How to get only the last part of a path in Python? How to get only the last part of a path in Python? python python

How to get only the last part of a path in Python?


Use os.path.normpath, then os.path.basename:

>>> os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))'folderD'

The first strips off any trailing slashes, the second gives you the last part of the path. Using only basename gives everything after the last slash, which in this case is ''.


With python 3 you can use the pathlib module (pathlib.PurePath for example):

>>> import pathlib>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/')>>> path.name'folderD'

If you want the last folder name where a file is located:

>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/file.py')>>> path.parent.name'folderD'


You could do

>>> import os>>> os.path.basename('/folderA/folderB/folderC/folderD')

UPDATE1: This approach works in case you give it /folderA/folderB/folderC/folderD/xx.py. This gives xx.py as the basename. Which is not what you want I guess. So you could do this -

>>> import os>>> path = "/folderA/folderB/folderC/folderD">>> if os.path.isdir(path):        dirname = os.path.basename(path)

UPDATE2: As lars pointed out, making changes so as to accomodate trailing '/'.

>>> from os.path import normpath, basename>>> basename(normpath('/folderA/folderB/folderC/folderD/'))'folderD'


matomo