What is the best way to sort this list?

Question

I have the following list:

my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']

And I would like sort the list and print it out in order, like this

name.1
name.2
name.4
name.13
name.32

What I have tried so far is:

print sorted(my_list)

name.1
name.13
name.2
name.32
name.4

The sorted() command obviously treats the string alphabetically. Maybe it would be better to sort numerically after detecting the . first?

Is there an good way to sort it properly? What would be the most efficient approach to take? How would I apply this is I had a list of tuples and wanted to sort it using the second elemnt of the tuples? For instance:

tuple_list = [('i','name.2'),('t','name.13'),('s','name.32'),('l','name.1'),('s','name.4')]

print tuple_list
'l','name.1'
'i','name.2'
's','name.4'
't','name.13'
's','name.32'

Thanks for your help and, as always, comment if you think the question can be improved/clarified.

Alex

Is the text before the period always the same (i.e. 'name')? If not, how do you want to sort items with different names? — Zero Piraeus, Jul 12 '13 at 22:14
possible duplicate of [How to sort alpha numeric set in python](http://stackoverflow.com/questions/2669059/how-to-sort-alpha-numeric-set-in-python) — tamasgal, Jul 12 '13 at 23:51

score 8 · Accepted Answer · edited May 23 '17 at 12:14

You could try it similarly to this answer:

>>> my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']
>>> sorted(my_list, key=lambda a: (int(a.split('.')[1])))
['name.1', 'name.2', 'name.4', 'name.13', 'name.32']

Or with tuples:

>>> tuple_list = [('i','name.2'),('t','name.13'),('s','name.32'),('l','name.1'),('s','name.4')]
>>> sorted(tuple_list, key=lambda (a,b): (int(b.split('.')[1])))
[('l', 'name.1'), ('i', 'name.2'), ('s', 'name.4'), ('t', 'name.13'), ('s', 'name.32')]

Edit to explain what this does:

The example passes a lambda function to sorted that splits the strings and then converts the second part to an integer. sorted then uses this integer to sort the list items.

This example completely ignores the strings to the left of the dot when sorting. Please let me know if you also need to sort by the first part.

This is a cool solution! Thanks - how would it be adapted to cope with the list of tuples situation posted in the question edit? — user1083734, Jul 12 '13 at 22:22

SethMMorton · Answer 2 · 2017-08-19T18:34:50.927

4

UPDATED ANSWER

As of natsort version 4.0.0, this works out of the box without having to specify any options:

>>> import natsort
>>> my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']
>>> natsort.natsorted(my_list)
['name.1','name.2', 'name.4','name.13', 'name.32']

OLD ANSWER for natsort < 4.0.0

If you do not object to external packages, try the natsort package (version >= 3.0.0):

>>> import natsort
>>> my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']
>>> natsort.natsorted(my_list, number_type=int)
['name.1','name.2', 'name.4','name.13', 'name.32']

The number_type argument is necessary in this case because natsort by default looks for floats, and the decimal point would make all these numbers be interpreted as floats.

Full disclosure: I am the natsort author.

edited Aug 19 '17 at 18:34

answered Jul 12 '13 at 22:26

SethMMorton

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I do not object, this is very useful! So to cope with the list of tuples situation, I would do `natsort.natsorted(my_list[1])` ? – user1083734 Jul 12 '13 at 22:29
I'm glad you like it! Yes, that should work for a list of tuples. – SethMMorton Jul 12 '13 at 22:53

karthikr · Answer 3 · 2013-07-12T22:26:55.090

3

Try this:

sorted(my_list, key=lambda x: int(x.split('.')[1]))

Demo:

>>> my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']
>>> sorted(my_list, key=lambda x: int(x.split('.')[1]))
['name.1', 'name.2', 'name.4', 'name.13', 'name.32']
>>>

Extending it to the tuple,

sorted(tuple_list, key = lambda x: int(x[1].split('.')[1]))

DEMO:

>>> sorted(tuple_list, key = lambda x: int(x[1].split('.')[1]))
[('l', 'name.1'), ('i', 'name.2'), ('s', 'name.4'), ('t', 'name.13'), ('s', 'name.32')]

edited Jul 12 '13 at 22:26

answered Jul 12 '13 at 22:07

karthikr

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@karthikr It doesn't sort numerically as per the question. – Zero Piraeus Jul 12 '13 at 22:12
Oh snap.. I misread the question. Shall delete the answer when the system allows me to. – karthikr Jul 12 '13 at 22:14
How would one adapt this to cope with the question edit? i.e. a list of tuples? – user1083734 Jul 12 '13 at 22:24

perreal · Answer 4 · 2013-07-12T22:36:23.647

3

This one uses more memory but does not split each item multiple times:

from operator import itemgetter
my_list = ['name.13','name.1', 'name.2','name.4', 'name.32']
my_nlist= [ (int(n.split('.')[1]), i) for i,n in enumerate(my_list)]
my_list = [ my_list[t[1]] for t in sorted(my_nlist, key=itemgetter(0))]

to deal with tuples:

my_list = [('i','name.2'),('t','name.13'),
           ('s','name.32'),('l','name.1'),('s','name.4')]
my_nlist= [ (int(n[1].split('.')[1]), i) for i,n in enumerate(my_list)]
my_list = [ my_list[t[1]] for t in sorted(my_nlist, key=itemgetter(0))]

edited Jul 12 '13 at 22:36

answered Jul 12 '13 at 22:12

perreal

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An interesting approach, how would it cope with the list of tuples situation in the question edit? – user1083734 Jul 12 '13 at 22:23

What is the best way to sort this list?

4 Answers4