Sort filenames in directory in ascending order

Question

I have a directory with jpgs and other files in it, the jpgs all have filenames with numbers in them. Some may have additional strings in the filename.

For example.

01.jpg

Or it could be

Picture 03.jpg

In Python I need a list of all the jpgs in ascending order. Here is the code snippet for this

import os
import numpy as np

myimages = [] #list of image filenames
dirFiles = os.listdir('.') #list of directory files
dirFiles.sort() #good initial sort but doesnt sort numerically very well
sorted(dirFiles) #sort numerically in ascending order

for files in dirFiles: #filter out all non jpgs
    if '.jpg' in files:
        myimages.append(files)
print len(myimages)
print myimages

What I get is this

['0.jpg', '1.jpg', '10.jpg', '11.jpg', '12.jpg', '13.jpg', '14.jpg',
 '15.jpg', '16.jpg', '17.jpg', '18.jpg', '19.jpg', '2.jpg', '20.jpg',
 '21.jpg', '22.jpg', '23.jpg', '24.jpg', '25.jpg', '26.jpg', '27.jpg',
 '28.jpg', '29.jpg', '3.jpg', '30.jpg', '31.jpg', '32.jpg', '33.jpg',
 '34.jpg', '35.jpg', '36.jpg', '37.jpg', '4.jpg', '5.jpg', '6.jpg',
 '7.jpg', '8.jpg', '9.jpg']

Clearly it sorts blindly the most significant number first. I tried using sorted() as you can see hoping that it would fix it but it makes no difference.

`fnames = sorted([fname for fname in os.listdir('.') if fname.endswith('.jpg')], key=lambda f: int(f.rsplit(os.path.extsep, 1)[0].rsplit(None,1)[-1]))` — inspectorG4dget, Oct 15 '15 at 21:53
[Sort a list containing alphanumeric strings](http://stackoverflow.com/questions/19366517/sorting-in-python-how-to-sort-a-list-containing-alphanumeric-values) — sam, Oct 15 '15 at 21:54
@inspectorG4dget, you could have also used generator comprehension. — Cristian Ciupitu, Oct 15 '15 at 22:07
@inspectorG4dget, indeed, but that's just an implementation detail. Other implementations might not listify the input. — Cristian Ciupitu, Oct 15 '15 at 22:11
@inspectorG4dget I'm also getting ValueError: invalid literal for int() with base 10 with yours — ChumbiChubaGo, Oct 15 '15 at 22:34
@inspectorG4dget's code is nice, btw in this specific example we don't need the `.rsplit(None,1)[-1]` - it's separating with whitespace preceding the number. — starriet, Aug 27 '22 at 13:09

score 122 · Accepted Answer · edited Aug 31 '23 at 12:38

122

Assuming there's just one number in each file name:

For Python 2X:

>>> dirFiles = ['Picture 03.jpg', '02.jpg', '1.jpg']
>>> dirFiles.sort(key=lambda f: int(filter(str.isdigit, f)))
>>> dirFiles
['1.jpg', '02.jpg', 'Picture 03.jpg']

For Python 3X:

>>> dirFiles = ['Picture 03.jpg', '02.jpg', '1.jpg']
>>> dirFiles.sort(key=lambda f: int(''.join(filter(str.isdigit, f))))
>>> dirFiles
['1.jpg', '02.jpg', 'Picture 03.jpg']

A version that also works in Python 3:

>>> import re
>>> dirFiles.sort(key=lambda f: int(re.sub('\D', '', f)))

edited Aug 31 '23 at 12:38

eduardosufan

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answered Oct 15 '15 at 22:39

Stefan Pochmann

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Works too! Very clean I like your solution the best. But it will fail if the list has an element with no numbers in it for some reason. But i just made sure to do the sort in myimages[] which are will have a number in the name for sure. – ChumbiChubaGo Oct 15 '15 at 23:14
@user3474042 Yeah, if all jpegs have numbers, then that works. Otherwise, maybe use `int(filter(str.isdigit, f) or -1))` (this puts digit-less filenames at the front). Btw, check out the `glob` module, which lets you directly ask for `*.jpg` files. – Stefan Pochmann Oct 15 '15 at 23:26
44

Lambda for python 3 gets uglier: `lambda f: int(''.join(filter(str.isdigit, f))))` (or else it'll complain that *int() argument must be a string, a bytes-like object or a number, not 'filter'*… I miss ruby's duck-typing…) – melissa_boiko May 06 '17 at 09:07
1

yeah , as @melboiko say, your answer is only for Python 2 – dorbodwolf Nov 24 '18 at 05:46
You would think python would do this automatically, especially with a set of filenames that only contains numbers. Good thinking using re's, works for me in python 3 – Gunner Stone Dec 17 '20 at 02:39
Seems Python 2 code and does not work in Python 3: `dirFiles.sort(key=lambda f: int(filter(str.isdigit, f)))`.. Ok just read the comment from @dorbodwolf, but still get an error with the **join..** – Timo Dec 24 '20 at 12:17
@melissa_boiko I get `None` with `entries = ['C:\\Users\\User\\pictures\\bd\\unbenannt1.txt', 'C:\\Users\\User\\pictures\\bd\\unbenannt22.PNG'].sort(key=lambda f: int(''.join(filter(str.isdigit, f))))` – Timo Dec 24 '20 at 14:35
2

@Timo Of course you get `None`. Like everybody else. So apply it properly, like I showed. Or use `sorted`, which does return a sorted list instead of sorting one in-place. – Stefan Pochmann Dec 24 '20 at 19:15

score 65 · Answer 2 · edited Sep 05 '18 at 06:35

there is a module natsort. Just do pip install natsort.

>>> import natsort 
>>> ll = ['Picture 13.jpg', 'Picture 14.jpg', 'Picture 15.jpg','Picture 0.jpg', 'Picture 1.jpg', 'Picture 10.jpg', 'Picture 11.jpg', 'Picture 12.jpg',  'Picture 16.jpg', 'Picture 17.jpg', 'Picture 18.jpg', 'Picture 19.jpg', 'Picture 2.jpg', 'Picture 20.jpg', 'Picture 21.jpg', 'Picture 22.jpg', 'Picture 23.jpg', 'Picture 24.jpg', 'Picture 25.jpg', 'Picture 26.jpg', 'Picture 27.jpg', 'Picture 28.jpg', 'Picture 29.jpg', 'Picture 3.jpg', 'Picture 30.jpg', 'Picture 31.jpg', 'Picture 32.jpg', 'Picture 33.jpg', 'Picture 34.jpg', 'Picture 35.jpg', 'Picture 36.jpg', 'Picture 37.jpg']         
>>> print(natsort.natsorted(ll,reverse=True))
['Picture 37.jpg', 'Picture 36.jpg', 'Picture 35.jpg', 'Picture 34.jpg', 'Picture 33.jpg', 'Picture 32.jpg', 'Picture 31.jpg', 'Picture 30.jpg', 'Picture 29.jpg', 'Picture 28.jpg', 'Picture 27.jpg', 'Picture 26.jpg', 'Picture 25.jpg', 'Picture 24.jpg', 'Picture 23.jpg', 'Picture 22.jpg', 'Picture 21.jpg', 'Picture 20.jpg', 'Picture 19.jpg', 'Picture 18.jpg', 'Picture 17.jpg', 'Picture 16.jpg', 'Picture 15.jpg', 'Picture 14.jpg', 'Picture 13.jpg', 'Picture 12.jpg', 'Picture 11.jpg', 'Picture 10.jpg', 'Picture 3.jpg', 'Picture 2.jpg', 'Picture 1.jpg', 'Picture 0.jpg']

I'm getting ValueError: invalid literal for int() with base 10: — ChumbiChubaGo, Oct 15 '15 at 22:30
I found a better solution. Now will update. But i did not get error on data. If you have only numbers as name is should work for you . But since there is general approach you may like it better. — LetzerWille, Oct 15 '15 at 22:40
FYI, looks like the install command is `pip install natsort`. Might have changed over the years, I am not entirely sure. — HighVoltage, Jul 16 '18 at 18:53

hft · Answer 3 · 2021-09-07T19:55:02.107

I have a directory with jpgs and other files in it.

[...]

['0.jpg', '1.jpg', '10.jpg', '11.jpg', '12.jpg', '13.jpg', '14.jpg', '15.jpg', '16.jpg', '17.jpg', '18.jpg', '19.jpg', '2.jpg', '20.jpg', '21.jpg', '22.jpg', '23.jpg', '24.jpg', '25.jpg', '26.jpg', '27.jpg', '28.jpg', '29.jpg', '3.jpg', '30.jpg', '31.jpg', '32.jpg', '33.jpg', '34.jpg', '35.jpg', '36.jpg', '37.jpg', '4.jpg', '5.jpg', '6.jpg', '7.jpg', '8.jpg', '9.jpg'] Clearly it sorts blindly the most significant number first. I tried using sorted() as you can see hoping that it would fix it but it makes no difference

You can use splitext to get the part without the extension and convert it to an int for the sorting. If the list is named 'l' and the sorted list is named 'lsorted' you can use:

lsorted = sorted(imgs_list, key=lambda x: int(os.path.splitext(x)[0]))

"imgs_list" here is the list of images. If you have a directory of images, simply obtain a list of these images by :

l = os.listdir('/path/to/directory/of/images')

Explanation: os.path.splitext on '10.jpg' returns ['10','.jpg'] so taking the int() of element zero will give you want you want as long as the filenames without the extention only contain strings that can be converted to integers with int(). Otherwise you will run into an Error.

Exactly what I needed! Thank you! – Sharpey Jun 03 '23 at 00:27 — Sharpey, Jun 03 '23 at 00:27

Sort filenames in directory in ascending order

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