CODEWITHSUNDEEP
pythonpython-2.7listdir
Glostas
Glostas

Reputation: 1180

reordering filenames in python

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:

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

Upvotes: 2

Views: 1585

Answers (3)

宏杰李
宏杰李

Reputation: 12168

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.

Upvotes: 7

Gustavo Bezerra
Gustavo Bezerra

Reputation: 11034

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']

Upvotes: 2

furas
furas

Reputation: 142651

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']

Upvotes: 3

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