get filenames in a directory without extension - Python [duplicate]

Question

I am trying to get the filenames in a directory without getting the extension. With my current code -

import os

path = '/Users/vivek/Desktop/buffer/xmlTest/' 
files = os.listdir(path) 
print (files)

I end up with an output like so:

['img_8111.jpg', 'img_8120.jpg', 'img_8127.jpg', 'img_8128.jpg', 'img_8129.jpg', 'img_8130.jpg']

However I want the output to look more like so:

['img_8111', 'img_8120', 'img_8127', 'img_8128', 'img_8129', 'img_8130']

How can I make this happen?

Neil · Accepted Answer · 2017-11-15 23:58:06Z

10

You can use os's splitext.

import os

path = '/Users/vivek/Desktop/buffer/xmlTest/' 
files = [os.path.splitext(filename)[0] for filename in os.listdir(path)]
print (files)

Just a heads up: basename won't work for this. basename doesn't remove the extension.

answered Nov 15, 2017 at 23:58

Neil

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Comments

kkj · Accepted Answer · 2017-11-15 23:58:35Z

2

Here are two options

import os
print(os.path.splitext("path_to_file")[0])

Or

from os.path import basename
print(basename("/a/b/c.txt"))

answered Nov 15, 2017 at 23:58

kkj

425 bronze badges

basename("/a/b/c.txt") returns c.txt, WITH extension.