Sort results non-lexicographically?

Question

I'm trying to display some results in a human-readable way. For the purposes of this question, some of them are numbers, some are letters, some are a combination of the two.

I'm trying to figure out how I could get them to sort like this:

input = ['1', '10', '2', '0', '3', 'Hello', '100', 'Allowance']
sorted_input = sorted(input)
print(sorted_input)

Desired Results:

['0', '1', '2', '3', '10', '100', 'Allowance', 'Hello']

Actual results:

['0', '1', '10', '100', '2', '3', 'Allowance', 'Hello']

I'm having trouble coming up with how to do this.

possible duplicate of [Does Python have a built in function for string natural sort?](http://stackoverflow.com/questions/4836710/does-python-have-a-built-in-function-for-string-natural-sort) — Steve Jessop, Oct 27 '13 at 23:20

Mingyu · Accepted Answer · 2013-10-29T21:09:59.890

15

1 - Install natsort module

pip install natsort

2 - Import natsorted

>>> input = ['1', '10', '2', '0', '3', 'Hello', '100', 'Allowance']

>>> from natsort import natsorted
>>> natsorted(input)
['0', '1', '2', '3', '10', '100', 'Allowance', 'Hello']

Source: https://pypi.python.org/pypi/natsort

edited Oct 29 '13 at 21:09

answered Oct 27 '13 at 23:24

Mingyu

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I'm glad to see that `natsort` is useful for people other than just me. Thanks for recommending it! – SethMMorton Oct 28 '13 at 00:02
Thanks for writing the `natsort` module, @SethMMorton. It's really handy. – Mingyu Oct 28 '13 at 00:12

score 3 · Answer 2 · answered Oct 27 '13 at 23:22

3

I have found the code in the following link about natural sorting order very useful in the past:

http://www.codinghorror.com/blog/2007/12/sorting-for-humans-natural-sort-order.html

answered Oct 27 '13 at 23:22

Ben

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This was really useful towards understanding how it works, I was torn between marking this and Mingyu's, I ended up going with Mingyu's since it actually has a usable solution, but this was a great read. – user2923558 Oct 27 '13 at 23:37

score 1 · Answer 3 · answered Oct 27 '13 at 23:23

1

This will do it. For purposes of comparison, it converts strings that can be converted to an integer to that integer, and leaves other strings alone:

def key(s):
    try:
        return int(s)
    except ValueError:
        return s

sorted_input = sorted(input, key=key)

answered Oct 27 '13 at 23:23

Tim Peters

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I think you missed this part "some are a combination of the two.". Also won't work in Python3 because "unorderable types" – John La Rooy Oct 27 '13 at 23:24
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Nope, I saw it but ignored it, because they didn't say a word about what they wanted in that case ;-) – Tim Peters Oct 27 '13 at 23:25
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"In the face of ambiguity, refuse the temptation to guess." I should have known better :) – John La Rooy Oct 27 '13 at 23:27

score 0 · Answer 4 · edited Oct 27 '13 at 23:33

0

For your specific case:

def mySort(l):
    numbers = []
    words = []
    for e in l:
        try:
            numbers.append(int(e))
        except:
            words.append(e)
    return [str(i) for i in sorted(numbers)] + sorted(words)

print mySort(input)

edited Oct 27 '13 at 23:33

JadedTuna

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answered Oct 27 '13 at 23:25

Christian Tapia

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score 0 · Answer 5 · answered Oct 27 '13 at 23:25

You could split up the list, sort, then put it back together. Try something like this:

numbers = sorted(int(i) for i in input_ if i.isdigit())
nonnums = sorted(i for i in input_ if not i.isdigit())
sorted_input = [str(i) for i in numbers] + nonnums

You'll have to do something more complicated than isdigit if you can have non-integers.

If this doesn't cover your "some are a combination of the two," please elaborate what that means and what output you expect from them.

(Not tested, but should convey the idea.)

Sort results non-lexicographically?

5 Answers5

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