A simpler way to calculate a numerical multiplier

Question

function num(n) {
  if (n >= 1000 && n < 10000) return (n/1000).toFixed(3) + "K";
  if (n >= 10000 && n < 100000) return (n/1000).toFixed(1) + "K";
  if (n >= 100000 && n < 1000000) return (n/1000).toFixed(0) + "K";
  if (n >= 1000000 && n < 10000000) return (n/1000000).toFixed(3) + "M";
  if (n >= 10000000 && n < 100000000) return (n/1000000).toFixed(1) + "M";
  if (n >= 100000000 && n < 1000000000) return (n/1000000).toFixed(0) + "M";
  if (n >= 1000000000 && n < 10000000000) return (n/1000000000).toFixed(3) + "B";
  if (n >= 10000000000 && n < 100000000000) return (n/1000000000).toFixed(1) + "B";
  if (n >= 100000000000 && n < 1000000000000) return (n/1000000000).toFixed(0) + "B";
  if (n >= 1000000000000 && n < 10000000000000) return (n/1000000000000).toFixed(3) + "T";
  if (n >= 10000000000000 && n < 100000000000000) return (n/1000000000000).toFixed(1) + "T";
  if (n >= 100000000000000 && n < 1000000000000000) return (n/1000000000000).toFixed(0) + "T";
  return n;
}

Since at some point I’m probably going to rise to the power of hundreds, is there an easier way to do this?

Answer 1

function formatNumber(number) {
    var i = 0; units = [ "", "K", "M", "B", "T" ]; // etc
    while (number > 1000) {
        number /= 1000;
        i += 1;
    }
    return Math.floor(number * 1000) / 1000 + units[i];
}

formatNumber(1234567); // 1.234M
formatNumber(1230567); // 1.23M

---

This can be faster for very large numbers:

function formatNumber(number) {
    var i; units = [ "", "K", "M", "B", "T" ]; // etc
    i = Math.round(Math.log(number) / Math.log(10) / 3);
    number /= Math.pow(10, i * 3);
    return Math.floor(number * 1000) / 1000 + units[i];
}

formatNumber(1234567); // 1.234M
formatNumber(1230567); // 1.23M

Answer 2

My approach is here:

function num(n) {
  if(n < 1000) {
    return n
  }

  var units = ['K', 'M', 'B', 'T']
  for(var i = -1; n >= 1000; i++) {
    n /= 1000
  }

  if(n >= 100) {
    return n.toFixed(0) + units[i]
  }
  if(n >= 10) {
    return n.toFixed(1) + units[i]
  }
  return n.toFixed(3) + units[i]
}

Answer 3

Halcyon.

function fnum(number) {
    var i, x, units = [ "", "K", "M", "B", "T", "Qa", "Qi", "Sx", "Sp", "Oc", "No", "Dc" ]; 
    i = Math.floor(Math.log(number) / Math.log(10) / 3);
    x = Math.floor(Math.log(number) / Math.log(10)) % 3;
    number /= Math.pow(10, i * 3);
    return ((number * 1000) / 1000).toFixed(77 % ((x * 2) + 3)) + units[i];
}

0 999 2 1 10 0 100 , . .

fnum(22000); >> 22.00K fnum(153000000); >> 153M fnum(2,230,000,000,000); >> 2.23T