Format a number as 2.5K if a thousand or more, otherwise 900

Question

A more generalized version:

function nFormatter(num, digits) {
  const lookup = [
    { value: 1, symbol: "" },
    { value: 1e3, symbol: "k" },
    { value: 1e6, symbol: "M" },
    { value: 1e9, symbol: "G" },
    { value: 1e12, symbol: "T" },
    { value: 1e15, symbol: "P" },
    { value: 1e18, symbol: "E" }
  ];
  const rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
  var item = lookup.slice().reverse().find(function(item) {
    return num >= item.value;
  });
  return item ? (num / item.value).toFixed(digits).replace(rx, "$1") + item.symbol : "0";
}

/*
 * Tests
 */
const tests = [
  { num: 0, digits: 1 },
  { num: 12, digits: 1 },
  { num: 1234, digits: 1 },
  { num: 100000000, digits: 1 },
  { num: 299792458, digits: 1 },
  { num: 759878, digits: 1 },
  { num: 759878, digits: 0 },
  { num: 123, digits: 1 },
  { num: 123.456, digits: 1 },
  { num: 123.456, digits: 2 },
  { num: 123.456, digits: 4 }
];
tests.forEach(function(test) {
  console.log("nFormatter(" + test.num + ", " + test.digits + ") = " + nFormatter(test.num, test.digits));
});

Sounds like this should work for you:

function kFormatter(num) {
    return Math.abs(num) > 999 ? Math.sign(num)*((Math.abs(num)/1000).toFixed(1)) + 'k' : Math.sign(num)*Math.abs(num)
}
    
console.log(kFormatter(1200)); // 1.2k
console.log(kFormatter(-1200)); // -1.2k
console.log(kFormatter(900)); // 900
console.log(kFormatter(-900)); // -900

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(Math.abs(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?

ES2020 adds support for this in Intl.NumberFormat Using notation as follows:

let formatter = Intl.NumberFormat('en', { notation: 'compact' });
// example 1
let million = formatter.format(1e6);
// example 2
let billion = formatter.format(1e9);
// print
console.log(million == '1M', billion == '1B');

Note as shown above, that the second example produces 1B instead of 1G. NumberFormat specs:

• https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat
• https://tc39.es/ecma402#numberformat-objects

Note that at the moment not all browsers support ES2020, so you may need this Polyfill: https://formatjs.io/docs/polyfills/intl-numberformat

Further improving Salman's Answer because it returns nFormatter(33000) as 33.0K

function nFormatter(num) {
     if (num >= 1000000000) {
        return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
     }
     if (num >= 1000000) {
        return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
     }
     if (num >= 1000) {
        return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
     }
     return num;
}

now nFormatter(33000) = 33K

A straight-forward approach has the best readability, and uses the least memory. No need to over-engineer with the use of regex, map objects, Math objects, for-loops, etc.

Formatting Cash value with K

const formatCash = n => {
  if (n < 1e3) return n;
  if (n >= 1e3) return +(n / 1e3).toFixed(1) + "K";
};

console.log(formatCash(2500));

Formatting Cash value with K M B T

const formatCash = n => {
  if (n < 1e3) return n;
  if (n >= 1e3 && n < 1e6) return +(n / 1e3).toFixed(1) + "K";
  if (n >= 1e6 && n < 1e9) return +(n / 1e6).toFixed(1) + "M";
  if (n >= 1e9 && n < 1e12) return +(n / 1e9).toFixed(1) + "B";
  if (n >= 1e12) return +(n / 1e12).toFixed(1) + "T";
};

console.log(formatCash(1235000));

Using negative numbers

let format;
const number = -1235000;

if (number < 0) {
  format = '-' + formatCash(-1 * number);
} else {
  format = formatCash(number);
}