[jhar131]

.. there can be other backslashes in the pattern. For example, from this path:

./usr/lib64/mpich2/lib/libmpichcxx.so.1

I just want to grab:

./usr/lib64/mpich2/lib

I'm sure this can be done, but my regex powers are not strong.

[Martyr2]

Well the first pattern you can try is... \..*\/. What this is saying is find the part of the text which starts with a period and has any number of characters but ends with a forward slash. This will match all the way to the last forward slash because regex is naturally "Greedy". It will always want the biggest match and pass up the smaller matches.

The result will typically contain the last forward slash, so if you need to chop that, you can simply remove the last character off the result match.

[jhar131]

Thanks for the response. + rep, and just for posterities sake, I had already written a working expression with that in it. To be specific, mine was:

.*\\/.*$

Although I guess the $ is redundant, as you show in your example.



ETA: I can't spell

This post has been edited by jhar131: 19 August 2011 - 03:45 PM

[Motoma]

No need for a regex, just use rsplit():

fullpath = './usr/lib64/mpich2/lib/libmpichcxx.so.1'

path, filename = fullpath.rsplit('/', 1)

To make it platform independent, use os.path.sep instead of '/'.