Convert digits into words with JavaScript

Question

I am making a code which converts the given amount into words, heres is what I have got after googling. But I think its a little lengthy code to achieve a simple task. Two Regular Expressions and two for loops, I want something simpler.

I am trying to make it as shorter as possible. and will post what I come up with

Any suggestions?

var th = ['','thousand','million', 'billion','trillion'];
var dg = ['zero','one','two','three','four', 'five','six','seven','eight','nine'];
 var tn = ['ten','eleven','twelve','thirteen', 'fourteen','fifteen','sixteen', 'seventeen','eighteen','nineteen'];
 var tw = ['twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];

function toWords(s) {
    s = s.toString();
    s = s.replace(/[\, ]/g,'');
    if (s != parseFloat(s)) return 'not a number';
    var x = s.indexOf('.');
    if (x == -1)
        x = s.length;
    if (x > 15)
        return 'too big';
    var n = s.split(''); 
    var str = '';
    var sk = 0;
    for (var i=0;   i < x;  i++) {
        if ((x-i)%3==2) { 
            if (n[i] == '1') {
                str += tn[Number(n[i+1])] + ' ';
                i++;
                sk=1;
            } else if (n[i]!=0) {
                str += tw[n[i]-2] + ' ';
                sk=1;
            }
        } else if (n[i]!=0) { // 0235
            str += dg[n[i]] +' ';
            if ((x-i)%3==0) str += 'hundred ';
            sk=1;
        }
        if ((x-i)%3==1) {
            if (sk)
                str += th[(x-i-1)/3] + ' ';
            sk=0;
        }
    }

    if (x != s.length) {
        var y = s.length;
        str += 'point ';
        for (var i=x+1; i<y; i++)
            str += dg[n[i]] +' ';
    }
    return str.replace(/\s+/g,' ');
}

Also, the above code converts to English numbering system like Million/Billion, I wan't South Asian numbering system. like in Lakhs and Crores

Answer 1

You might want to try it recursive. It works for numbers between 0 and 999999. Keep in mind that (~~) does the same as Math.floor

var num = "zero one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen".split(" ");
var tens = "twenty thirty forty fifty sixty seventy eighty ninety".split(" ");

function number2words(n){
    if (n < 20) return num[n];
    var digit = n%10;
    if (n < 100) return tens[~~(n/10)-2] + (digit? "-" + num[digit]: "");
    if (n < 1000) return num[~~(n/100)] +" hundred" + (n%100 == 0? "": " " + number2words(n%100));
    return number2words(~~(n/1000)) + " thousand" + (n%1000 != 0? " " + number2words(n%1000): "");
}

Answer 2

I spent a while developing a better solution to this. It can handle very big numbers but once they get over 16 digits you have pass the number in as a string. Something about the limit of JavaScript numbers.

	function numberToEnglish( n ) {
		
		var string = n.toString(), units, tens, scales, start, end, chunks, chunksLen, chunk, ints, i, word, words, and = 'and';

		/* Remove spaces and commas */
		string = string.replace(/[, ]/g,"");

		/* Is number zero? */
		if( parseInt( string ) === 0 ) {
			return 'zero';
		}
		
		/* Array of units as words */
		units = [ '', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen' ];
		
		/* Array of tens as words */
		tens = [ '', '', 'twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety' ];
		
		/* Array of scales as words */
		scales = [ '', 'thousand', 'million', 'billion', 'trillion', 'quadrillion', 'quintillion', 'sextillion', 'septillion', 'octillion', 'nonillion', 'decillion', 'undecillion', 'duodecillion', 'tredecillion', 'quatttuor-decillion', 'quindecillion', 'sexdecillion', 'septen-decillion', 'octodecillion', 'novemdecillion', 'vigintillion', 'centillion' ];
		
		/* Split user arguemnt into 3 digit chunks from right to left */
		start = string.length;
		chunks = [];
		while( start > 0 ) {
			end = start;
			chunks.push( string.slice( ( start = Math.max( 0, start - 3 ) ), end ) );
		}
		
		/* Check if function has enough scale words to be able to stringify the user argument */
		chunksLen = chunks.length;
		if( chunksLen > scales.length ) {
			return '';
		}
		
		/* Stringify each integer in each chunk */
		words = [];
		for( i = 0; i < chunksLen; i++ ) {
			
			chunk = parseInt( chunks[i] );
			
			if( chunk ) {
				
				/* Split chunk into array of individual integers */
				ints = chunks[i].split( '' ).reverse().map( parseFloat );
			
				/* If tens integer is 1, i.e. 10, then add 10 to units integer */
				if( ints[1] === 1 ) {
					ints[0] += 10;
				}
				
				/* Add scale word if chunk is not zero and array item exists */
				if( ( word = scales[i] ) ) {
					words.push( word );
				}
				
				/* Add unit word if array item exists */
				if( ( word = units[ ints[0] ] ) ) {
					words.push( word );
				}
				
				/* Add tens word if array item exists */
				if( ( word = tens[ ints[1] ] ) ) {
					words.push( word );
				}
				
				/* Add 'and' string after units or tens integer if: */
				if( ints[0] || ints[1] ) {
					
					/* Chunk has a hundreds integer or chunk is the first of multiple chunks */
					if( ints[2] || ! i && chunksLen ) {
						words.push( and );
					}
				
				}
				
				/* Add hundreds word if array item exists */
				if( ( word = units[ ints[2] ] ) ) {
					words.push( word + ' hundred' );
				}
				
			}
			
		}
		
		return words.reverse().join( ' ' );
		
	}


// - - - - - Tests - - - - - -
function test(v) {
  var sep = ('string'==typeof v)?'"':'';
  console.log("numberToEnglish("+sep + v.toString() + sep+") = "+numberToEnglish(v));
}
test(2);
test(721);
test(13463);
test(1000001);
test("21,683,200,000,621,384");

Answer 3

This is in response to @LordZardeck's comment to @naomik's excellent answer above. Sorry, I would've commented directly but I've never posted before so I don't have the privilege to do so, so I am posting here instead.

Anyhow, I just happened to translate the ES5 version to a more readable form this past weekend so I'm sharing it here. This should be faithful to the original (including the recent edit) and I hope the naming is clear and accurate.

function int_to_words(int) {
  if (int === 0) return 'zero';

  var ONES  = ['','one','two','three','four','five','six','seven','eight','nine','ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen'];
  var TENS  = ['','','twenty','thirty','fourty','fifty','sixty','seventy','eighty','ninety'];
  var SCALE = ['','thousand','million','billion','trillion','quadrillion','quintillion','sextillion','septillion','octillion','nonillion'];

  // Return string of first three digits, padded with zeros if needed
  function get_first(str) {
    return ('000' + str).substr(-3);
  }

  // Return string of digits with first three digits chopped off
  function get_rest(str) {
    return str.substr(0, str.length - 3);
  }

  // Return string of triplet convereted to words
  function triplet_to_words(_3rd, _2nd, _1st) {
    return (_3rd == '0' ? '' : ONES[_3rd] + ' hundred ') + (_1st == '0' ? TENS[_2nd] : TENS[_2nd] && TENS[_2nd] + '-' || '') + (ONES[_2nd + _1st] || ONES[_1st]);
  }

  // Add to words, triplet words with scale word
  function add_to_words(words, triplet_words, scale_word) {
    return triplet_words ? triplet_words + (scale_word && ' ' + scale_word || '') + ' ' + words : words;
  }

  function iter(words, i, first, rest) {
    if (first == '000' && rest.length === 0) return words;
    return iter(add_to_words(words, triplet_to_words(first[0], first[1], first[2]), SCALE[i]), ++i, get_first(rest), get_rest(rest));
  }

  return iter('', 0, get_first(String(int)), get_rest(String(int)));
}

Answer 4

This is also in response to naomik's excellent post! Unfortunately I don't have the rep to post in the correct place but I leave this here in case it can help anyone.

If you need British English written form you need to make some adaptions to the code. British English differs from the American in a couple of ways. Basically you need to insert the word 'and' in two specific places.

1. After a hundred assuming there are tens and ones. E.g One hundred and ten. One thousand and seventeen. NOT One thousand one hundred and.
2. In certain edges, after a thousand, a million, a billion etc. when there are no smaller units. E.g. One thousand and ten. One million and forty four. NOT One million and one thousand.

The first situation can be addressed by checking for 10s and 1s in the makeGroup method and appending 'and' when they exist.

makeGroup = ([ones,tens,huns]) => {
var adjective = this.num(ones) ? ' hundred and ' : this.num(tens) ? ' hundred and ' : ' hundred';
return [
  this.num(huns) === 0 ? '' : this.a[huns] + adjective,
  this.num(ones) === 0 ? this.b[tens] : this.b[tens] && this.b[tens] + '-' || '',
  this.a[tens+ones] || this.a[ones]
].join('');

};

The second case is more complicated. It is equivalent to

• add 'and' to 'a million, a thousand', or 'a billion' if the antepenultimate number is zero. e.g.

1,100,057 one million one hundred thousand and fifty seven. 5,000,006 five million and six

I think this could be implemented in @naomik's code through the use of a filter function but I wasn't able to work out how. In the end I settled on hackily looping through the returned array of words and using indexOf to look for instances where the word 'hundred' was missing from the final element.

Answer 5

Another conversion that uses remainders and supports different languages:

function numberToWords(number) {
  var result = [];

  var fraction = number.toFixed(2).split('.');
  var integer_part = parseInt(fraction[0]);
  // var fractional_part = parseInt(fraction[1]); -- not handled here

  var previousNumber = null;
  for (var i = 0; i < fraction[0].length; i++) {
    var reminder = Math.floor(integer_part % 10);
    integer_part /= 10;
    var name = getNumberName(reminder, i, fraction[0].length, previousNumber);
    previousNumber = reminder;
    if (name)
      result.push(name);
  }

  result.reverse();
  return result.join(' ');
}

The getNumberName function is language-dependent and handles numbers up to 9999 (but it is easy to extend it to handle larger numbers):

function getNumberName(number, power, places, previousNumber) {
  var result = "";
  if (power == 1) {
    result = handleTeensAndTys(number, previousNumber);
  } else if (power == 0 && places != 1 || number == 0) {
    // skip number that was handled in teens and zero
  } else {
    result = locale.numberNames[number.toString()] + locale.powerNames[power.toString()];
  }

  return result;
}

handleTeensAndTys handles multiples of ten:

function handleTeensAndTys(number, previousNumber) {
  var result = "";
  if (number == 1) { // teens
    if (previousNumber in locale.specialTeenNames) {
      result = locale.specialTeenNames[previousNumber];    
    } else if (previousNumber in locale.specialTyNames) {
      result = locale.specialTyNames[previousNumber] + locale.teenSuffix;
    } else {
      result = locale.numberNames[previousNumber] + locale.teenSuffix;    
    }
  } else if (number == 0) { // previousNumber was not handled in teens
    result = locale.numberNames[previousNumber.toString()];
  } else { // other tys
    if (number in locale.specialTyNames) {
      result = locale.specialTyNames[number];
    } else {
      result = locale.numberNames[number];
    }
    result += locale.powerNames[1];
    if (previousNumber != 0) {
      result += " " + locale.numberNames[previousNumber.toString()];
    }
  }
  return result;
}

Finally, locale examples:

var locale = { // English
  numberNames: {1: "one", 2: "two", 3: "three", 4: "four", 5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine" },
  powerNames: {0: "", 1: "ty", 2: " hundred", 3: " thousand" },
  specialTeenNames: {0: "ten", 1: "eleven", 2: "twelve" },
  specialTyNames: {2: "twen", 3: "thir", 5: "fif" },
  teenSuffix: "teen"
};

var locale = { // Estonian
  numberNames: {1: "üks", 2: "kaks", 3: "kolm", 4: "neli", 5: "viis", 6: "kuus", 7: "seitse", 8: "kaheksa", 9: "üheksa"},
  powerNames: {0: "", 1: "kümmend", 2: "sada", 3: " tuhat" },
  specialTeenNames: {0: "kümme"},
  specialTyNames: {},
  teenSuffix: "teist"
};

Here's a JSFiddle with tests: https://jsfiddle.net/rcrxna7v/15/

Answer 6

Below are the translations from

• integer to word
• float to word
• money to word

Test cases are at the bottom

var ONE_THOUSAND = Math.pow(10, 3);
var ONE_MILLION = Math.pow(10, 6);
var ONE_BILLION = Math.pow(10, 9);
var ONE_TRILLION = Math.pow(10, 12);
var ONE_QUADRILLION = Math.pow(10, 15);
var ONE_QUINTILLION = Math.pow(10, 18);

function integerToWord(integer) {
  var prefix = '';
  var suffix = '';

  if (!integer){ return "zero"; }
  
  if(integer < 0){
    prefix = "negative";
    suffix = integerToWord(-1 * integer);
    return prefix + " " + suffix;
  }
  if(integer <= 90){
    switch (integer) {
      case integer < 0:
        prefix = "negative";
        suffix = integerToWord(-1 * integer);
        return prefix + " "  + suffix;
      case 1: return "one";
      case 2: return "two";
      case 3: return "three";
      case 4:  return "four";
      case 5: return "five";
      case 6: return "six";
      case 7: return "seven";
      case 8: return "eight";
      case 9: return "nine";
      case 10: return "ten";
      case 11: return "eleven";
      case 12: return "twelve";
      case 13: return "thirteen";
      case 14: return "fourteen";
      case 15: return "fifteen";
      case 16: return "sixteen";
      case 17: return "seventeen";
      case 18: return "eighteen";
      case 19: return "nineteen";
      case 20: return "twenty";
      case 30: return "thirty";
      case 40: return "forty";
      case 50: return "fifty";
      case 60: return "sixty";
      case 70: return "seventy";
      case 80: return "eighty";
      case 90: return "ninety";
      default: break;
    }
  }

  if(integer < 100){
    prefix = integerToWord(integer - integer % 10);
    suffix = integerToWord(integer % 10);
    return prefix + "-"  + suffix;
  }

  if(integer < ONE_THOUSAND){
    prefix = integerToWord(parseInt(Math.floor(integer / 100), 10) )  + " hundred";
    if (integer % 100){ suffix = " and "  + integerToWord(integer % 100); }
    return prefix + suffix;
  }

  if(integer < ONE_MILLION){
    prefix = integerToWord(parseInt(Math.floor(integer / ONE_THOUSAND), 10))  + " thousand";
    if (integer % ONE_THOUSAND){ suffix = integerToWord(integer % ONE_THOUSAND); }
  }
  else if(integer < ONE_BILLION){
    prefix = integerToWord(parseInt(Math.floor(integer / ONE_MILLION), 10))  + " million";
    if (integer % ONE_MILLION){ suffix = integerToWord(integer % ONE_MILLION); }
  }
  else if(integer < ONE_TRILLION){
    prefix = integerToWord(parseInt(Math.floor(integer / ONE_BILLION), 10))  + " billion";
    if (integer % ONE_BILLION){ suffix = integerToWord(integer % ONE_BILLION); }
  }
  else if(integer < ONE_QUADRILLION){
    prefix = integerToWord(parseInt(Math.floor(integer / ONE_TRILLION), 10))  + " trillion";
    if (integer % ONE_TRILLION){ suffix = integerToWord(integer % ONE_TRILLION); }
  }
  else if(integer < ONE_QUINTILLION){
    prefix = integerToWord(parseInt(Math.floor(integer / ONE_QUADRILLION), 10))  + " quadrillion";
    if (integer % ONE_QUADRILLION){ suffix = integerToWord(integer % ONE_QUADRILLION); }
  } else {
    return '';
  }
  return prefix + " "  + suffix;
}

function moneyToWord(value){
  var decimalValue = (value % 1);
  var integer = value - decimalValue;
  decimalValue = Math.round(decimalValue * 100);
  var decimalText = !decimalValue? '': integerToWord(decimalValue) + ' cent' + (decimalValue === 1? '': 's');
  var integerText= !integer? '': integerToWord(integer) + ' dollar' + (integer === 1? '': 's');
  return (
    integer && !decimalValue? integerText:
    integer && decimalValue? integerText + ' and ' + decimalText:
    !integer && decimalValue? decimalText:
    'zero cents'
  );
}

function floatToWord(value){
  var decimalValue = (value % 1);
  var integer = value - decimalValue;
  decimalValue = Math.round(decimalValue * 100);
  var decimalText = !decimalValue? '':
    decimalValue < 10? "point o' " + integerToWord(decimalValue):
    decimalValue % 10 === 0? 'point ' + integerToWord(decimalValue / 10):
    'point ' + integerToWord(decimalValue);
  return (
    integer && !decimalValue? integerToWord(integer):
    integer && decimalValue? [integerToWord(integer),  decimalText].join(' '):
    !integer && decimalValue? decimalText:
    integerToWord(0)
  );
}

// test
(function(){
  console.log('integerToWord ==================================');
  for(var i = 0; i < 101; ++i){
    console.log('%s=%s', i, integerToWord(i));
  }
  console.log('floatToWord ====================================');
  i = 131;
  while(i--){
    console.log('%s=%s', i / 100, floatToWord(i / 100));
  }
  console.log('moneyToWord ====================================');
  for(i = 0; i < 131; ++i){
    console.log('%s=%s', i / 100, moneyToWord(i / 100));
  }
}());