加上上面的答案，这将给出1000的1k而不是1.0k

function kFormatter(num) {
    return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}

ES2020在Intl中增加了对此的支持。使用如下表示法:

let formatter = Intl。NumberFormat('en'，{符号:'紧凑'}); //示例1 让million = formatter.format(1e6); //示例2 Let billion = formatter.format(1e9); / /打印 console.log(million == '1M'， billion == '1B');

注意如上所示，第二个示例生成1B而不是1G。 NumberFormat规格:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat https://tc39.es/ecma402#numberformat-objects

注意，目前并不是所有的浏览器都支持ES2020，所以你可能需要这个 Polyfill: https://formatjs.io/docs/polyfills/intl-numberformat

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"

通过消除@martin-sznapka解决方案中的循环，您将减少40%的执行时间。

function formatNum(num,digits) {
    let units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'];
    let floor = Math.floor(Math.abs(num).toString().length / 3);
    let value=+(num / Math.pow(1000, floor))
    return value.toFixed(value > 1?digits:2) + units[floor - 1];

}

速度测试(200000随机样本)从这个线程不同的解决方案

Execution time: formatNum          418  ms
Execution time: kFormatter         438  ms it just use "k" no "M".."T" 
Execution time: beautify           593  ms doesnt support - negatives
Execution time: shortenLargeNumber 682  ms    
Execution time: Intl.NumberFormat  13197ms 

最多支持数量。MAX_SAFE_INTEGER到Number。MIN_SAFE_INTEGER

function abbreviateThousands(value) { const num = Number(value) const absNum = Math.abs(num) const sign = Math.sign(num) const numLength = Math.round(absNum).toString().length const symbol = ['K', 'M', 'B', 'T', 'Q'] const symbolIndex = Math.floor((numLength - 1) / 3) - 1 const abbrv = symbol[symbolIndex] || symbol[symbol.length - 1] let divisor = 0 if (numLength > 15) divisor = 1e15 else if (numLength > 12) divisor = 1e12 else if (numLength > 9) divisor = 1e9 else if (numLength > 6) divisor = 1e6 else if (numLength > 3) divisor = 1e3 else return num return `${((sign * absNum) / divisor).toFixed(divisor && 1)}${abbrv}` } console.log(abbreviateThousands(234523452345)) // 234.5b (billion) console.log(abbreviateThousands(Number.MIN_SAFE_INTEGER)) // -9.0q (quadrillion)

不满足任何张贴的解决方案，所以这是我的版本:

Supports positive and negative numbers Supports negative exponents Rounds up to next exponent if possible Performs bounds checking (doesn't error out for very large/small numbers) Strips off trailing zeros/spaces Supports a precision parameter function abbreviateNumber(number,digits=2) { var expK = Math.floor(Math.log10(Math.abs(number)) / 3); var scaled = number / Math.pow(1000, expK); if(Math.abs(scaled.toFixed(digits))>=1000) { // Check for rounding to next exponent scaled /= 1000; expK += 1; } var SI_SYMBOLS = "apμm kMGTPE"; var BASE0_OFFSET = SI_SYMBOLS.indexOf(' '); if (expK + BASE0_OFFSET>=SI_SYMBOLS.length) { // Bound check expK = SI_SYMBOLS.length-1 - BASE0_OFFSET; scaled = number / Math.pow(1000, expK); } else if (expK + BASE0_OFFSET < 0) return 0; // Too small return scaled.toFixed(digits).replace(/(\.|(\..*?))0+$/,'$2') + SI_SYMBOLS[expK+BASE0_OFFSET].trim(); } ////////////////// const tests = [ [0.0000000000001,2], [0.00000000001,2], [0.000000001,2], [0.000001,2], [0.001,2], [0.0016,2], [-0.0016,2], [0.01,2], [1,2], [999.99,2], [999.99,1], [-999.99,1], [999999,2], [999999999999,2], [999999999999999999,2], [99999999999999999999,2], ]; for (var i = 0; i < tests.length; i++) { console.log(abbreviateNumber(tests[i][0], tests[i][1]) ); }