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