Question

我有一个旧的pickle文件，当我尝试加载它时，我收到错误

ImportError: No module named oldname.submodule

这是因为很久以前oldname已重命名为newname。

如何加载此pickle文件？

我可以创建从oldname到newname的符号链接，但我想知道是否有办法让模块oldname.*引用newname.*而不触及文件系统？

Answer 1

不是创建第二个并行模块和外部依赖项的符号链接，而是在sys.modules中添加一个条目。为了测试，我挑选了一个类实例，然后重命名了模块

td@mintyfresh ~/tmp $ cat oldname.py
class Foo(object):
    def __init__(self):
        self.name = 'bar'

td@mintyfresh ~/tmp $ python
Python 2.7.6 (default, Oct 26 2016, 20:30:19) 
[GCC 4.8.4] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import oldname
>>> foo = oldname.Foo()
>>> foo.name
'bar'
>>> import pickle
>>> pickle.dump(foo, open('test.pkl', 'wb'), 2)
>>> exit()
td@mintyfresh ~/tmp $ mv oldname.py newname.py
td@mintyfresh ~/tmp $ rm oldname.*

现在加载失败

td@mintyfresh ~/tmp $ python
Python 2.7.6 (default, Oct 26 2016, 20:30:19) 
[GCC 4.8.4] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import newname
>>> import pickle
>>> foo = pickle.load(open('test.pkl', 'rb'))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.7/pickle.py", line 1378, in load
    return Unpickler(file).load()
  File "/usr/lib/python2.7/pickle.py", line 858, in load
    dispatch[key](self)
  File "/usr/lib/python2.7/pickle.py", line 1090, in load_global
    klass = self.find_class(module, name)
  File "/usr/lib/python2.7/pickle.py", line 1124, in find_class
    __import__(module)
ImportError: No module named oldname

但是如果我在sys.modules中复制模块，它就可以了。有趣的是，类实例具有新的模块名称。

>>> import sys
>>> sys.modules['oldname'] = sys.modules['newname']
>>> foo = pickle.load(open('test.pkl', 'rb'))
>>> foo
<newname.Foo object at 0x7f6b837d3310>

警告：如果课程本身发生了变化，那么无论如何都会失败。

如何加载旧的pickle文件？

1 个答案: