Producing ordinal numbers

Question

Is it possible to write this in fewer lines of code?

If you input an integer it will output it as an ordinal number if it is less than 100. The below code works perfectly, but I'm wondering if it could be written more succinctly.

def ordinal(self, num):
    """
      Returns ordinal number string from int, e.g. 1, 2, 3 becomes 1st, 2nd, 3rd, etc.
    """
    self.num = num
    n = int(self.num)
    if 4 <= n <= 20:
      suffix = 'th'
    elif n == 1 or (n % 10) == 1:
      suffix = 'st'
    elif n == 2 or (n % 10) == 2:
      suffix = 'nd'
    elif n == 3 or (n % 10) == 3:
      suffix = 'rd'
    elif n < 100:
      suffix = 'th'
    ord_num = str(n) + suffix
    return ord_num

Answer 1

def ordinal(self, num):
    """
      Returns ordinal number string from int, e.g. 1, 2, 3 becomes 1st, 2nd, 3rd, etc.
    """

Its suspicious that this seems to be a method rather than a free standing function.

    self.num = num

Why are you storing the input here? Given the purpose of this function that seems odd.

    n = int(self.num)

Its doubtful that this is a good idea. What are you converting from? Converting to int should be really be done closer to whether this number came from.

    if 4 <= n <= 20:

You've made this case larger than necessary, many of those would be correct even with out this test, and its not clear what so special about the range 4-20.

      suffix = 'th'
    elif n == 1 or (n % 10) == 1:

You don't need the or. If n == 1, then that the second condition will be true anyways.

      suffix = 'st'
    elif n == 2 or (n % 10) == 2:
      suffix = 'nd'
    elif n == 3 or (n % 10) == 3:
      suffix = 'rd'
    elif n < 100:
      suffix = 'th'

What happens if suffix is >= 100? You'll get an error.

    ord_num = str(n) + suffix
    return ord_num

You don't need to split this across two lines.

Here is my version:

# much code can be improved by using a datastructe.
SUFFIXES = {1: 'st', 2: 'nd', 3: 'rd'}
def ordinal(num):
    # I'm checking for 10-20 because those are the digits that
    # don't follow the normal counting scheme. 
    if 10 <= num % 100 <= 20:
        suffix = 'th'
    else:
        # the second parameter is a default.
        suffix = SUFFIXES.get(num % 10, 'th')
    return str(num) + suffix

Answer 2

You can simplify the repeated n == 1 or (n % 10) == 1, as well as special-case test for 11th, 12th, and 13th, by using a ternary expression;

So:

i = n if (n < 20) else (n % 10)
if i == 1:
  suffix = 'st'
elif i == 2:
  suffix = 'nd'
elif i == 3:
  suffix = 'rd'
elif n < 100:
  suffix = 'th'

I think you can also use a dictionary:

suffixes = { 1: "st", 2: "nd", 3: "rd" }
i = n if (n < 20) else (n % 10)
if 0 < i <= 3:
  suffix = suffixes[i]
elif n < 100:
  suffix = 'th'

Or use dictionary get which lets you supply a default value:

suffixes = { 1: "st", 2: "nd", 3: "rd" }
i = n if (n < 20) else (n % 10)
suffix = suffixes.get(i, 'th')

Maybe you can reduce it to one line of code:

suffix = { 1: "st", 2: "nd", 3: "rd" }.get(n if (n < 20) else (n % 10), 'th')

(Beware the above may be buggy because I don't know Python).