Sort String List By A Number In String?

I have the following string list. Then, I want to sort it by a number in each element. sorted failed because it cannot handle the order such as between 10 and 3. I can imagine if I

Solution 1:

the key is ... the key

sorted(names, key=lambda x: int(x.partition('-')[2].partition('.')[0]))

Getting that part of the string recognized as the sort order by separating it out and transforming it to an int.

Solution 2:

Some alternatives:

(1) Slicing by position:

sorted(names, key=lambda x: int(x[5:-6]))

(2) Stripping substrings:

sorted(names, key=lambda x: int(x.replace('Test-', '').replace('.model', '')))

(3) Splitting characters (also possible via str.partition):

sorted(names, key=lambda x: int(x.split('-')[1].split('.')[0]))

(4) Map with np.argsort on any of (1)-(3):

list(map(names.__getitem__, np.argsort([int(x[5:-6]) for x in names])))

Solution 3:

I found a similar question and a solution by myself. Nonalphanumeric list order from os.listdir() in Python

import re
defsorted_alphanumeric(data):
    convert = lambda text: int(text) if text.isdigit() else text.lower()
    alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ] 
    returnsorted(data, key=alphanum_key, reverse=True)

Solution 4:

You can use re.findall in with the key of the sort function:

import re
names = [
 'Test-1.model',
 'Test-4.model',
 'Test-6.model',
 'Test-8.model',
 'Test-10.model',
 'Test-20.model'
]
final_data = sorted(names, key=lambda x:int(re.findall('(?<=Test-)\d+', x)[0]), reverse=True)

Output:

['Test-20.model', 'Test-10.model', 'Test-8.model', 'Test-6.model', 'Test-4.model', 'Test-1.model']

Solution 5:

def find_between( s, first, last ):
    try:
        start= s.index( first ) + len( first )
        end= s.index( last, start )
        return s[start:end]
    except ValueError:
        return ""

and then do something like

sorted(names, key=lambda x: int(find_between(x, 'Test-', '.model')))