this might be
# useful if you have unpicklable objects that need removing, or perhaps
# an external resource that can be reloaded in __setstate__ instead of
# pickling inside the stream
del out["z"]
return out
# The default __setstate__ will update Foo's __dict__;
# so no need for a custom implementation here if __getstate__ returns a dict;
# Be aware that __init__ is not called by default; Foo.__new__ gets called,
# and the empty object is modified by __setstate__
f = Foo()
f_str = pickle.dumps(f)
f2 = pickle.loads(f_str)
print("Pre-pickle:", f.x, f.y, hasattr(f,"z"))
print("Post-pickle:", f2.x, f2.y, hasattr(f2,"z"))