What is the best way of creating an alphabetically sorted list in Python?

# python – How to sort a list of strings?

## The Question :

*430 people think this question is useful*

*The Question Comments :*

- Use
`locale`

and it’s string collation methods to sort naturally according to current locale.

## The Answer 1

*531 people think this answer is useful*

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)

## The Answer 2

*56 people think this answer is useful*

It is also worth noting the `sorted()`

function:

for x in sorted(list): print x

This returns a new, sorted version of a list without changing the original list.

## The Answer 3

*38 people think this answer is useful*

list.sort()

It really is that simple ðŸ™‚

## The Answer 4

*18 people think this answer is useful*

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.

## The Answer 5

*14 people think this answer is useful*

Please use sorted() function in Python3

items = ["love", "like", "play", "cool", "my"] sorted(items2)

## The Answer 6

*10 people think this answer is useful*

But how does this handle language specific sorting rules? Does it take locale into account?

No, `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.

## The Answer 7

*1 people think this answer is useful*

Old question, but if you want to do **locale-aware sorting without setting** `locale.LC_ALL`

you can do so by using the PyICU library as suggested by this answer:

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.)

## The Answer 8

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Suppose `s = "ZWzaAd"`

To sort above string the simple solution will be below one.

print ''.join(sorted(s))

## The Answer 9

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Or maybe:

names = ['Jasmine', 'Alberto', 'Ross', 'dig-dog'] print ("The solution for this is about this names being sorted:",sorted(names, key=lambda name:name.lower()))

## The Answer 10

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l =['abc' , 'cd' , 'xy' , 'ba' , 'dc'] l.sort() print(l1)

Result

[‘abc’, ‘ba’, ‘cd’, ‘dc’, ‘xy’]

## The Answer 11

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It is simple: https://trinket.io/library/trinkets/5db81676e4

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): print(x)