腌制链接的对象 [英] Pickle linked objects

问题描述

我想腌制一个对象和引用第一个的第二个对象.当我天真地腌制/解开两个对象时，引用将成为副本.如何保留两个对象foo和bar.foo_ref之间的链接?

I want to pickle an object and a second object that references the first. When I naively pickle/unpickle the two objects, the reference becomes a copy. How do I preserve the link between the two objects foo and bar.foo_ref?

import pickle

class Foo(object):
    pass

foo = Foo()
bar = Foo()
bar.foo_ref = foo

with open('tmp.pkl', 'wb') as f:
    pickle.dump(foo, f)
    pickle.dump(bar, f)
with open('tmp.pkl', 'rb') as f:
    foo2 = pickle.load(f)
    bar2 = pickle.load(f)

print id(foo) == id(bar.foo_ref) # True
print id(foo2) == id(bar2.foo_ref) # False
# want id(foo2) == id(bar2.foo_ref)

推荐答案

我以前的答案没有说明您的意思.代码的问题是您没有使用Pickler和Unpickler对象.这是具有多个转储调用的工作版本:

My previous answer was missing your point. The problem with your code is that you're not using the Pickler and Unpickler objects. Here's a working version with multiple dump calls:

import pickle

class Foo(object):
    pass

foo = Foo()
bar = Foo()
bar.foo_ref = foo

f = open('tmp.pkl', 'wb')
p = pickle.Pickler(f)
p.dump(foo)
p.dump(bar)
f.close()

f = open('tmp.pkl', 'rb')
up = pickle.Unpickler(f)
foo2 = up.load()
bar2 = up.load()

print id(foo) == id(bar.foo_ref) # True
print id(foo2) == id(bar2.foo_ref) # True

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