javascript - number - ¿Hay alguna manera de redondear los números en un formato amigable para el lector?(por ejemplo, $ 1.1k)

redondear float javascript

Question

Al igual que el redondeo de la reputación de Stackoverlow, espero hacer lo mismo con la moneda

$ 1,000 => 1k

$ 1,000,000 => 1m

¿Ya hay una biblioteca que hace esto? (preferiblemente en jQuery)

Answer 1

Aquí hay una función simple para hacerlo:

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;
}

Productos:

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

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

Answer 2

var pow=Math.pow, floor=Math.floor, abs=Math.abs, log=Math.log;

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 = ''kmb''[base-1];
    return suffix ? round(n/pow(1000,base),2)+suffix : ''''+n;
}

Manifestación:

> 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 "1000000000000"