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]) ); }

我决定在这里扩展@Novellizator的答案，以满足我的需求。我想要一个灵活的函数来处理我的大部分格式化需求，而不需要外部库。

特性

选择使用顺序后缀(k, M等) 选项指定要使用的订单后缀的自定义列表 选项来约束最小和最大顺序 控制小数点后的位数 自动顺序分隔逗号 可选百分比或美元格式 控制在非数字输入的情况下返回什么 适用于负数和无穷数

例子

let x = 1234567.8;
formatNumber(x);  // '1,234,568'
formatNumber(x, {useOrderSuffix: true});  // '1M'
formatNumber(x, {useOrderSuffix: true, decimals: 3, maxOrder: 1});  // '1,234.568k'
formatNumber(x, {decimals: 2, style: '$'});  // '$1,234,567.80'

x = 10.615;
formatNumber(x, {style: '%'});  // '1,062%'
formatNumber(x, {useOrderSuffix: true, decimals: 1, style: '%'});  // '1.1k%'
formatNumber(x, {useOrderSuffix: true, decimals: 5, style: '%', minOrder: 2});  // '0.00106M%'

formatNumber(-Infinity);  // '-∞'
formatNumber(NaN);  // ''
formatNumber(NaN, {valueIfNaN: NaN});  // NaN

函数

/*
 * Return the given number as a formatted string.  The default format is a plain
 * integer with thousands-separator commas.  The optional parameters facilitate
 * other formats:
 *   - decimals = the number of decimals places to round to and show
 *   - valueIfNaN = the value to show for non-numeric input
 *   - style
 *     - '%': multiplies by 100 and appends a percent symbol
 *     - '$': prepends a dollar sign
 *   - useOrderSuffix = whether to use suffixes like k for 1,000, etc.
 *   - orderSuffixes = the list of suffixes to use
 *   - minOrder and maxOrder allow the order to be constrained.  Examples:
 *     - minOrder = 1 means the k suffix should be used for numbers < 1,000
 *     - maxOrder = 1 means the k suffix should be used for numbers >= 1,000,000
 */
function formatNumber(number, {
    decimals = 0,
    valueIfNaN = '',
    style = '',
    useOrderSuffix = false,
    orderSuffixes = ['', 'k', 'M', 'B', 'T'],
    minOrder = 0,
    maxOrder = Infinity
  } = {}) {

  let x = parseFloat(number);

  if (isNaN(x))
    return valueIfNaN;

  if (style === '%')
    x *= 100.0;

  let order;
  if (!isFinite(x) || !useOrderSuffix)
    order = 0;
  else if (minOrder === maxOrder)
    order = minOrder;
  else {
    const unboundedOrder = Math.floor(Math.log10(Math.abs(x)) / 3);
    order = Math.max(
      0,
      minOrder,
      Math.min(unboundedOrder, maxOrder, orderSuffixes.length - 1)
    );
  }

  const orderSuffix = orderSuffixes[order];
  if (order !== 0)
    x /= Math.pow(10, order * 3);

  return (style === '$' ? '$' : '') +
    x.toLocaleString(
      'en-US',
      {
        style: 'decimal',
        minimumFractionDigits: decimals,
        maximumFractionDigits: decimals
      }
    ) +
    orderSuffix +
    (style === '%' ? '%' : '');
}

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?

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

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 }

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
}