What is the best way of creating an alphabetically sorted list in Python?
It really is that simple 🙂
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
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):
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)
But how does this handle language specific sorting rules? Does it take locale into account?
list.sort() is a generic sorting function. If you want to sort according to the Unicode rules, you’ll have to define a custom sort key function. You can try using the pyuca module, but I don’t know how complete it is.
It is also worth noting the
for x in sorted(list): print x
This returns a new, sorted version of a list without changing the original list.
The proper way to sort strings is:
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'), cmp=locale.strcoll) == [u'aa', u'Ab', u'ad'] # Without using locale.strcoll you get: assert sorted((u'Ab', u'ad', u'aa')) == [u'Ab', u'aa', u'ad']
The previous example of
mylist.sort(key=lambda x: x.lower()) will work fine for ASCII-only contexts.
s = "ZWzaAd"
To sort above string the simple solution will be below one.
Please use sorted() function in Python3
items = ["love", "like", "play", "cool", "my"] sorted(items2)
names = ['Jasmine', 'Alberto', 'Ross', 'dig-dog'] print ("The solution for this is about this names being sorted:",sorted(names, key=lambda name:name.lower()))
import icu # PyICU def sorted_strings(strings, locale=None): if locale is None: return sorted(strings) collator = icu.Collator.createInstance(icu.Locale(locale)) return sorted(strings, key=collator.getSortKey)
Then call with e.g.:
new_list = sorted_strings(list_of_strings, "de_DE.utf8")
This worked for me without installing any locales or changing other system settings.
(This was already suggested in a comment above, but I wanted to give it more prominence, because I missed it myself at first.)
l =['abc' , 'cd' , 'xy' , 'ba' , 'dc'] l.sort() print(l1)
[‘abc’, ‘ba’, ‘cd’, ‘dc’, ‘xy’]
It is simple:
scores = '54 - Alice,35 - Bob,27 - Carol,27 - Chuck,05 - Craig,30 - Dan,27 - Erin,77 - Eve,14 - Fay,20 - Frank,48 - Grace,61 - Heidi,03 - Judy,28 - Mallory,05 - Olivia,44 - Oscar,34 - Peggy,30 - Sybil,82 - Trent,75 - Trudy,92 - Victor,37 - Walter'
scores = scores.split(‘,’)
for x in sorted(scores):