Format a number as 2.5K if a thousand or more, oth

Question

I need to show a currency value in the format of 1K of equal to one thousand, or 1.1K, 1.2K, 1.9K etc, if its not an even thousands, otherwise if under a thousand, display normal 500, 100, 250 etc, using javascript to format the number?

Answer 1

Here's a simple solution that avoids all the if statements (with the power of Math).

var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];

function abbreviateNumber(number){

    // what tier? (determines SI symbol)
    var tier = Math.log10(number) / 3 | 0;

    // if zero, we don't need a suffix
    if(tier == 0) return number;

    // get suffix and determine scale
    var suffix = SI_SYMBOL[tier];
    var scale = Math.pow(10, tier * 3);

    // scale the number
    var scaled = number / scale;

    // format number and add suffix
    return scaled.toFixed(1) + suffix;
}

Bonus Meme

What does SI stand for?

Answer 2

You can use the d3-format package modeled after Python Advanced String Formatting PEP3101 :

var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"

Answer 3

Give Credit to Waylon Flinn if you like this

This was improved from his more elegant approach to handle negative numbers and ".0" case.

The fewer loops and "if" cases you have, the better IMO.

function abbreviateNumber(number) {
    var SI_POSTFIXES = ["", "k", "M", "G", "T", "P", "E"];
    var tier = Math.log10(Math.abs(number)) / 3 | 0;
    if(tier == 0) return number;
    var postfix = SI_POSTFIXES[tier];
    var scale = Math.pow(10, tier * 3);
    var scaled = number / scale;
    var formatted = scaled.toFixed(1) + '';
    if (/\.0$/.test(formatted))
      formatted = formatted.substr(0, formatted.length - 2);
    return formatted + postfix;
}

jsFiddle with test cases -> https://jsfiddle.net/xyug4nvz/7/

Answer 4

this is is quite elegant.

function formatToUnits(number, precision) {
  const abbrev = ['', 'k', 'm', 'b', 't'];
  const unrangifiedOrder = Math.floor(Math.log10(Math.abs(number)) / 3)
  const order = Math.max(0, Math.min(unrangifiedOrder, abbrev.length -1 ))
  const suffix = abbrev[order];

  return (number / Math.pow(10, order * 3)).toFixed(precision) + suffix;
}

formatToUnits(12345, 2)
==> "12.35k"
formatToUnits(0, 3)
==> "0.000"

Answer 5

Further improving @Yash's answer with negative number support:

function nFormatter(num) {
    isNegative = false
    if (num < 0) {
        isNegative = true
    }
    num = Math.abs(num)
    if (num >= 1000000000) {
        formattedNumber = (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
    } else if (num >= 1000000) {
        formattedNumber =  (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
    } else  if (num >= 1000) {
        formattedNumber =  (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
    } else {
        formattedNumber = num;
    }   
    if(isNegative) { formattedNumber = '-' + formattedNumber }
    return formattedNumber;
}

nFormatter(-120000)
"-120K"
nFormatter(120000)
"120K"

Answer 6

Sounds like this should work for you:

function kFormatter(num) {
    return num > 999 ? (num/1000).toFixed(1) + 'k' : num
}

console.log(kFormatter(1200));
console.log(kFormatter(900));