javascript - Easier way to calculate numeric multiplier suffix -

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 @ point i'm going going upwards power of hundreds, there easier way this?

function formatnumber(number) {     var = 0; units = [ "", "k", "m", "b", "t" ]; // etc     while (number > 1000) {         number /= 1000;         += 1;     }     return math.floor(number * 1000) / 1000 + units[i]; }  formatnumber(1234567); // 1.234m formatnumber(1230567); // 1.23m 

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this might faster large numbers:

function formatnumber(number) {     var i; units = [ "", "k", "m", "b", "t" ]; // etc     = math.round(math.log(number) / math.log(10) / 3);     number /= math.pow(10, * 3);     return math.floor(number * 1000) / 1000 + units[i]; }  formatnumber(1234567); // 1.234m formatnumber(1230567); // 1.23m