reordering filenames in python

Question

I have the following problem:

I need to load several data files. The files are named by my device like:

meas98.dat
meas99.dat
meas100.dat
meas101.dat

With other words, there are no leading zeros. Therefore, if I get the filenames via

os.listdir

they are ordered alphabetically, meaning "meas100.dat" will be the first one. This is obviously not what I want to achieve. The question is what is the most elegant way of doing this?

The (unelegant) way I came up with is:

load the filenames
extract the filenumber
order the filenumber (get the indices)
order the filenames with those indices

I am pretty sure python has something build in that can do this while loading the files...

Possible duplicate of [Does Python have a built in function for string natural sort?](http://stackoverflow.com/questions/4836710/does-python-have-a-built-in-function-for-string-natural-sort) — Mureinik, Jan 02 '17 at 10:29
Write some custom lambda function to sort function. sort by filename.split('.')[-1].replace('meas') — Murali Mopuru, Jan 02 '17 at 10:38

宏杰李 · Answer 1 · 2017-01-02T11:21:52.853

7

l = ['meas98.dat',
    'meas99.dat',
    'meas100.dat',
    'meas101.dat']
l.sort(key=lambda i: int(i.strip('meas.dat')))

There is a pythonic way to do this by using pathlib module:

this is the files in my ternimal:

~/so$ ls
meas100.dat  meas98.dat  meas99.dat

this is the files in python:

from pathlib import Path
p = Path('/home/li/so/')
list(p.iterdir())
[PosixPath('/home/li/so/meas99.dat'),
 PosixPath('/home/li/so/meas98.dat'),
 PosixPath('/home/li/so/meas100.dat')]

looks like the pathlib has do this sort for you, you can take a try.

edited Jan 02 '17 at 11:21

answered Jan 02 '17 at 10:48

宏杰李

11,820
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@Glostas please accept an answer to close this question, this will save other people's time. – 宏杰李 Jan 06 '17 at 12:08
As of today [the documentation says](https://docs.python.org/3/library/pathlib.html#pathlib.Path.iterdir): "The children are yielded in arbitrary order" – Mikhail Gerasimov Dec 06 '20 at 14:15

score 3 · Answer 2 · answered Jan 02 '17 at 10:47

Using slicing [4:-4] to get only numbers from filename - and sorted() will use them to sort filenames.

# random order
l = [
    'meas98.dat',
    'meas100.dat',
    'meas99.dat',
    'meas101.dat',
    'meas1.dat',
]

sorted(l, key=lambda x: int(x[4:-4]))

print(l)

result

['meas1.dat', 'meas98.dat', 'meas99.dat', 'meas100.dat', 'meas101.dat']

Gustavo Bezerra · Accepted Answer · 2017-01-02T10:55:50.590

2

Perhaps this will suit your problem:

import re

l = ['meas100.dat',
     'meas101.dat',
     'meas98.dat',
     'meas99.dat']


sorted(l, key=lambda x: int(re.match('\D*(\d+)', x).group(1)))

Output:

['meas98.dat', 'meas99.dat', 'meas100.dat', 'meas101.dat']

edited Jan 02 '17 at 10:55

answered Jan 02 '17 at 10:42

Gustavo Bezerra

9,984
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I have no idea why you got downvoted. I selected it as accepted, since it is in my opinion the most elegant solution – Glostas Mar 27 '18 at 13:58

reordering filenames in python

3 Answers3