How to get a list of filenames(without extension) in directory in python?

Question

Assume my folder structure to be

+Data
  -abc.jpg
  -db.jpg
  -ap.jpg

Input is 'path/to/Data'

Expected output is ['abc','db','ap']

I saw many similar questions but did not get what exactly I wanted. I prefer to use os module in python.

Possible duplicate of [How do I list all files of a directory?](https://stackoverflow.com/questions/3207219/how-do-i-list-all-files-of-a-directory) — Christian König, Jul 28 '17 at 06:59
For extracting the file name have a look here: [Extracting extension from filename in Python](https://stackoverflow.com/questions/541390/extracting-extension-from-filename-in-python) — Christian König, Jul 28 '17 at 07:01

Mohamed Thasin ah · Answer 1 · 2018-06-28T08:10:03.287

14

simply try this,

l=os.listdir('path')
li=[x.split('.')[0] for x in l]

First list your directory files
split file by. and take first argument.

edited Jun 28 '18 at 08:10

answered Jul 28 '17 at 07:13

Mohamed Thasin ah

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What happens if the name of the file itself contains a period? (macOS does allow this.) Is there a way to make sure the file is split using the last period? – Peter Schorn Dec 29 '19 at 06:18
Yeah, you can, remove last word after split then perform merge of rest of the values. – Mohamed Thasin ah Dec 29 '19 at 07:38

Keatinge · Accepted Answer · 2017-07-28T07:06:44.053

10

import os    
files_no_ext = [".".join(f.split(".")[:-1]) for f in os.listdir() if os.path.isfile(f)]
print(files_no_ext)

edited Jul 28 '17 at 07:06

answered Jul 28 '17 at 07:02

Keatinge

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[`os.listdir`](https://docs.python.org/3.6/library/os.html#os.listdir) does not make a difference between directories and files, so your list will have (mostly empty) entries for subdirectories. – Christian König Jul 28 '17 at 07:04
3

You could furthermore exchange your `join/split` by [`os.path.splitext(f)[0]`](https://docs.python.org/3.6/library/os.path.html#os.path.splitext), which takes care of some special cases (leading periods on the basename). – Christian König Jul 28 '17 at 07:58

score 3 · Answer 3 · edited Mar 06 '23 at 06:08

3

import glob
from pathlib import Path

for f in glob.glob("*.*"):
    print(Path(f).stem)

edited Mar 06 '23 at 06:08

Sreeragh A R

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answered Nov 07 '19 at 14:30

LetzerWille

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2

Thanks for the answer, just a top-up on the answer, it is supposed `import pathlib`, then `pathlib.Path(f).stem`. – Yeo Keat Jan 17 '23 at 05:57

score 2 · Answer 4 · answered Jul 28 '17 at 07:21

2

import os
filenames=next(os.walk(os.getcwd()))[2]
efn=[f.split('.')[0] for f in filenames]

os.getcwd()   #for get current directory path

answered Jul 28 '17 at 07:21

Ahmad

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Darkaird · Answer 5 · 2017-07-28T08:51:54.993

You can use os.listdir which take path as argument and return a list of files and directories in it.
>>> list_ = os.listdir("path/to/Data")
>>> list_
>>> ['abc.jpg', 'dn.jpg', 'ap.jpg']

With that list, you only have to do a comprehension list which split each element on '.' (dot), take all the elements except the last one an join them with '.' (dot) and check if the element is a file using os.path.file().

>>> list_ = ['.'.join(x.split('.')[:-1]) for x in os.listdir("path/to/Data") if os.path.isfile(os.path.join('path/to/Data', x))]
>>> list_
>>> ['abc', 'dn', 'ap']

grimur82 · Answer 6 · 2017-07-28T08:17:07.997

0

    import os
    from os import listdir
    from os.path import isfile, join
    def filesMinusExtension(path):
        # os.path.splitext(f)[0] map with filename without extension with checking if file exists.
        files = [os.path.splitext(f)[0] for f in listdir(path) if isfile(join(path, f))];
        return files;

edited Jul 28 '17 at 08:17

answered Jul 28 '17 at 07:10

grimur82

149
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There are some problems with you code... But, what bother me is : semi-colons ? You are in Python guy, just throw them away ! – Darkaird Jul 28 '17 at 07:15
Thanks for the feedback. Yes the java background seems to affect my python typing of semicolons at times. Anyway can you share what problems could arise with this code ? – grimur82 Jul 28 '17 at 07:19
Except semi-colons, why importing isfile and join that way without doing it for os.path.splittext ? Furthermore, you use a comprehension list for filtering (the 'if' at the right) but not for mapping. You just have to do `[os.path.splitext(f)[0] for f in listdir(path) if isfile(join(path, f))]`. – Darkaird Jul 28 '17 at 07:23
Okay I've done the research and you are right. My modified solution removed redundant and unnecessary code. – grimur82 Jul 28 '17 at 08:20
You can even not use an intermediate variable and directly return the comprehension list if you want. – Darkaird Jul 28 '17 at 08:49

How to get a list of filenames(without extension) in directory in python?

6 Answers6