Date Ordinal Output?

Question

I\'m wondering if there is a quick and easy way to output ordinals given a number in python.

For example, given the number 1, I\'d like to output

Answer 1

These days I'd use Arrow http://arrow.readthedocs.io/en/latest/ (which definately wasn't around in '09)

>>> import arrow
>>> from datetime import datetime
>>> arrow.get(datetime.utcnow()).format('Do')
'27th'

Answer 2

Except for 1st, 2nd, and 3rd, I think they all just add th... 4th, 5th, 6th, 11th, 21st ... oh, oops ;-)

I think this might work:

def ordinal(num):
     ldig = num % 10
     l2dig = (num // 10) % 10
     if l2dig == 1:
         suffix = 'th'
     elif ldig == 1:
         suffix = 'st'
     elif ldig == 2:
         suffix = 'nd'
     elif ldig == 3:
         suffix = 'rd'
     else: 
         suffix = 'th'
     return '%d%s' % (num, suffix)

Answer 3

def ordinal(n):
    return ["th", "st", "nd", "rd"][n%10 if n%10<4 and not (10<n%100<14) else 0]

Answer 4

A more general and shorter solution (as a function):

def get_ordinal(num)
    ldig = num % 10
    l2dig = (num // 10) % 10

    if (l2dig == 1) or (ldig > 3):
        return '%d%s' % (num, 'th')
    else:
        return '%d%s' % (num, {1: 'st', 2: 'nd', 3: 'rd'}.get(ldig))

I just combined David's solutions and libraries (as deegeedubs did). You can even replace the variables (ldig, l2dig) for the real math (since l2dig is used only once), then you get four lines of code.

Answer 5

Here is an even shorter general solution:

def foo(n):
    return str(n) + {1: 'st', 2: 'nd', 3: 'rd'}.get(4 if 10 <= n % 100 < 20 else n % 10, "th")

Although the other solutions above are probably easier to understand at first glance, this works just as well while using a bit less code.

Answer 6

I wanted to use ordinals for a project of mine and after a few prototypes I think this method although not small will work for any positive integer, yes any integer.

It works by determiniting if the number is above or below 20, if the number is below 20 it will turn the int 1 into the string 1st , 2 , 2nd; 3, 3rd; and the rest will have "st" added to it.

For numbers over 20 it will take the last and second to last digits, which I have called the tens and unit respectively and test them to see what to add to the number.

This is in python by the way, so I'm not sure if other languages will be able to find the last or second to last digit on a string if they do it should translate pretty easily.

def o(numb):
    if numb < 20: #determining suffix for < 20
        if numb == 1: 
            suffix = 'st'
        elif numb == 2:
            suffix = 'nd'
        elif numb == 3:
            suffix = 'rd'
        else:
            suffix = 'th'  
    else:   #determining suffix for > 20
        tens = str(numb)
        tens = tens[-2]
        unit = str(numb)
        unit = unit[-1]
        if tens == "1":
           suffix = "th"
        else:
            if unit == "1": 
                suffix = 'st'
            elif unit == "2":
                suffix = 'nd'
            elif unit == "3":
                suffix = 'rd'
            else:
                suffix = 'th'
    return str(numb)+ suffix

I called the function "o" for ease of use and can be called by importing the file name which I called "ordinal" by import ordinal then ordinal.o(number).

Let me know what you think :D

P.S. I've posted this answer on another ordinals question but realised this one is more applicable considering it's python.