Spelling of ordinal numbers
Ordinal numbers (as used in this Rosetta Code task), are numbers that describe the position of something in a list.
It is this context that ordinal numbers will be used, using an English-spelled name of an ordinal number.
The ordinal numbers are (at least, one form of them):
1st 2nd 3rd 4th 5th 6th 7th ··· 99th 100th ··· 1000000000th ··· etc
sometimes expressed as:
1st 2nd 3rd 4th 5th 6th 7th ··· 99th 100th ··· 1000000000th ···
For this task, the following (English-spelled form) will be used:
first second third fourth fifth sixth seventh ninety-nineth one hundredth one billionth
Furthermore, the American version of numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
- Task
Write a driver and a function (subroutine/routine ···) that returns the English-spelled ordinal version of a specified number (a positive integer).
Optionally, try to support as many forms of an integer that can be expressed: 123 00123.0 1.23e2 all are forms of the same integer.
Show all output here.
- Test cases
Use (at least) the test cases of:
1 2 3 4 5 11 65 100 101 272 23456 8007006005004003
- Related tasks
Julia
This makes use of code posted on this site by MichaeLeroy for a similar task at https://rosettacode.org/wiki/Number_names#Julia. The function num2text is used (but not included here) as posted from that location. <lang Julia> const irregular = Dict("one" => "first", "two" => "second", "three" => "third",
"five" => "fifth", "nine" => "ninth", "twelve" => "twelfth")
const suffix = "th" const ysuffix = "ieth"
function numtext2ordinal(s)
lastword = split(s)[end]
redolast = split(lastword, "-")[end]
if redolast != lastword
lastsplit = "-"
word = redolast
else
lastsplit = " "
word = lastword
end
firstpart = reverse(split(reverse(s), lastsplit, limit=2)[end])
firstpart = (firstpart == word) ? "": firstpart * lastsplit
if haskey(irregular, word)
word = irregular[word]
elseif word[end] == 'y'
word = word[1:end-1] * ysuffix
else
word = word * suffix
end
firstpart * word
end
const testcases = [1 2 3 4 5 11 65 100 101 272 23456 8007006005004003] for n in testcases
println("$n => $(numtext2ordinal(num2text(n)))")
end </lang>
- Output:
1 => first 2 => second 3 => third 4 => fourth 5 => fifth 11 => eleventh 65 => sixty-fifth 100 => one hundredth 101 => one hundred and first 272 => two hundred and seventy-second 23456 => twenty-three thousand four hundred and fifty-sixth 8007006005004003 => eight quadrillion seven trillion six billion five million four thousand and third
Kotlin
This makes use of the code at https://rosettacode.org/wiki/Number_names#Kotlin, which I also wrote, and adjusts it to output the corresponding ordinal. Although, for good measure, the program can deal with negative integers, zero and UK-style numbers (the insertion of 'and' at strategic places, no 'milliards' I promise!) none of these are actually tested in line with the task's requirements. <lang scala>// version 1.1.4-3
typealias IAE = IllegalArgumentException
val names = mapOf(
1 to "one", 2 to "two", 3 to "three", 4 to "four", 5 to "five", 6 to "six", 7 to "seven", 8 to "eight", 9 to "nine", 10 to "ten", 11 to "eleven", 12 to "twelve", 13 to "thirteen", 14 to "fourteen", 15 to "fifteen", 16 to "sixteen", 17 to "seventeen", 18 to "eighteen", 19 to "nineteen", 20 to "twenty", 30 to "thirty", 40 to "forty", 50 to "fifty", 60 to "sixty", 70 to "seventy", 80 to "eighty", 90 to "ninety"
) val bigNames = mapOf(
1_000L to "thousand", 1_000_000L to "million", 1_000_000_000L to "billion", 1_000_000_000_000L to "trillion", 1_000_000_000_000_000L to "quadrillion", 1_000_000_000_000_000_000L to "quintillion"
)
val irregOrdinals = mapOf(
"one" to "first", "two" to "second", "three" to "third", "five" to "fifth", "eight" to "eighth", "nine" to "ninth", "twelve" to "twelfth"
)
fun String.toOrdinal(): String {
val splits = this.split(' ', '-')
var last = splits[splits.lastIndex]
return if (irregOrdinals.containsKey(last)) this.dropLast(last.length) + irregOrdinals[last]!!
else if (last.endsWith("y")) this.dropLast(1) + "ieth"
else this + "th"
}
fun numToOrdinalText(n: Long, uk: Boolean = false): String {
if (n == 0L) return "zeroth" // or alternatively 'zeroeth'
val neg = n < 0L
val maxNeg = n == Long.MIN_VALUE
var nn = if (maxNeg) -(n + 1) else if (neg) -n else n
val digits3 = IntArray(7)
for (i in 0..6) { // split number into groups of 3 digits from the right
digits3[i] = (nn % 1000).toInt()
nn /= 1000
}
fun threeDigitsToText(number: Int) : String {
val sb = StringBuilder()
if (number == 0) return ""
val hundreds = number / 100
val remainder = number % 100
if (hundreds > 0) {
sb.append(names[hundreds], " hundred")
if (remainder > 0) sb.append(if (uk) " and " else " ")
}
if (remainder > 0) {
val tens = remainder / 10
val units = remainder % 10
if (tens > 1) {
sb.append(names[tens * 10])
if (units > 0) sb.append("-", names[units])
}
else sb.append(names[remainder])
}
return sb.toString()
}
val strings = Array<String>(7) { threeDigitsToText(digits3[it]) }
var text = strings[0]
var andNeeded = uk && digits3[0] in 1..99
var big = 1000L
for (i in 1..6) {
if (digits3[i] > 0) {
var text2 = strings[i] + " " + bigNames[big]
if (text.length > 0) {
text2 += if (andNeeded) " and " else ", "
andNeeded = false
}
else andNeeded = uk && digits3[i] in 1..99
text = text2 + text
}
big *= 1000
}
if (maxNeg) text = text.dropLast(5) + "eight"
if (neg) text = "minus " + text
return text.toOrdinal()
}
fun numToOrdinalText(s: String, uk: Boolean = false): String {
val d = s.toDoubleOrNull() ?: throw IAE("String is not numeric")
if (d !in Long.MIN_VALUE.toDouble() .. Long.MAX_VALUE.toDouble())
throw IAE("Double is outside the range of a Long Integer")
val n = d.toLong()
if (n.toDouble() != d) throw IAE("String does not represent a Long Integer")
return numToOrdinalText(n, uk)
}
fun main(args: Array<String>) {
val la = longArrayOf(1, 2, 3, 4, 5, 11, 65, 100, 101, 272, 23456, 8007006005004003)
println("Using US representation:")
for (i in la) println("${"%16d".format(i)} = ${numToOrdinalText(i)}")
val sa = arrayOf("123", "00123.0", "1.23e2")
for (s in sa) println("${"%16s".format(s)} = ${numToOrdinalText(s)}")
}</lang>
- Output:
Using US representation:
1 = first
2 = second
3 = third
4 = fourth
5 = fifth
11 = eleventh
65 = sixty-fifth
100 = one hundredth
101 = one hundred first
272 = two hundred seventy-second
23456 = twenty-three thousand, four hundred fifty-sixth
8007006005004003 = eight quadrillion, seven trillion, six billion, five million, four thousand, third
123 = one hundred twenty-third
00123.0 = one hundred twenty-third
1.23e2 = one hundred twenty-third
Perl 6
This would be pretty simple to implement from scratch; it would be straightforward to do a minor modification of the Number names task code. Much simpler to just use the Lingua::EN::Numbers::Cardinal module from the Perl 6 ecosystem though. It easily handles ordinal numbers even though that is not its primary focus.
We need to be slightly careful of terminology. In Perl 6, 123, 00123.0, & 1.23e2 are not all integers. They are respectively an Int (integer), a Rat (rational number) and a Num (floating point number). For this task it doesn't much matter as the ordinal routine coerces its argument to an Int, but to Perl 6 they are different things. We can further abuse allomorphic types for some somewhat non-intuitive results as well.
It is not really clear what is meant by "Write a driver and a function...". Well, the function part is clear enough; driver not so much. Perhaps this will suffice.
<lang perl6>use Lingua::EN::Numbers::Cardinal;
printf( "\%16s : %s\n", $_, ordinal($_) ) for
- Required tests
|<1 2 3 4 5 11 65 100 101 272 23456 8007006005004003>,
- Optional tests
|<123 00123.0 1.23e2 123+0i 0b1111011 0o173 0x7B 861/7>;</lang>
- Output:
1 : first
2 : second
3 : third
4 : fourth
5 : fifth
11 : eleventh
65 : sixty-fifth
100 : one hundredth
101 : one hundred first
272 : two hundred seventy-second
23456 : twenty-three thousand, four hundred fifty-sixth
8007006005004003 : eight quadrillion, seven trillion, six billion, five million, four thousand third
123 : one hundred twenty-third
00123.0 : one hundred twenty-third
1.23e2 : one hundred twenty-third
123+0i : one hundred twenty-third
0b1111011 : one hundred twenty-third
0o173 : one hundred twenty-third
0x7B : one hundred twenty-third
861/7 : one hundred twenty-third
REXX
<lang REXX>/*REXX programs spells out ordinal numbers (in English, using the American system). */ numeric digits 3000 /*just in case the user uses gihugic #s*/ parse arg n /*obtain optional arguments from the CL*/
if n= | n="," then n= 1 2 3 4 5 11 65 100 101 272 23456 8007006005004003
pgmOpts= 'ordinal quiet' /*define options needed for $SPELL#.REX*/
do j=1 for words(n) /*process each of the specified numbers*/
x=word(n, j) /*obtain a number from the input list. */
os=$spell#(x pgmOpts) /*invoke REXX routine to spell ordinal#*/
say right(x, max(20, length(x) ) ) ' spelled ordinal number ───► ' os
end /*j*/</lang>
- output when using the default inputs:
1 spelled ordinal number ───► first
2 spelled ordinal number ───► second
3 spelled ordinal number ───► third
4 spelled ordinal number ───► fourth
5 spelled ordinal number ───► fifth
11 spelled ordinal number ───► eleventh
65 spelled ordinal number ───► sixty-fifth
100 spelled ordinal number ───► one hundredth
101 spelled ordinal number ───► one hundred first
272 spelled ordinal number ───► two hundred seventy-second
23456 spelled ordinal number ───► twenty-three thousand four hundred fifty-sixth
8007006005004003 spelled ordinal number ───► eight quadrillion seven trillion six billion five million four thousand third
The $SPELL#.REX routine can be found here ───► $SPELL#.REX.
zkl
<lang zkl>fcn nth(n,th=True){
var [const]
nmsth=T("","first","second","third","fourth","fifth","sixth","seventh","eighth","ninth"),
nms1=T("","one","two","three","four","five","six","seven","eight","nine"),
nms10=T("ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"),
nms10th=T("tenth","eleventh","twelfth","thirteenth","fourteenth","fifteenth","sixteenth","seventeenth","eighteenth","nineteenth"),
nms20=T("twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"),
nms1000=T("thousand","million","billion","trillion","quadrillion"); // 3,6,9,12,15
if (n<0) String("negative ",nth(-n,th));
else if(n<10) th and nmsth[n] or nms1[n];
else if(n<20) th and nms10th[n-10] or nms10[n-10];
else if(n<10) th and nmsth[n] or nms1[n];
else if(n<100){
m,txt := n%10,nms20[n/10-2];
if(m) String(txt,dash(n%10,"-",th));
else String(txt[0,-1],"ieth");
}
else if(n<1000) String(nms1[n/100]," hundred",dash(n%100," ",th));
else{
n=n.toInt(); // yuck, only here to handle floats, 1.23-->"first"
ds:=(n.numDigits()-1)/3*3; // 1e3->3, 1e4-->3, 1e5-->3, 1e6-->6, 1e7-->6
z:=(10).pow(ds); // 1234-->1000, 12345-->10000
thou:=ds/3 - 1; // 1000-->0, 10000-->0, 2,000,000-->1
nnn,ys := n/z, n%z;
String((if(nnn<10) nms1[nnn] else nth(nnn,False)),
" ",nms1000[thou], if(ys==0) "th" else String(" ",nth(ys,th)));
}
} fcn dash(n,d,th){ if(n) String(d,nth(n,th)) else (th and "th" or "") }</lang> <lang zkl>testNs:=L(1,2,3,4,5,11,65,100,101,272,23456,8007006005004003,
123,00123.0,1.23e2,);
foreach n in (testNs){
if(n.isType(Float)) println("%16.2f --> %s".fmt(n,nth(n)));
else println("%16d --> %s".fmt(n,nth(n)));
}</lang>
- Output:
1 --> first
2 --> second
3 --> third
4 --> fourth
5 --> fifth
11 --> eleventh
65 --> sixty-fifth
100 --> one hundredth
101 --> one hundred first
272 --> two hundred seventy-second
23456 --> twenty-three thousand four hundred fifty-sixth
8007006005004003 --> eight quadrillion seven trillion six billion five million four thousand third
123 --> one hundred twenty-third
123.00 --> one hundred twenty-third
123.00 --> one hundred twenty-third