case-insensitive list sorting, without lowercasing the result?
case-insensitive, list, python, sorting, string
Solution
In Python 3.3+ there is the `str.casefold` method that's specifically designed for caseless matching:
sorted_list = sorted(unsorted_list, key=str.casefold)
In Python 2 use `lower()`:
sorted_list = sorted(unsorted_list, key=lambda s: s.lower())
It works for both normal and unicode strings, since they both have a `lower` method.
In Python 2 it works for a mix of normal and unicode strings, since values of the two types can be compared with each other. Python 3 doesn't work like that, though: you can't compare a byte string and a unicode string, so in Python 3 you should do the sane thing and only sort lists of one type of string.
>>> lst = ['Aden', u'abe1']
>>> sorted(lst)
['Aden', u'abe1']
>>> sorted(lst, key=lambda s: s.lower())
[u'abe1', 'Aden']
Problem
I have a list of strings like this: ``` ['Aden', 'abel'] ``` I want to sort the items, case-insensitive. So I want to get: ``` ['abel', 'Aden'] ``` But I get the opposite with `sorted()` or `list.sort()`, because uppercase appears before lowercase. How can I ignore the case? I've seen solutions which involves lowercasing all list items, but I don't want to change the case of the list items.