Python snippet: ordinalth(n)

Blogged so I don’t lose it (blogging as external memory):

def ordinalth(n):
    """ 1 => 1st, 2 => 2nd etc """
    last = n - n / 10 * 10
    if last == 1:
        return '%dst' % n
    if last == 2:
        return '%dnd' % n
    if last == 3:
        return '%drd' % n
    return '%dth' % n

Update: See comments for improved version.

Posted 8th October 2003 at 3:03 pm

11 comments

Nice trick Simon, but i would remove if's completely:

def ordinalth2(n):

t = '%dth', '%dst', '%dnd', '%drd', '%dth', '%dth', '%dth' , '%dth' , '%dth', '%dth'

last = n - (n / 10 * 10)

return t[last] % n
if you pass 21 to ordinalth and ordinalth they both retiurn "21st", mmm...
deelan - 8th October 2003 15:42 - #
Watch out for 11st, 12nd, 13rd.
talon - 8th October 2003 15:44 - #
last = n % 10
Sam Ruby - 8th October 2003 15:58 - #

Improved version, taking the above comments in to account:

def ordinalth(n):
    t = 'th st nd rd th th th th th th'.split()
    if n in (11, 12, 13): #special case
        return '%dth' % n
    return str(n) + t[n % 10]

Test code:

for i in range(1, 101):
    print ordinalth(i)

(I've finally enabled the <pre> tag)

Simon Willison - 8th October 2003 16:11 - #

That should be
if (n % 100) in (11, 12, 13):

Curioso - 8th October 2003 16:24 - #
You're absolutely right, I hadn't spotted that.
Simon Willison - 8th October 2003 16:30 - #

You missed 111th. Untested, but should work (just need to mod 100 for the 11th..13th):

def ordinalth(n):
    t = 'th st nd rd th th th th th th'.split()
    if n % 100 in (11, 12, 13): #special case
        return '%dth' % n
    return str(n) + t[n % 10]

Brian Harnish - 8th October 2003 16:32 - #

Nice! "blogging as external memory" HAH! What is the computer analogous to that incredible display of rapid-fire collaboration following a simple "note to self"?
DeanG - 8th October 2003 17:11 - #

I whipped this up last week:

def inttoord(i):
    ordinal = [ ('th', 'st', 'nd', 'rd', 'th', 'th', 'th', 'th', 'th', 'th'),
                ('th', 'th', 'th', 'th', 'th', 'th', 'th', 'th', 'th', 'th')]
    ones = i % 10
    tens = (i % 100) / 10
    return str(i) + ordinal[tens==1][ones]

Bill - 8th October 2003 17:32 - #

whoa, you guys are geniuses. i don't know what this is but it looks complicated. im bad at math and shit.
ryandjess@aaaaaaatt.net - 29th December 2003 15:40 - #

I like to avoid all procedural statements and program functionally (as a lisp programmer might), avoiding repetitious code as much as possible. Here's my implementation.

def ordinalth(n):
	"""Return the ordinal with 'st', 'th', or 'nd' appended as appropriate.
	>>> map( ordinalth, xrange( -5, 22 ) )
	['-5th', '-4th', '-3rd', '-2nd', '-1st', '0th', '1st', '2nd', '3rd', '4th', '5th', '6th', '7th', '8th', '9th', '10th', '11th', '12th', '13th', '14th', '15th', '16th', '17th', '18th', '19th', '20th', '21st']
	"""
	# zero through three map to 'th', 'st', 'nd', 'rd'
	t = ( 'th', 'st', 'nd', 'rd' )
	# special case: ones digit is > 3
	ones = abs( n ) % 10
	forceth = ones > 3
	# special case: n ends in 11, 12, 13
	forceth |= abs( n ) % 100 in (11, 12, 13)
	index = [ ones, 0 ][forceth]
	return '%d%s' % ( n, t[index] )

Jason R. Coombs - 14th May 2004 17:22 - #

Comments are closed.

Previously hosted at http://simon.incutio.com/archive/2003/10/08/externalMemory

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Python snippet: ordinalth(n)

11 comments