Convert long number into abbreviated string in JavaScript, with a special shortness requirement

Question

In JavaScript, how would one write a function that converts a given [edit: positive integer] number (below 100 billion) into a 3-letter abbreviation -- where 0-9 an

Answer 1

Use as a Number Prototype

For easy and direct you. Simply make a prototype of it. Here is an example.

Number.prototype.abbr = function (decimal = 2): string {
  const notations = ['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
    i = Math.floor(Math.log(this) / Math.log(1000));
  return `${parseFloat((this / Math.pow(1000, i)).toFixed(decimal))}${notations[i]}`;
};

Answer 2

Indian Currency format to (K, L, C) Thousand, Lakh, Crore

const formatCash = n => {
  if (n < 1e3) return n;
  if (n >= 1e3 && n < 1e5) return +(n / 1e3).toFixed(1) + "K";
  if (n >= 1e5 && n <= 1e6) return +(n / 1e5).toFixed(1) + "L";
  if (n >= 1e6 && n <= 1e9) return +(n / 1e7).toFixed(1) + "C";
};

Answer 3

I'm using this function to get these values.

function Converter(number, fraction) {
    let ranges = [
      { divider: 1, suffix: '' },
      { divider: 1e3, suffix: 'K' },
      { divider: 1e6, suffix: 'M' },
      { divider: 1e9, suffix: 'G' },
      { divider: 1e12, suffix: 'T' },
      { divider: 1e15, suffix: 'P' },
      { divider: 1e18, suffix: 'E' },
    ]
    //find index based on number of zeros
    let index = (Math.abs(number).toString().length / 3).toFixed(0)
    return (number / ranges[index].divider).toFixed(fraction) + ranges[index].suffix
}

Each 3 digits has different suffix, that's what i'm trying to find firstly.

So, remove negative symbol if exists, then find how many 3 digits in this number.

after that find appropriate suffix based on previous calculation added to divided number.

Converter(1500, 1)

Will return:

1.5K

Answer 4

This handles very large values as well and is a bit more succinct and efficient.

abbreviate_number = function(num, fixed) {
  if (num === null) { return null; } // terminate early
  if (num === 0) { return '0'; } // terminate early
  fixed = (!fixed || fixed < 0) ? 0 : fixed; // number of decimal places to show
  var b = (num).toPrecision(2).split("e"), // get power
      k = b.length === 1 ? 0 : Math.floor(Math.min(b[1].slice(1), 14) / 3), // floor at decimals, ceiling at trillions
      c = k < 1 ? num.toFixed(0 + fixed) : (num / Math.pow(10, k * 3) ).toFixed(1 + fixed), // divide by power
      d = c < 0 ? c : Math.abs(c), // enforce -0 is 0
      e = d + ['', 'K', 'M', 'B', 'T'][k]; // append power
  return e;
}

Results:

for(var a='', i=0; i < 14; i++){ 
    a += i; 
    console.log(a, abbreviate_number(parseInt(a),0)); 
    console.log(-a, abbreviate_number(parseInt(-a),0)); 
}

0 0
-0 0
01 1
-1 -1
012 12
-12 -12
0123 123
-123 -123
01234 1.2K
-1234 -1.2K
012345 12.3K
-12345 -12.3K
0123456 123.5K
-123456 -123.5K
01234567 1.2M
-1234567 -1.2M
012345678 12.3M
-12345678 -12.3M
0123456789 123.5M
-123456789 -123.5M
012345678910 12.3B
-12345678910 -12.3B
01234567891011 1.2T
-1234567891011 -1.2T
0123456789101112 123.5T
-123456789101112 -123.5T
012345678910111213 12345.7T
-12345678910111212 -12345.7T