Question

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

Answer 1

The problem is not well-defined. What do you want to have in this case:

/usr/bin/a
/usr/bin/b
/usr/etc
/usr/local

Shall it be one /usr or two: /usr/bin /usr, or three?

In either case, the algorithm will be like this:

1. sort the list
2. take the first element and do os.path.commonprefix() with 2nd, 3rd, …, i-th until common prefix is not /; that will be your first group
3. repeat step 2, starting from (i+1)th