How to sort an int from a file

Question

I have this code:

import glob
import os

for file in glob.glob("./data/*.jpg"):
    lel=(os.path.basename(file))
    lel = (os.path.splitext(lel)[0])
    list1 = lel
    list1 = [int(x) for x in list1]
    list1.sort()
    print(list1)

and it gives this output:

[0]
[1]
[0, 1]
[1, 1]
[1, 2]
[1, 3]
[1, 4]
[1, 5]
[1, 6]
[1, 7]
[1, 8]
[1, 9]
[2]
[0, 2]
[1, 2]
[2, 2]
[2, 3]
[2, 4]
[2, 5]
[2, 6]
[2, 7]
[2, 8]
[2, 9]
[3]
[0, 3]
[1, 3]
[2, 3]
[3, 3]
[3, 4]
[4]
[5]
[6]
[7]
[8]
[9]

is there a way to make it output this?:

0
1
2
3
4
5
6
7
8
9
10
etc etc..

Please expalin better your desired output. – logi-kal Feb 02 '21 at 20:28 — logi-kal, Feb 02 '21 at 20:28

TheEagle · Answer 1 · 2021-02-02T20:46:14.707

This code should work:

import glob
import os

list1 = []

for file in glob.glob("./data/*.jpg"): # iterate over filename
    list1.append(
                 os.path.splitext(
                                  os.path.basename(file)
                                 )[0]
    ) # strip away path and extension and append to the list1
print(*sorted(list1)) # print it out

You were iterating over the letters of the filenames, and converting every letter to int. You can even do it with a list comprehension:

import glob
import os

list1 = [os.path.splitext(os.path.basename(filename))[0] for filename in glob.glob("./data/*.jpg")]
print(*sorted(list1))

score 0 · Accepted Answer · answered Feb 02 '21 at 20:39

From what I can tell, you have files named 0.jpg, up to 34.jpg

You need to remove [int(x) for x in list1] because this gets a list per file

If you want to sort the numbers in ascending order rather than lexicographic, then make a list outside the loop, e.g.

import glob
import os

files = []
for file in glob.glob("./data/*.jpg"):
    lel = os.path.basename(file)
    lel = os.path.splitext(lel)[0]
    files.append(int(lel))
files.sort()
for f in files:
    print(f)

How to sort an int from a file

2 Answers2