Python Sort Strings With Leading Numbers Alphabetically

I have a list of filenames, each of them beginning with a leading number: 10_file 11_file 1_file 20_file 21_file 2_file ... I need to put it in this order: 1_file 10_file 11_file

Solution 1:

Sorting, splitting and list comprehensions work well here.

lst = ['10_file', '11_file', '1_file', '20_file', '21_file', '2_file']

lst_split = ['_'.join(x) for x in sorted(i.split('_') for i in lst)]

# ['1_file', '10_file', '11_file', '2_file', '20_file', '21_file'] 

Solution 2:

Edited with what the OP really wanted:

>>> from functools import partial
>>> lst = ['10_file', '11_file', '1_file', '20_file', '21_file', '2_file']
>>> sorted(lst, key=partial(str.split, sep='_', maxsplit=1))
['1_file', '10_file', '11_file', '2_file', '20_file', '21_file']

Solution 3:

How about this:

flist = ['10_file',
        '11_file',
        '1_file',
        '20_file',
        '21_file',
        '2_file']

tempdict = {}

for item in flist:
    num = item.split('_')[0]
    tempdict[num] = item

output = []

# for truly numeric sorting#for k in sorted([int(k) for k in tempdict.keys()]):#output.append(tempdict[str(k)])# for alphabetical sorting:for k in sorted(tempdict.keys()):
    output.append(tempdict[k])

print('\n'.join(output))

Result

1_file
10_file
11_file
2_file
20_file
21_file

Solution 4:

The simple way. Just extract the digits and then sort it as string:

sorted(l, key=lambda s: s.split("_")[0] )

That's all you need... try:

l=['2_file', '10_file', '11_file', '1_file', '20_file', '21_file']
print"\n".join(sorted(l, key=lambda s: s.split("_")[0] ))
1_file
10_file
11_file
2_file
20_file
21_file