支持负数 检查!isFinite 如果你想要最大单位是M，将' K M G T P E Z Y'改为' K M' 基数选项(1K = 1000 / 1K = 1024)

Number.prototype.prefix = function (precision, base) { var units = ' K M G T P E Z Y'.split(' '); if (typeof precision === 'undefined') { precision = 2; } if (typeof base === 'undefined') { base = 1000; } if (this == 0 || !isFinite(this)) { return this.toFixed(precision) + units[0]; } var power = Math.floor(Math.log(Math.abs(this)) / Math.log(base)); // Make sure not larger than max prefix power = Math.min(power, units.length - 1); return (this / Math.pow(base, power)).toFixed(precision) + units[power]; }; console.log('0 = ' + (0).prefix()) // 0.00 console.log('10000 = ' + (10000).prefix()) // 10.00K console.log('1234000 = ' + (1234000).prefix(1)) // 1.2M console.log('-10000 = ' + (-10240).prefix(1, 1024)) // -10.0K console.log('-Infinity = ' + (-Infinity).prefix()) // -Infinity console.log('NaN = ' + (NaN).prefix()) // NaN

简单通用的方法

可以将COUNT_FORMATS配置对象设置为您想要的长度或长度，这取决于您测试的值范围。

// Configuration const COUNT_FORMATS = [ { // 0 - 999 letter: '', limit: 1e3 }, { // 1,000 - 999,999 letter: 'K', limit: 1e6 }, { // 1,000,000 - 999,999,999 letter: 'M', limit: 1e9 }, { // 1,000,000,000 - 999,999,999,999 letter: 'B', limit: 1e12 }, { // 1,000,000,000,000 - 999,999,999,999,999 letter: 'T', limit: 1e15 } ]; // Format Method: function formatCount(value) { const format = COUNT_FORMATS.find(format => (value < format.limit)); value = (1000 * value / format.limit); value = Math.round(value * 10) / 10; // keep one decimal number, only if needed return (value + format.letter); } // Test: const test = [274, 1683, 56512, 523491, 9523489, 5729532709, 9421032489032]; test.forEach(value => console.log(`${ value } >>> ${ formatCount(value) }`));

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;
}

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

您可以使用模仿Python高级字符串格式化PEP3101的d3格式包:

var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"

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
}