Pickle 链接对象

Question

我想腌制一个对象和引用第一个对象的第二个对象。当我天真地腌制/解开这两个对象时，引用就变成了副本。如何保留foo 和bar.foo_ref 两个对象之间的链接？

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)

Answer 1

我之前的回答没有抓住你的意思。您的代码的问题在于您没有使用 Pickler 和 Unpickler 对象。这是一个带有多个转储调用的工作版本：

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

Answer 2

如果将它们腌制在一起，pickle 模块会跟踪引用并且只腌制一次foo 变量。你能像这样把foo 和bar 一起腌制吗？

import pickle

class Foo(object):
    pass

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

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

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

Answer 3

好吧，你能做到吗：

bar2 = pickle.load(f)
foo2 = bar2.foo_ref

让pickle为你处理链接。