泡椒鏈接對象

Question

我想泡菜一個對象和第二個對象，引用第一個。當我天真地醃製/取下這兩個物體時，參考變成了一個副本。如何保留兩個對象foo和bar.foo_ref之間的鏈接？

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

我以前的回答錯過了你的觀點。你的代碼的問題是你沒有使用Pickler和Unpickler對象。下面是與多個轉儲調用一個工作版本：

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

嗯，你可以這樣做：

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

讓泡菜處理環節爲您服務。

Answer 3

如果你把它們醃在一起，醃泡模塊將跟蹤參考，只醃foo變量一次。你可以把foo和bar醃在一起，像這樣嗎？

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