Add st, nd, rd and th (ordinal) suffix to a number

Intl.PluralRules, the standard method.

I would just like to drop the canonical way of doing this in here, as nobody seems to know it.

If you want your code to be

  • self-documenting
  • easy to localize
  • with the modern standard

― this is the way to go.

const english_ordinal_rules = new Intl.PluralRules("en", {type: "ordinal"});
const suffixes = {
    one: "st",
    two: "nd",
    few: "rd",
    other: "th"
};
function ordinal(number/*: number */) {
    const category = english_ordinal_rules.select(number);
    const suffix = suffixes[category];
    return (number + suffix);
} // -> string

const test = Array(201)
    .fill()
    .map((_, index) => index - 100)
    .map(ordinal)
    .join(" ");
console.log(test);
  • The Intl.PluralRules constructor (Draft ECMA-402)
  • Unicode’s six plurality categories

Code-golf

While I do not recommend golfing with your code and killing the readability, I came up with one for those golfers (92 bytes):

n=>n+{e:"st",o:"nd",w:"rd",h:"th"}[new Intl.PluralRules("en",{type:"ordinal"}).select(n)[2]]

The rules are as follows:

  • st is used with numbers ending in 1 (e.g. 1st, pronounced first)
  • nd is used with numbers ending in 2 (e.g. 92nd, pronounced ninety-second)
  • rd is used with numbers ending in 3 (e.g. 33rd, pronounced thirty-third)
  • As an exception to the above rules, all the "teen" numbers ending with 11, 12 or 13 use -th (e.g. 11th, pronounced eleventh, 112th, pronounced one hundred [and] twelfth)
  • th is used for all other numbers (e.g. 9th, pronounced ninth).

The following JavaScript code (rewritten in Jun '14) accomplishes this:

function ordinal_suffix_of(i) {
    var j = i % 10,
        k = i % 100;
    if (j == 1 && k != 11) {
        return i + "st";
    }
    if (j == 2 && k != 12) {
        return i + "nd";
    }
    if (j == 3 && k != 13) {
        return i + "rd";
    }
    return i + "th";
}

Sample output for numbers between 0-115:

  0  0th
  1  1st
  2  2nd
  3  3rd
  4  4th
  5  5th
  6  6th
  7  7th
  8  8th
  9  9th
 10  10th
 11  11th
 12  12th
 13  13th
 14  14th
 15  15th
 16  16th
 17  17th
 18  18th
 19  19th
 20  20th
 21  21st
 22  22nd
 23  23rd
 24  24th
 25  25th
 26  26th
 27  27th
 28  28th
 29  29th
 30  30th
 31  31st
 32  32nd
 33  33rd
 34  34th
 35  35th
 36  36th
 37  37th
 38  38th
 39  39th
 40  40th
 41  41st
 42  42nd
 43  43rd
 44  44th
 45  45th
 46  46th
 47  47th
 48  48th
 49  49th
 50  50th
 51  51st
 52  52nd
 53  53rd
 54  54th
 55  55th
 56  56th
 57  57th
 58  58th
 59  59th
 60  60th
 61  61st
 62  62nd
 63  63rd
 64  64th
 65  65th
 66  66th
 67  67th
 68  68th
 69  69th
 70  70th
 71  71st
 72  72nd
 73  73rd
 74  74th
 75  75th
 76  76th
 77  77th
 78  78th
 79  79th
 80  80th
 81  81st
 82  82nd
 83  83rd
 84  84th
 85  85th
 86  86th
 87  87th
 88  88th
 89  89th
 90  90th
 91  91st
 92  92nd
 93  93rd
 94  94th
 95  95th
 96  96th
 97  97th
 98  98th
 99  99th
100  100th
101  101st
102  102nd
103  103rd
104  104th
105  105th
106  106th
107  107th
108  108th
109  109th
110  110th
111  111th
112  112th
113  113th
114  114th
115  115th

Minimal one-line approach for ordinal suffixes

function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}

(this is for positive integers, see below for other variations)

Explanation

Start with an array with the suffixes ["st", "nd", "rd"]. We want to map integers ending in 1, 2, 3 (but not ending in 11, 12, 13) to the indexes 0, 1, 2.

Other integers (including those ending in 11, 12, 13) can be mapped to anything else—indexes not found in the array will evaluate to undefined. This is falsy in javascript and with the use of logical or (|| "th") the expression will return "th" for these integers, which is exactly what we want.

The expression ((n + 90) % 100 - 10) % 10 - 1 does the mapping. Breaking it down:

  • (n + 90) % 100: This expression takes the input integer − 10 mod 100, mapping 10 to 0, ... 99 to 89, 0 to 90, ..., 9 to 99. Now the integers ending in 11, 12, 13 are at the lower end (mapped to 1, 2, 3).
  • - 10: Now 10 is mapped to −10, 19 to −1, 99 to 79, 0 to 80, ... 9 to 89. The integers ending in 11, 12, 13 are mapped to negative integers (−9, −8, −7).
  • % 10: Now all integers ending in 1, 2, or 3 are mapped to 1, 2, 3. All other integers are mapped to something else (11, 12, 13 are still mapped to −9, −8, −7).
  • - 1: Subtracting one gives the final mapping of 1, 2, 3 to 0, 1, 2.

Verifying that it works

function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}

//test integers from 1 to 124
for(var r = [], i = 1; i < 125; i++) r.push(i + nth(i));

//output result
document.getElementById('result').innerHTML = r.join('<br>');
<div id="result"></div>

Variations

Allowing negative integers:

function nth(n){return["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"}


function nth(n){return["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"}

//test integers from 15 to -124
for(var r = [], i = 15; i > -125; i--) r.push(i + nth(i));

//output result
document.getElementById('result').innerHTML = r.join('<br>');
<div id="result"></div>

In ES6 fat arrow syntax (anonymous function):

n=>["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"

Update

An even shorter alternative for positive integers is the expression

[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'

See this post for explanation.

Update 2

[,'st','nd','rd'][n/10%10^1&&n%10]||'th'

From Shopify

function getNumberWithOrdinal(n) {
  var s = ["th", "st", "nd", "rd"],
      v = n % 100;
  return n + (s[(v - 20) % 10] || s[v] || s[0]);
}

[-4,-1,0,1,2,3,4,10,11,12,13,14,20,21,22,100,101,111].forEach(
  n => console.log(n + ' -> ' + getNumberWithOrdinal(n))
);

Tags:

Javascript

Jquery

Numbers

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