Ordinal numbers replacement

Here's a terse solution taken from Gareth on codegolf:

ordinal = lambda n: "%d%s" % (n,"tsnrhtdd"[(n//10%10!=1)*(n%10<4)*n%10::4])

Works on any number:

print([ordinal(n) for n in range(1,32)])

['1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th',
 '11th', '12th', '13th', '14th', '15th', '16th', '17th', '18th', '19th',
 '20th', '21st', '22nd', '23rd', '24th', '25th', '26th', '27th', '28th',
 '29th', '30th', '31st']

If you don't want to pull in an additional dependency on an external library (as suggested by luckydonald) but also don't want the future maintainer of the code to haunt you down and kill you (because you used golfed code in production) then here's a short-but-maintainable variant:

def make_ordinal(n):
    '''
    Convert an integer into its ordinal representation::

        make_ordinal(0)   => '0th'
        make_ordinal(3)   => '3rd'
        make_ordinal(122) => '122nd'
        make_ordinal(213) => '213th'
    '''
    n = int(n)
    if 11 <= (n % 100) <= 13:
        suffix = 'th'
    else:
        suffix = ['th', 'st', 'nd', 'rd', 'th'][min(n % 10, 4)]
    return str(n) + suffix

How about this:

suf = lambda n: "%d%s"%(n,{1:"st",2:"nd",3:"rd"}.get(n if n<20 else n%10,"th"))
print [suf(n) for n in xrange(1,32)]

['1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th',
 '11th', '12th', '13th', '14th', '15th', '16th', '17th', '18th', '19th',
 '20th', '21st', '22nd', '23rd', '24th', '25th', '26th', '27th', '28th',
 '29th', '30th', '31st']

Another solution is the num2words library (pip | github). It especially offers different languages, so localization/internationalization (aka. l10n/i18n) is a no-brainer.

Usage is easy after you installed it with pip install num2words:

from num2words import num2words
# english is default
num2words(4458, to="ordinal_num")
'4458th'

# examples for other languages
num2words(4458, lang="en", to="ordinal_num")
'4458th'

num2words(4458, lang="es", to="ordinal_num")
'4458º'

num2words(4458, lang="de", to="ordinal_num")
'4458.'

num2words(4458, lang="id", to="ordinal_num")
'ke-4458'

Bonus:

num2words(4458, lang="en", to="ordinal")
'four thousand, four hundred and fifty-eighth'

The accepted answer to a previous question has an algorithm for half of this: it turns "first" into 1. To go from there to "1st", do something like:

suffixes = ["th", "st", "nd", "rd", ] + ["th"] * 16
suffixed_num = str(num) + suffixes[num % 100]

This only works for numbers 0-19.