Reputation: 1905
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'
Upvotes: 1
Views: 177
Reputation: 403278
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.
Upvotes: 2
Reputation: 71620
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']
>>>
Upvotes: 1
Reputation: 48
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
Upvotes: 1