How to sort each individual string in a list of strings?

Question

I want to sort each string in a list of string to be in alphabetical order.

For example,

l1 = ["bba", "yxx", "abc"]

I want to sort it to be

l1 = ["abb", "xxy", "abc"]

I can do it in a for loop but wonder if this is a more pythonic way using python list comprehension. Thanks :D

falsetru · Accepted Answer · 2015-02-15 07:51:31Z

4

>>> l1 = ["bba", "yxx", "abc"] >>> [''.join(sorted(s)) for s in l1] ['abb', 'xxy', 'abc']

answered Feb 15, 2015 at 7:51

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Comments

Vivek Sable · Accepted Answer · 2015-02-15 08:00:54Z

by List compression with sorted method and string join method

>>> l1 = ["bba", "yxx", "abc"] >>> [''.join(sorted(i)) for i in l1] ['abb', 'xxy', 'abc'] >>>

by lambda

>>> l1 ['bba', 'yxx', 'abc'] >>> map(lambda x:"".join(sorted(x)),l1) ['abb', 'xxy', 'abc']

For Python beginner

e.g.

>>> l1 ['bba', 'yxx', 'abc'] >>> l2 = [] >>> for i in l1: ... a = sorted(i) ... b = "".join(a) ... l2.append(b) ... >>> l2 ['abb', 'xxy', 'abc']

the lambda one needs to be converted in to a list if we want to be fair and compare times.

Eithos · Accepted Answer · 2015-02-15 09:29:46Z

Just use the sorted built-in function.

l1 = ["bba", "yxx", "abc"] print [''.join(sorted(i)) for i in l1]

Another way of using list comprehensions (if your strings have a fixed length of 3) would be:

print ['{0}{1}{2}'.format(*sorted(i)) for i in l1]