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)); // 1.2kconsole.log(kFormatter(-1200)); // -1.2kconsole.log(kFormatter(900)); // 900console.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;}