Is there a way to round numbers into a reader friendly format? (e.g. $1.1k) [closed]

Question

Much like the Stackoverlow reputation rounding, I'm hoping to do the same thing with currency

$1,000 => 1k

$1,000,000 => 1m

How can I achieve this in JavaScript (preferably in jQuery)?

Answer 1

Here is a simple function to do it:

function abbrNum(number, decPlaces) {     // 2 decimal places => 100, 3 => 1000, etc     decPlaces = Math.pow(10,decPlaces);      // Enumerate number abbreviations     var abbrev = [ "k", "m", "b", "t" ];      // Go through the array backwards, so we do the largest first     for (var i=abbrev.length-1; i>=0; i--) {          // Convert array index to "1000", "1000000", etc         var size = Math.pow(10,(i+1)*3);          // If the number is bigger or equal do the abbreviation         if(size <= number) {              // Here, we multiply by decPlaces, round, and then divide by decPlaces.              // This gives us nice rounding to a particular decimal place.              number = Math.round(number*decPlaces/size)/decPlaces;               // Handle special case where we round up to the next abbreviation              if((number == 1000) && (i < abbrev.length - 1)) {                  number = 1;                  i++;              }               // Add the letter for the abbreviation              number += abbrev[i];               // We are done... stop              break;         }     }      return number; } 

Outputs:

abbrNum(12 , 1)          => 12 abbrNum(0 , 2)           => 0 abbrNum(1234 , 0)        => 1k abbrNum(34567 , 2)       => 34.57k abbrNum(918395 , 1)      => 918.4k abbrNum(2134124 , 2)     => 2.13m abbrNum(47475782130 , 2) => 47.48b 

Demo: http://jsfiddle.net/jtbowden/SbqKL/

Answer 2

var pow = Math.pow, floor = Math.floor, abs = Math.abs, log = Math.log; var abbrev = 'kmb'; // could be an array of strings: [' m', ' Mo', ' Md']  function round(n, precision) {     var prec = Math.pow(10, precision);     return Math.round(n*prec)/prec; }  function format(n) {     var base = floor(log(abs(n))/log(1000));     var suffix = abbrev[Math.min(2, base - 1)];     base = abbrev.indexOf(suffix) + 1;     return suffix ? round(n/pow(1000,base),2)+suffix : ''+n; } 

Demo:

> tests = [-1001, -1, 0, 1, 2.5, 999, 1234,             1234.5, 1000001, Math.pow(10,9), Math.pow(10,12)] > tests.forEach(function(x){ console.log(x,format(x)) })  -1001 "-1k" -1 "-1" 0 "0" 1 "1" 2.5 "2.5" 999 "999" 1234 "1.23k" 1234.5 "1.23k" 1000001 "1m" 1000000000 "1b" 1000000000000 "1000b"