Sorting list of strings like [3a, 13a, 5a, 11a] by numeric value

Question

I'm trying to sort a list with numbers and letter by numeric value.

list = ['3a', '13a', '5a', '11a']

When I use the .sort function it sorts it like this:

['11a', '13a', '3a', '5a']

Like it's only looking at the first number.

Answer 1

If you just use an unmodified sorted it will sort on the ASCII codes on each character in the string:

>>> sorted(['2','11','01000'])
['01000', '11', '2']

vs adding a key value that tells sorted to use the integer value:

>>> sorted(['2','11','01000'], key=int)
['2', '11', '01000']

For strings that are a mixture of integers and letters, use a natural sort where letters are sorted lexicographically and numbers are treated to simple integer comparison:

import re

def natural_sort(l): 
    convert = lambda text: int(text) if text.isdigit() else text.lower() 
    alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ] 
    return sorted(l, key = alphanum_key)

>>> li = ['3a', '13a', '5a', '11a']
>>> natural_sort(li)
['3a', '5a', '11a', '13a']

The advantage of this version is it handles multiple groups of numbers and letters:

>>> li = ['3a33', '3a2', '3b1', '3a1', '5a', '11a']
>>> natural_sort(li)
['3a1', '3a2', '3a33', '3b1', '5a', '11a']

As well as what you would expect with the letters in front:

>>> li = ['a33', 'a2', 'b1', 'a1', 'a', 'a00004']
>>> natural_sort(li)
['a', 'a1', 'a2', 'a00004', 'a33', 'b1']

(PS: Best to not call a python list list because it stomps on the name of the function...)

Answer 2

You can pass in a lambda for a sorting key.

list = ['3a', '13a', '5a', '11a']
list.sort(key=lambda x: int(x[:-1]))

The important part is:

lambda x: int(x[:-1])

which takes in a string, removes the last character, and converts it to an integer.

Try it online!

Answer 3

Your list is a list of strings. So list.sort will sort the elements lexicographically. You can change this by passing a key to your sorting function. You can use re.match (import re first) to extract the digits and compare. Something like this:

In [30]: list_ = ['3a', '13a', '5a', '11a']

In [31]: matcher = re.compile('\d+')

In [32]: sorted(list_, key=lambda x: int(matcher.match(x).group()))
Out[32]: ['3a', '5a', '11a', '13a']

This works best if your elements have something more than a at the end of them.

Answer 4

There is package called natsort

from natsort import natsorted
LL = ['3a', '13a', '5a', '11a']
natsorted(LL)

Out[296]: ['3a', '5a', '11a', '13a']

@dawg's example using natsort:

li = ['3a33', '3a2', '3a1', '13a', '5a', '11a']

natsorted(li)
Out[298]: ['3a1', '3a2', '3a33', '5a', '11a', '13a']