How to sort a version list?

Question

I have a method that list some versions of an application on my computer, the list following has this format and the content are strings:

['1.15.1.23', '1.10.1.34', '1.9.2.3', ...]

When I call sorted(mylist), the output does not appear to sort some items, I'm getting this:

['1.15.1.51', '1.15.1.9', '1.15.2.11', '1.15.2.6']

What I'm doing wrong? I expected the output as:

['1.15.1.9', '1.15.1.51', '1.15.2.6', '1.15.2.11']

Strings are compared character-by-character, so `'10'` comes before `'2'`. — interjay, Jan 13 '17 at 13:13
It does work, you must not have used it correctly. But keep making rude posts instead of trying to actually learn. — interjay, Jan 13 '17 at 13:27

a_guest · Accepted Answer · 2017-01-13T13:17:20.040

5

Supply a key to the sorted function:

sorted(mylist, key=lambda v: map(int, v.split('.')))

edited Jan 13 '17 at 13:17

answered Jan 13 '17 at 13:12

a_guest

1

Did you try `mylist.sort(key=...)`? Because that sorts the list in-place and returns `None`. `sorted` definitely returns the result. – a_guest Jan 13 '17 at 13:18

1 Answers1