Sorting filepaths by extension in python

Question

I want to sort this list in a way that .log should be the first file and .gz file should be in a descending order

my_list = [
     '/abc/a.log.1.gz',
     '/abc/a.log',
     '/abc/a.log.gz',
     '/abc/a.log.30.gz',
     '/abc/a.log.2.gz',
     '/abc/a.log.5.gz',
     '/abc/a.log.3.gz',
     '/abc/a.log.6.gz',
     '/abc/a.log.4.gz',
     '/abc/a.log.12.gz',
     '/abc/a.log.10.gz',
     '/abc/a.log.8.gz',
     '/abc/a.log.14.gz',
     '/abc/a.log.29.gz'
]

Expected Result:

my_list = ['/abc/a.log',
        '/abc/a.log.gz',
        '/abc/a.log.30.gz',
        '/abc/a.log.29.gz',
        '/abc/a.log.29.gz',
        '/abc/a.log.14.gz',
        '/abc/a.log.12.gz',
        '/abc/a.log.10.gz',
        '/abc/a.log.8.gz',
        '/abc/a.log.6.gz',
        '/abc/a.log.5.gz',
        '/abc/a.log.4.gz',
        '/abc/a.log.3.gz',
        '/abc/a.log.2.gz'
        '/abc/a.log.1.gz']

My solution: import os

def get_sort_keys(filepath):
    split_file_path = os.path.splitext(filepath)
    sort_key = (split_file_path[1], *os.path.splitext(split_file_path[0]))
    return (sort_key[0], sort_key[1], int(sort_key[2].strip(".")) if sort_key[2] else 0)

print(sorted(my_list, key=get_sort_keys, reverse=True))

Getting error:

ValueError: invalid literal for int() with base 10: 'log'

cs95 · Accepted Answer · 2019-02-18 03:08:15Z

2

You can use sorted with a custom function that does some try-except checking.

def try_convert(x):
    y = x.rsplit('.', 2)[-2]
    return ('log' not in x, int(y) if y.isdigit() else float('inf'), x)

sorted(my_list, key=try_convert, reverse=True)

['/abc/a.log.gz',
 '/abc/a.log',
 '/abc/a.log.30.gz',
 '/abc/a.log.29.gz',
 '/abc/a.log.14.gz',
 '/abc/a.log.12.gz',
 '/abc/a.log.10.gz',
 '/abc/a.log.8.gz',
 '/abc/a.log.6.gz',
 '/abc/a.log.5.gz',
 '/abc/a.log.4.gz',
 '/abc/a.log.3.gz',
 '/abc/a.log.2.gz',
 '/abc/a.log.1.gz']

The function ensures that filenames without the integer component are to be ordered last (first, if you sort in descending). Additionally, all ".log" files come first.

edited Feb 18, 2019 at 3:08

answered Feb 18, 2019 at 2:57

cs95

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2 Comments

user15051990 Over a year ago

I have edited question, could you please update your answer accordingly?

cs95 Over a year ago

@user15051990 I can get you a solution that has the first and second list items interchanged. Is this ok?

JBurt · Accepted Answer · 2019-02-18 02:57:15Z

1

You're trying to pass the string 'log' into int().

It cant then convert this to an int and raises ValueError: invalid literal for int() with base 10: 'log'

This is happening in here return (sort_key[0], sort_key[1], int(sort_key[2].strip(".")) if sort_key[2] else 0)

Try using a try catch block on the conversion

answered Feb 18, 2019 at 2:57

JBurt

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Comments

U13-Forward · Accepted Answer · 2019-02-18 03:06:26Z

1

Or use:

>>> sorted(my_list,key=lambda x: int(x.split('.')[2]) if x.split('.')[2].isdigit() else 31,reverse=True)
['/abc/a.log.gz', '/abc/a.log.30.gz', '/abc/a.log.29.gz', '/abc/a.log.14.gz', '/abc/a.log.12.gz', '/abc/a.log.10.gz', '/abc/a.log.8.gz', '/abc/a.log.6.gz', '/abc/a.log.5.gz', '/abc/a.log.4.gz', '/abc/a.log.3.gz', '/abc/a.log.2.gz', '/abc/a.log.1.gz']
>>>

Updated question:

>>> sorted(my_list,key=lambda x: int(x.split('.')[-2]) if x.split('.')[-2].isdigit() else 31,reverse=True)
['/abc/a.log', '/abc/a.log.gz', '/abc/a.log.30.gz', '/abc/a.log.29.gz', '/abc/a.log.14.gz', '/abc/a.log.12.gz', '/abc/a.log.10.gz', '/abc/a.log.8.gz', '/abc/a.log.6.gz', '/abc/a.log.5.gz', '/abc/a.log.4.gz', '/abc/a.log.3.gz', '/abc/a.log.2.gz', '/abc/a.log.1.gz']
>>>

edited Feb 18, 2019 at 3:06

answered Feb 18, 2019 at 3:04

U13-Forward

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2 Comments

user15051990 Over a year ago

I have updated the question can you please update the answer..\

U13-Forward Over a year ago

@user15051990 Edited mine.

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