91

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

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11 Answers 11

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151

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).

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4 Comments

Hi Jeff, Thank you very much. That is exactly what I was looking for. Good luck.
Is it possible to modify this for a list of tuples based on the first value in the tuple? Example: [('b', 0), ('0', 1), ('a', 2)] is sorted to [('0', 1), ('a', 2), ('b', 0)]
This function is case sensitive. Upper case strings will take precedence. To fix this add .lower() to key in re.split.
@paragbaxi Add [0] after key in the alphanum_key lambda function: lambda key: [ convert(c) for c in re.split('([0-9]+)', key[0]) ]
63

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)
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3 Comments

chokes on 4a sheets but who cares? to fix this you'd need a real function instead of a lambda.
That might work for this trivial example but not for instance a list like ["1. bla", "2. blub"]. Probably the split should be a regex instead, and also sort by the second part afterwards, so ["1 bcd", "2 abc", "1 xyz"] comes out correctly.
Unfortunately, @FrankyBoy is correct, this does not work for sorting sets of version numbers alphanumerically; e.g., v1.0.1, v3.5.3, v3.2.4
28

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
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13

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)
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9

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)]
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4

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

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3 
>>> 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']
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2

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']
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0

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

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3 Comments

Not true - the sorted() built-in will take any sequence and return a sorted list.
So instead of creating a list and sorting it, you instead use a builtin to create a sorted list.... Yeah, I was way off.
sets implemented a SortedSets (and not HashSets) are inherently ordered
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']
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-1

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.

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