我有一个旧的 pickle 文件,当我尝试加载它时出现错误
ImportError: No module named oldname.submodule
这是因为oldname
早已被重命名为newname
。
如何加载这个 pickle 文件?
我可以创建一个从 oldname
到 newname
的符号链接(symbolic link),但我想知道是否有办法使模块 oldname.*
引用到 newname.*
而不触及文件系统?
最佳答案
在 sys.modules
中添加条目,而不是创建第二个并行模块和外部依赖项的符号链接(symbolic link)。为了测试,我 pickle 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>
注意买者:如果类本身发生了很大变化,pickle 无论如何都会失败。
关于python - 如何加载旧的 pickle 文件?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41554738/