Outputting ordinal numbers (1st, 2nd, 3rd)

Question

I would like to generate (as a return result of a function, or simply as the output of a program) the ordinal suffix of a positive integer concatenated to the number.

Samples:

1st  
2nd  
3rd  
4th  
...  
11th  
12th  
13th  
...  
20th  
21st
22nd
23rd
24th

And so on, with the suffix repeating the initial 1-10 subpattern every 10 until 100, where the pattern ultimately starts over.

The input would be the number and the output the ordinal string as shown above.

What is the smallest algorithm for this?

Answer 1

Perl, 37 + 1 characters

s/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg

This is a regexp substitution that appends the appropriate ordinal suffix to any numbers in $_ that are not already followed by a letter. To apply it to file input, use the p command line switch, like this:

perl -pe 's/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg'

This is a complete Perl program that reads input from stdin and writes the processed output to stdout. The actual code is 37 chars long, but the p switch counts as one extra character.

Sample input:

This is the 1 line of the sample input...
...and this is the 2.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
101 102 103 104 105 106 107 108 109 110

Output:

This is the 1st line of the sample input...
...and this is the 2nd.
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
101st 102nd 103rd 104th 105th 106th 107th 108th 109th 110th

Numbers already followed by letters will be ignored, so feeding the output again through the filter won't change it. Spaces, commas and periods between numbers are not treated specially, so they're assumed to separate numbers like any other punctuation. Thus, e.g. 3.14159 becomes 3rd.14159th.

How does it work?

• First, this is a global regexp replacement (s///g). The regexp being matched is 1?\d\b, where \d matches any digit and \b is a zero-width assertion matching the boundary between an alphanumeric and a non-alphanumeric character. Thus, 1?\d\b matches the last digit of any number, plus the previous digit if it happens to be 1.

• In the substitution, which is evaluated as Perl code due to the /e switch, we take the matched string segment ($&) and append (.) to it the suffix obtained by using $& itself as an integer index to the list (0,st,nd,rd); if this suffix is zero or undefined (i.e. when $& is zero or greater than three), the || operator replaces it with th.

---

Edit: If the input is restricted to a single integer, then this 35 character solution will suffice:

s/1?\d$/$&.((0,st,nd,rd)[$&]||th)/e

Answer 2

Ruby, 60

It's not as good as the Perl entry, but I figured I'd work on my Ruby skills.

def o(n)n.to_s+%w{th st nd rd}[n/10%10==1||n%10>3?0:n%10]end

Function takes one integer argument, n, and returns a string as the ordinal form.

Works according to the following logic:
If the tens digit is a 1 or the ones digit is greater than 3 use the suffix 'th'; otherwise find the suffix from the array ['th', 'st', 'nd', 'rd'] using the final digit as the index.

Answer 3

Python, 68 characters

i=input()
k=i%10
print"%d%s"%(i,"tsnrhtdd"[(i/10%10!=1)*(k<4)*k::4])

Answer 4

Golfscript, 34 characters

~.10/10%1=!1$10%*.4<*'thstndrd'2/=

Answer 5

Mathematica 39

StringSplit[SpokenString[p[[#]]]][[2]] &

Usage

StringSplit[SpokenString[p[[#]]]][[2]] &[31]

31st

StringSplit[SpokenString[p[[#]]]][[2]] &/@Range[21]

{"1st", "2nd", "3rd", "4th", "5th", "6th", "7th", "8th", "9th", "10th", "11th", "12th", "13th", "14th", "15th", "16th", "17th", "18th", "19th", "20th", "21st"}

---

How it works

SpokenString writes out an any valid Mathematica expression as it might be spoken. Below are two examples from the documentation for SpokenString,

SpokenString[Sqrt[x/(y + z)]]

"square root of the quantity x over the quantity y plus z" *)

SpokenString[Graphics3D[Sphere[{{0, 0, 0}, {1, 1, 1}}]], "DetailedGraphics" -> True]

"a three-dimensional graphic consisting of unit spheres centered at 0, 0, 0 and 1, 1, 1"

---

Now, for the example at hand,

SpokenString[p[[#]]] &[31]

"the 31st element of p"

Let's represent the above string as a list of words:

StringSplit[%]

{"the", "31st", "element", "of", "p"}

and take the second element...

%[[2]]

31st

Answer 6

Ocaml

I'm pretty new to ocaml, but this is the shortest i could get.

let n x =
   let v = x mod 100 and string_v = string_of_int x in
   let z = v mod 10 in
   if v=11 || v=12 || v=13 then string_v^"th" 
   else if v = 1 || z = 1 then string_v^"st" else if v = 2 || z = 2 then string_v^"nd" else if v = 3 || z = 3 then string_v^"rd" else string_v^"th";;

I created a function n that takes a number as a parameter and does the work. Its long but thought it'd be great to have a functional example.

Answer 7

PHP, 151

I know that this program is not comparable to the others. Just felt like giving a solution.

<?$s=explode(' ',trim(fgets(STDIN)));foreach($s as$n){echo$n;echo(int)(($n%100)/10)==1?'th':($n%10==1?'st':($n%10==2?'nd':($n%10==3?'rd':'th')))."\n";}

Answer 8

Python 2.7, 137 chars

def b(n):
 for e,o in[[i,'th']for i in['11','12','13']+list('4567890')]+[['2','nd'],['1','st'],['3','rd']]:
  if n.endswith(e):return n+o

n should be a string

I know I'm beaten by the competition here already, but I thought I'd provide my idea anyways

this just basically generates a list of key, value pairs with number (as a string) ending e and the ordinal o. It attempts to match 'th' first (hence why I didn't use a dictionary), so that it won't accidentally return 'st', for example, when it should be 'th'. This will work for any positive integer

Answer 9

PowerShell, 92

process{"$_$(switch -r($_){"(?<!1)1$"{'st'}"(?<!1)2$"{'nd'}"(?<!1)3$"{'rd'}default{'th'}})"}

Works with one number per line of input. Input is given through the pipeline. Making it work for only a single number doesn't reduce the size.

Answer 10

Scala 86

def x(n:Int)=n+{if(n%100/10==1)"th"else(("thstndrd"+"th"*6).sliding(2,2).toSeq(n%10))}

Scala 102:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".take(n+1)%5*2.drop(n%5*2))else n+"th"

102 as well:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".sliding(2,2).toSeq(n%10))else n+"th"

ungolfed:

def x (n: Int) =
  n + { if (((n % 100) / 10) == 1) "th" 
        else (("thstndrd"  + ("th"  * 6)).sliding (2, 2).toSeq (n % 10))
      }

Answer 11

C: 95 characters

A ridiculously long solution:

n;main(){scanf("%d",&n);printf("%d%s",n,n/10%10-1&&(n=n%10)<4&&n?n>2?"rd":n<2?"st":"nd":"th");}

It needs to be mangled more.

Answer 12

Javascript, 75

function o(s){return s+((0|s/10%10)==1?"th":[,"st","nd","rd"][s%10]||"th")}

Answer 13

Haskell, 95 chars

h=foldr g"th".show
g '1'"th"="1st"
g '2'"th"="2nd"
g '3'"th"="3rd"
g '1'[x,_,_]='1':x:"th"
g a s=a:s

Testing:

*Main> map h [1..40]
["1st","2nd","3rd","4th","5th","6th","7th","8th","9th","10th","11th","12th","13t
h","14th","15th","16th","17th","18th","19th","20th","21st","22nd","23rd","24th",
"25th","26th","27th","28th","29th","30th","31st","32nd","33rd","34th","35th","36
th","37th","38th","39th","40th"]

Must be loaded with -XNoMonomorphismRestriction.