Given a file path
/path/to/some/file.jpg How would I get
/path/to/some I'm currently doing
fullpath = '/path/to/some/file.jpg' filepath = '/'.join(fullpath.split('/')[:-1]) But I think it is open to errors
4 Answers
With os.path.split:
dirname, fname = os.path.split(fullpath) Per the docs:
Split the pathname path into a pair,
(head, tail)where tail is the last pathname component and head is everything leading up to that. The tail part will never contain a slash; if path ends in a slash, tail will be empty. If there is no slash in path, head will be empty.
os.path is always the module suitable for the platform that the code is running on.
Comments
Please try this
fullpath = '/path/to/some/file.jpg' import os os.path.dirname(fullpath) 
Comments
Using pathlib you can get the path without the file name using the .parent attribute:
from pathlib import Path fullpath = Path("/path/to/some/file.jpg") filepath = str(fullpath.parent) # /path/to/some/ This handles both UNIX and Windows paths correctly.
Comments
With String rfind method.
fullpath = '/path/to/some/file.jpg' index = fullpath.rfind('/') fullpath[0:index] 4 Comments
Selcuk
What about this:
'/path/to/some//file.jpg' or 'C:\path\to\some\file.jpg'?
St1id3r
Based on the users given and then criteria, this should be good. But I agree this would not work in case of different OS.
Selcuk
It doesn't work correctly with my first example either. We should not suggest fail-prone methods when there is a documented, platform agnostic built-in method of doing something, such as the
os.path.split method suggested by @LevLevitsky. The OP has suggested a similar method themselves but they are looking for a better method because they think it is open to errors.St1id3r
Okay, makes sense.
/if you are planning to run this on different operating systems.pathlibmodule.