Pickle 自定义类和 unpickle 而无需访问这些类

Question

是否可以将对象的类定义包含到 pickled 文件中，以便在不向这些类提供源代码的情况下对对象进行 unpickled？或者也许可以使用另一种序列化方法/标准......

酸洗：

>>> class A(object):
...   def __init__(self, foo):
...     self.foo = foo
...
>>>
>>> a = A('abc')
>>> a
<__main__.A object at 0x7f78ce7f3550>
>>> a.foo
'abc'
>>> import pickle
>>> with open('test.pickle', 'wb') as handle:
...     pickle.dump(a, handle, protocol=pickle.HIGHEST_PROTOCOL)
...
>>>

Unpickle 无法访问 A 类：

>>> import pickle
>>>
>>>
>>> with open('test.pickle', 'rb') as handle:
...     b = pickle.load(handle)
...
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
AttributeError: Can't get attribute 'A' on <module '__main__' (built-in)>

我正在尝试为一些测试用例创建测试数据，并且想知道这些测试用例是否可以与生成这些数据的代码分开执行。

Answer 1

定义类：

import pickle

class A(object):
    def __init__(self, foo):
        self.foo = foo

创建对象：

a = A('abc')

将对象序列化为字节串：

serialized = pickle.dumps(a, protocol=pickle.HIGHEST_PROTOCOL)

将字节串写入文件：

with open('test.pickle', 'wb') as handle:
    handle.write(serialized)

从文件中读取字节串并反序列化：

with open('test.pickle', 'rb') as handle:
    serialized = handle.read()

b = pickle.loads(serialized)
print(b.foo)

输出：abc