进一步改进Salman's Answer，因为它将nFormatter(33000)返回为33.0K

function nFormatter(num) {
     if (num >= 1000000000) {
        return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
     }
     if (num >= 1000000) {
        return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
     }
     if (num >= 1000) {
        return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
     }
     return num;
}

now nFormatter(33000) = 33K

如果你喜欢，就把功劳归于韦伦·弗林

这比他处理负数和“。0”的情况。

循环和“如果”情况越少，IMO就越好。

function abbreviateNumber(number) {
    const SI_POSTFIXES = ["", "k", "M", "G", "T", "P", "E"];
    const sign = number < 0 ? '-1' : '';
    const absNumber = Math.abs(number);
    const tier = Math.log10(absNumber) / 3 | 0;
    // if zero, we don't need a prefix
    if(tier == 0) return `${absNumber}`;
    // get postfix and determine scale
    const postfix = SI_POSTFIXES[tier];
    const scale = Math.pow(10, tier * 3);
    // scale the number
    const scaled = absNumber / scale;
    const floored = Math.floor(scaled * 10) / 10;
    // format number and add postfix as suffix
    let str = floored.toFixed(1);
    // remove '.0' case
    str = (/\.0$/.test(str)) ? str.substr(0, str.length - 2) : str;
    return `${sign}${str}${postfix}`;
}

jsFiddle测试用例-> https://jsfiddle.net/qhbrz04o/9/

这是非常优雅的。

function formatToUnits(number, precision) {
  const abbrev = ['', 'k', 'm', 'b', 't'];
  const unrangifiedOrder = Math.floor(Math.log10(Math.abs(number)) / 3)
  const order = Math.max(0, Math.min(unrangifiedOrder, abbrev.length -1 ))
  const suffix = abbrev[order];

  return (number / Math.pow(10, order * 3)).toFixed(precision) + suffix;
}

formatToUnits(12345, 2)
==> "12.35k"
formatToUnits(0, 3)
==> "0.000"

Waylon flynn的答案的修改版本，支持负指数:

function metric(number) { const SI_SYMBOL = [ ["", "k", "M", "G", "T", "P", "E"], // + ["", "m", "μ", "n", "p", "f", "a"] // - ]; const tier = Math.floor(Math.log10(Math.abs(number)) / 3) | 0; const n = tier < 0 ? 1 : 0; const t = Math.abs(tier); const scale = Math.pow(10, tier * 3); return { number: number, symbol: SI_SYMBOL[n][t], scale: scale, scaled: number / scale } } function metric_suffix(number, precision) { const m = metric(number); return (typeof precision === 'number' ? m.scaled.toFixed(precision) : m.scaled) + m.symbol; } for (var i = 1e-6, s = 1; i < 1e7; i *= 10, s *= -1) { // toggles sign in each iteration console.log(metric_suffix(s * (i + i / 5), 1)); } console.log(metric(0));

预期的输出:

   1.2μ
 -12.0μ
 120.0μ
  -1.2m
  12.0m
-120.0m
   1.2
 -12.0
 120.0
  -1.2k
  12.0k
-120.0k
   1.2M
{ number: 0, symbol: '', scale: 1, scaled: 0 }

这里有一个简单的解决方案，可以避免所有的if语句(借助Math的力量)。

var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];

function abbreviateNumber(number){

    // what tier? (determines SI symbol)
    var tier = Math.log10(Math.abs(number)) / 3 | 0;

    // if zero, we don't need a suffix
    if(tier == 0) return number;

    // get suffix and determine scale
    var suffix = SI_SYMBOL[tier];
    var scale = Math.pow(10, tier * 3);

    // scale the number
    var scaled = number / scale;

    // format number and add suffix
    return scaled.toFixed(1) + suffix;
}

奖金模因

SI代表什么?

/**
 * Shorten number to thousands, millions, billions, etc.
 * http://en.wikipedia.org/wiki/Metric_prefix
 *
 * @param {number} num Number to shorten.
 * @param {number} [digits=0] The number of digits to appear after the decimal point.
 * @returns {string|number}
 *
 * @example
 * // returns '12.5k'
 * shortenLargeNumber(12543, 1)
 *
 * @example
 * // returns '-13k'
 * shortenLargeNumber(-12567)
 *
 * @example
 * // returns '51M'
 * shortenLargeNumber(51000000)
 *
 * @example
 * // returns 651
 * shortenLargeNumber(651)
 *
 * @example
 * // returns 0.12345
 * shortenLargeNumber(0.12345)
 */
function shortenLargeNumber(num, digits) {
    var units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
        decimal;

    for(var i=units.length-1; i>=0; i--) {
        decimal = Math.pow(1000, i+1);

        if(num <= -decimal || num >= decimal) {
            return +(num / decimal).toFixed(digits) + units[i];
        }
    }

    return num;
}

谢谢@Cos的评论，我删除了Math。round10依赖。