How do you sort alphanumeric values in python?

I have a set

set(['booklet', '4 sheets', '48 sheets', '12 sheets'])

After sorting I want it to look like

4 sheets,
12 sheets,
48 sheets,
booklet

Any idea please

SilentGhost

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asked Apr 19, 2010 at 16:21

Jeff Atwood talks about natural sort and gives an example of one way to do it in Python. Here is my variation on it:

import re 

def sorted_nicely( l ): 
    """ Sort the given iterable in the way that humans expect.""" 
    convert = lambda text: int(text) if text.isdigit() else text 
    alphanum_key = lambda key: [ convert(c) for c in re.split('([0-9]+)', key) ] 
    return sorted(l, key = alphanum_key)

Use like this:

s = set(['booklet', '4 sheets', '48 sheets', '12 sheets'])
for x in sorted_nicely(s):
    print(x)

Output:

4 sheets
12 sheets
48 sheets
booklet

One advantage of this method is that it doesn't just work when the strings are separated by spaces. It will also work for other separators such as the period in version numbers (for example 1.9.1 comes before 1.10.0).

answered Apr 19, 2010 at 16:31

Mark ByersMark Byers

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4

Short and sweet:

sorted(data, key=lambda item: (int(item.partition(' ')[0])
                               if item[0].isdigit() else float('inf'), item))

This version:

• Works in Python 2 and Python 3, because:
  • It does not assume you compare strings and integers (which won't work in Python 3)
  • It doesn't use the cmp parameter to sorted (which doesn't exist in Python 3)
• Will sort on the string part if the quantities are equal

If you want printed output exactly as described in your example, then:

data = set(['booklet', '4 sheets', '48 sheets', '12 sheets'])
r = sorted(data, key=lambda item: (int(item.partition(' ')[0])
                                   if item[0].isdigit() else float('inf'), item))
print ',\n'.join(r)

answered Apr 19, 2010 at 17:32

Daniel StutzbachDaniel Stutzbach

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3

You should check out the third party library natsort. Its algorithm is general so it will work for most input.

>>> import natsort
>>> your_list = set(['booklet', '4 sheets', '48 sheets', '12 sheets'])
>>> print ',\n'.join(natsort.natsorted(your_list))
4 sheets,
12 sheets,
48 sheets,
booklet

answered Jul 17, 2014 at 22:01

SethMMortonSethMMorton

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A simple way is to split up the strings to numeric parts and non-numeric parts and use the python tuple sort order to sort the strings.

import re
tokenize = re.compile(r'(\d+)|(\D+)').findall
def natural_sortkey(string):          
    return tuple(int(num) if num else alpha for num, alpha in tokenize(string))

sorted(my_set, key=natural_sortkey)

answered Apr 19, 2010 at 16:37

Ants AasmaAnts Aasma

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It was suggested that I repost this answer over here since it works nicely for this case also

from itertools import groupby
def keyfunc(s):
    return [int(''.join(g)) if k else ''.join(g) for k, g in groupby(s, str.isdigit)]

sorted(my_list, key=keyfunc)

Demo:

>>> my_set = {'booklet', '4 sheets', '48 sheets', '12 sheets'}
>>> sorted(my_set, key=keyfunc)
['4 sheets', '12 sheets', '48 sheets', 'booklet']

For Python3 it's necessary to modify it slightly (this version works ok in Python2 too)

def keyfunc(s):
    return [int(''.join(g)) if k else ''.join(g) for k, g in groupby('\0'+s, str.isdigit)]

answered Jun 6, 2013 at 7:35

John La RooyJohn La Rooy

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Generic answer to sort any numbers in any position in an array of strings. Works with Python 2 & 3.

def alphaNumOrder(string):
   """ Returns all numbers on 5 digits to let sort the string with numeric order.
   Ex: alphaNumOrder("a6b12.125")  ==> "a00006b00012.00125"
   """
   return ''.join([format(int(x), '05d') if x.isdigit()
                   else x for x in re.split(r'(\d+)', string)])

Sample:

s = ['a10b20','a10b1','a3','b1b1','a06b03','a6b2','a6b2c10','a6b2c5']
s.sort(key=alphaNumOrder)
s ===> ['a3', 'a6b2', 'a6b2c5', 'a6b2c10', 'a06b03', 'a10b1', 'a10b20', 'b1b1']

Part of the answer is from there

answered Oct 5, 2016 at 23:23

Le DroidLe Droid

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>>> a = set(['booklet', '4 sheets', '48 sheets', '12 sheets'])
>>> def ke(s):
    i, sp, _ = s.partition(' ')
    if i.isnumeric():
        return int(i)
    return float('inf')

>>> sorted(a, key=ke)
['4 sheets', '12 sheets', '48 sheets', 'booklet']

answered Apr 19, 2010 at 16:25

SilentGhostSilentGhost

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Based on SilentGhost's answer:

In [4]: a = set(['booklet', '4 sheets', '48 sheets', '12 sheets'])

In [5]: def f(x):
   ...:     num = x.split(None, 1)[0]
   ...:     if num.isdigit():
   ...:         return int(num)
   ...:     return x
   ...: 

In [6]: sorted(a, key=f)
Out[6]: ['4 sheets', '12 sheets', '48 sheets', 'booklet']

answered Apr 19, 2010 at 16:31

sets are inherently un-ordered. You'll need to create a list with the same content and sort that.

answered Apr 19, 2010 at 16:25

RakisRakis

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3

For people stuck with a pre-2.4 version of Python, without the wonderful sorted() function, a quick way to sort sets is:

l = list(yourSet)
l.sort() 

This does not answer the specific question above (12 sheets will come before 4 sheets), but it might be useful to people coming from Google.

answered Oct 22, 2013 at 16:11

0

b = set(['booklet', '10-b40', 'z94 boots', '4 sheets', '48 sheets',
         '12 sheets', '1 thing', '4a sheets', '4b sheets', '2temptations'])

numList = sorted([x for x in b if x.split(' ')[0].isdigit()],
                 key=lambda x: int(x.split(' ')[0]))

alphaList = sorted([x for x in b if not x.split(' ')[0].isdigit()])

sortedList = numList + alphaList

print(sortedList)

Out: ['1 thing',
      '4 sheets',
      '12 sheets',
      '48 sheets',
      '10-b40',
      '2temptations',
      '4a sheets',
      '4b sheets',
      'booklet',
      'z94 boots']

answered Apr 1, 2021 at 17:23

tldrtldr

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