Question

为什么pickle会重用现有的Python类'C'而不是从pickled字节中重建该类？有没有办法可以让我腌制和腌制而没有副作用？

这是我的代表会话：

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

Answer 1

我认为问题在于您正在腌制类本身，而不是特定的对象。

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'))

Answer 2

这是因为它们共享相同的标识，即，它们指向相同的内存位置。我们的行为与列表相同

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

您的问题：

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

解决方案。制作副本

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

如何在不产生副作用的情况下腌制和腌制？

2 个答案: