不同路径之间的多个公共前缀
抱歉标题，我的问题如下。我有一个路径列表，我想获得多个公共前缀。例如，假设我有：
['/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/pkg_name',
'/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/EGG-INFO',
'/usr/bin/pkg_name']
我想要有：
['/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/',
'/usr/bin/pkg_name']
因为前两个有一个共同的前缀，即目录。希望我说清楚了，
rubik
编辑：我的路径是 Python Egg 和一些可执行文件。我想删除整个 Egg，而不是其中的目录，例如 EGG-INFO 或 pkg_name。所以它必须是 /usr/.../dist-packages/pkg_name-0.1-py2.7.egg/。另一个路径，因为它是可执行文件，所以保持原样。
谢谢
Sorry for the title, my problem is as follows. I have a list of paths and I want to get multiple common prefixes. For example, given I have:
['/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/pkg_name',
'/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/EGG-INFO',
'/usr/bin/pkg_name']
I want to have:
['/usr/local/lib/python2.7/dist-packages/pkg_name-0.1-py2.7.egg/',
'/usr/bin/pkg_name']
Because the first two have a common prefix that is a directory. Hope I made this clear,
rubik
EDIT: The paths I have are Python eggs and some executables. I want to remove the entire egg, not the directories inside it, like EGG-INFO or pkg_name. So it has to be /usr/.../dist-packages/pkg_name-0.1-py2.7.egg/. The other path, since it is an executable remains as it is.
Thank you
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问题没有明确定义。在这种情况下，您想要什么：
是一个
/usr还是两个：/usr/bin/usr还是三个？无论哪种情况，算法都将如下所示：
/；这将是您的第一组The problem is not well-defined. What do you want to have in this case:
Shall it be one
/usror two:/usr/bin/usr, or three?In either case, the algorithm will be like this:
os.path.commonprefix()with 2nd, 3rd, ..., i-th until common prefix is not/; that will be your first group