不要使用符号链接来创建第二个并行模块和外部依赖项,而是在 sys.modules
中添加一个条目。为了测试,我对一个类实例进行了pickling,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中复制该模块,它就可以工作。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>
买方自负:如果类本身发生了很大的变化,腌菜(pickle)仍将失败。
sys.modules['oldname'] = sys.modules['newname']
。 - tdelaney