What is the best way of creating an alphabetically sorted list in Python?
Basic answer:
mylist = ["b", "C", "A"]
mylist.sort()
This modifies your original list (i.e. sorts in-place). To get a sorted copy of the list, without changing the original, use the sorted()
function:
for x in sorted(mylist):
print x
However, the examples above are a bit naive, because they don't take locale into account, and perform a case-sensitive sorting. You can take advantage of the optional parameter key
to specify custom sorting order (the alternative, using cmp
, is a deprecated solution, as it has to be evaluated multiple times - key
is only computed once per element).
So, to sort according to the current locale, taking language-specific rules into account (cmp_to_key
is a helper function from functools):
sorted(mylist, key=cmp_to_key(locale.strcoll))
And finally, if you need, you can specify a custom locale for sorting:
import locale
locale.setlocale(locale.LC_ALL, 'en_US.UTF-8') # vary depending on your lang/locale
assert sorted((u'Ab', u'ad', u'aa'),
key=cmp_to_key(locale.strcoll)) == [u'aa', u'Ab', u'ad']
Last note: you will see examples of case-insensitive sorting which use the lower()
method - those are incorrect, because they work only for the ASCII subset of characters. Those two are wrong for any non-English data:
# this is incorrect!
mylist.sort(key=lambda x: x.lower())
# alternative notation, a bit faster, but still wrong
mylist.sort(key=str.lower)