进一步改进Salman's Answer，因为像nFormatter(9999999,1)这样的情况返回1000K。

function formatNumberWithMetricPrefix(num, digits = 1) {
  const si = [
    {value: 1e18, symbol: 'E'},
    {value: 1e15, symbol: 'P'},
    {value: 1e12, symbol: 'T'},
    {value: 1e9, symbol: 'G'},
    {value: 1e6, symbol: 'M'},
    {value: 1e3, symbol: 'k'},
    {value: 0, symbol: ''},
  ];
  const rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
  function divideNum(divider) {
    return (num / (divider || 1)).toFixed(digits);
  }

  let i = si.findIndex(({value}) => num >= value);
  if (+divideNum(si[i].value) >= 1e3 && si[i - 1]) {
    i -= 1;
  }
  const {value, symbol} = si[i];
  return divideNum(value).replace(rx, '$1') + symbol;
}

这个函数可以将巨大的数字(正数和负数)转换为读者友好的格式，而不会失去其精度:

function abbrNum(n) { if (!n || (n && typeof n !== 'number')) { return ''; } const ranges = [ { divider: 1e12 , suffix: 't' }, { divider: 1e9 , suffix: 'b' }, { divider: 1e6 , suffix: 'm' }, { divider: 1e3 , suffix: 'k' } ]; const range = ranges.find(r => Math.abs(n) >= r.divider); if (range) { return (n / range.divider).toString() + range.suffix; } return n.toString(); } /* test cases */ let testAry = [99, 1200, -150000, 9000000]; let resultAry = testAry.map(abbrNum); console.log("result array: " + resultAry);

哇，这里有这么多答案。我想我会告诉你我是如何解决它的，因为它似乎是最容易阅读的，处理负数，并且在JavaScript的千位数范围内。它也很容易改变到你想要的或进一步扩展。

const symbols = [ { value: 1, symbol: '' }, { value: 1e3, symbol: 'k' }, { value: 1e6, symbol: 'M' }, { value: 1e9, symbol: 'G' }, { value: 1e12, symbol: 'T' }, { value: 1e15, symbol: 'P' }, { value: 1e18, symbol: 'E' } ]; function numberFormatter(num, digits) { const numToCheck = Math.abs(num); for (let i = symbols.length - 1; i >= 0; i--) { if (numToCheck >= symbols[i].value) { const newNumber = (num / symbols[i].value).toFixed(digits); return `${newNumber}${symbols[i].symbol}`; } } return '0'; } const tests = [ { num: 1234, digits: 1 }, { num: 100000000, digits: 1 }, { num: 299792458, digits: 1 }, { num: 759878, digits: 1 }, { num: -759878, digits: 0 }, { num: 123, digits: 1 }, { num: 123.456, digits: 1 }, { num: -123.456, digits: 2 }, { num: 123.456, digits: 4 } ]; for (let i = 0; i < tests.length; i++) { console.log(`numberFormatter(${tests[i].num}, ${tests[i].digits})=${numberFormatter(tests[i].num, tests[i].digits)}`); }

/*including negative values*/    
function nFormatter(num) {
      let neg = false;
       if(num < 0){
         num = num * -1;
         neg = true;
       }
       if (num >= 1000000000) {
         if(neg){
           return -1 * (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';  
         }
         return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
       }
       if (num >= 1000000) {
         if(neg){
           return -1 * (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';  
         }
         return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
       }
       if (num >= 1000) {
         if(neg){
           return -1 * (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';  
         }
         return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
       }
       return num;
    }

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 === '%' ? '%' : '');
}

function   transform(value,args) {
    const suffixes = ['K', 'M', 'B', 'T', 'P', 'E'];

    if (!value) {
      return null;
    }

    if (Number.isNaN(value)) {
      return null;
    }

    if (value < 1000) {
      return value;
    }

    const exp = Math.floor(Math.log(value) / Math.log(1000));

    const returnValue = (value / Math.pow(1000, exp)).toFixed(args) + suffixes[exp - 1];

    return returnValue;
  }

变换(9999 2)

/ /“10.00 k”