Varun Patro
Varun Patro

Reputation: 2179

How do I pickle and unpickle without side effects?

Why does pickle reuse the existing Python class 'C' instead of reconstructing the class from the pickled bytes? Is there a way I can pickle and unpickle without side effects?

Here's my repl session:

In [1]: import pickle

In [2]: class C:
   ...:     pass
   ...:

In [3]: hasattr(C, 'foo')
Out[3]: False

In [4]: pickled = pickle.dumps(C)

In [5]: C.foo = 'bar'

In [6]: hasattr(C, 'foo')
Out[6]: True

In [7]: C_without_foo = pickle.loads(pickled)

In [8]: hasattr(C_without_foo, 'foo')
Out[8]: True

In [9]: hasattr(C, 'foo')
Out[9]: True

Upvotes: 3

Views: 119

Answers (2)

Prayson W. Daniel
Prayson W. Daniel

Reputation: 15568

That is because they share the same identity, namely, they point to the same memory location. We have the same behaviour with list

a = [1, 2, 3]
b = a
print(id(a) == id(b)) # True

b.append(4)
print(a) # [1, 2, 3, 4]

c = a.copy()
a.append(5)
print(c)  # [1, 2, 3, 4]
print(id(c) == id(a)) # False

Your issue:

In [1]: import pickle

In [2]: class C:
   ...:     pass
   ...:

In [3]: hasattr(C,'foo')
Out[3]: False

In [4]: pickled = pickle.dumps(C)

In [5]: id(C)
Out[5]: 2887019170312

In [6]: C.foo = 'bar'

In [7]: id(C)
Out[7]: 2887019170312

In [8]: C_ = pickle.loads(pickled)

In [9]: id(C_)
Out[9]: 2887019170312

Solution. Make a copy

In [3]: C_ = type('C_', C.__bases__, dict(C.__dict__))

In [4]: pickled = pickle.dumps(C_)

In [5]: hasattr(C, 'foo')
Out[5]: False

In [6]:  C.foo = 'bar'

In [7]: C_without_foo = pickle.loads(pickled)

In [8]: hasattr(C_without_foo, 'foo')
Out[8]: False

Upvotes: 0

Taw
Taw

Reputation: 429

I think the problem is that you are pickle-ing the class itself, as opposed to a specific object.

import pickle
class C:
    pass

myC = C() #pickle myC object, not the C class
print(hasattr(myC, 'foo'))

pickled = pickle.dumps(myC)

myC.foo = 'bar'
print(hasattr(myC, 'foo'))

C_without_foo = pickle.loads(pickled)
print(hasattr(C_without_foo, 'foo'))
print(hasattr(myC, 'foo'))

Upvotes: 1

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