Smart sorting in python

Question

I am trying to sort tuples in list that have field of mixed nature: LABEL.NUMBER. For example:

 aaaa.143
 aadf.23
 aaaa.8
 ..

So, I would like to sort first by LABEL as strings, and at the same by NUMBER as numbers, i.e. after sorting the should come:

 aaaa.8
 aaaa.143
 aadf.23
 ..

I have now the following:

for i in sorted(v_distribution.items(), key=lambda x: x[0]):

which sorts using the whole filed as a string, so I get:

 aaaa.143
 aaaa.8
 aadf.23
 ..

How should I modify my lambda function to do the task?

Have you seen: http://stackoverflow.com/questions/11150239/python-natural-sorting — Jon Clements, Sep 11 '13 at 11:29
@JonClements No, thank you for the link, I need to check the solution, maybe my question is to be duplicated... — pmod, Sep 11 '13 at 11:33
@JonClements It turned out that solution posted here solves my specific task instead of general solution from your link. So I think it's not fully duplicate but a special case regarding the similar issue. — pmod, Sep 11 '13 at 12:05

score 2 · Answer 1 · answered Sep 11 '13 at 11:35

2

Something like:

>>> s = ['aaaa.143', 'aadf.23', 'aaaa.8']
>>> def key_f(x):
...     head, tail = x.split('.', 1)
...     return (head, int(tail))
...
>>> sorted(s, key=key_f)
['aaaa.8', 'aaaa.143', 'aadf.23']

Although this can be done with lambda, it is better to separate key calculation into separate function.

answered Sep 11 '13 at 11:35

Roman Bodnarchuk

29,461
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Thank you, since it was tagged with lambda and tuples was in question, I will accept another answer, +1 anyway – pmod Sep 11 '13 at 11:58

score 1 · Accepted Answer · answered Sep 11 '13 at 11:42

1

You can do this with lambda as:

for i in sorted(v_distribution.items(), key=lambda x: (x.split('.', 1)[0], int(x.split('.', 1)[1]):

answered Sep 11 '13 at 11:42

Paul Evans

27,315
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Yes, this works now and turned out to be very easy modification, thank you – pmod Sep 11 '13 at 11:55

Smart sorting in python

2 Answers2