直接的方法具有最好的可读性，并且使用最少的内存。不需要过多地使用regex、map对象、Math对象、for-loops等。

使用K格式化现金值

const formatCash = n => { 如果(n < 1e3)返回n; if (n >= 1e3) return +(n / 1e3).toFixed(1) +“K”; }; console.log (formatCash (2500));

使用K M B T格式化现金值

const formatCash = n => { 如果(n < 1e3)返回n; 如果1 e3 & & n (n > = < 1 e6)返回+ (n / 1 e3) .toFixed(1) +“K”; 如果1 e6 & & n (n > = < 1 e9) + 1 (n / e6)返回.toFixed(1) +“M”; if (n >= 1e9 && n < 1e12) return +(n / 1e9).toFixed(1) + "B"; if (n >= 1e12) return +(n / 1e12).toFixed(1) + "T"; }; console.log (formatCash (1235000));

使用负数

let format;
const number = -1235000;

if (number < 0) {
  format = '-' + formatCash(-1 * number);
} else {
  format = formatCash(number);
}

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 }

听起来这应该对你有用:

函数 kFormatter（num） { 返回 Math.abs（num） > 999 ？Math.sign（num）*（（Math.abs（num）/1000）.toFixed（1）） + 'k' ： Math.sign（num）*Math.abs（num） } console.log（kFormatter（1200））;1.2k console.log（kFormatter（-1200））;-1.2k console.log（kFormatter（900））;900 console.log（kFormatter（-900））;-900

韦伦·弗林解决方案的2020版。

const SI_SYMBOLS = ["", "k", "M", "G", "T", "P", "E"];

const abbreviateNumber = (number, minDigits, maxDigits) => {
    if (number === 0) return number;

    // determines SI symbol
    const tier = Math.floor(Math.log10(Math.abs(number)) / 3);

    // get suffix and determine scale
    const suffix = SI_SYMBOLS[tier];
    const scale = 10 ** (tier * 3);

    // scale the number
    const scaled = number / scale;

    // format number and add suffix
    return scaled.toLocaleString(undefined, {
        minimumFractionDigits: minDigits,
        maximumFractionDigits: maxDigits,
    }) + suffix;
};

Tests and examples: const abbreviateNumberFactory = (symbols) => ( (number, minDigits, maxDigits) => { if (number === 0) return number; // determines SI symbol const tier = Math.floor(Math.log10(Math.abs(number)) / 3); // get suffix and determine scale const suffix = symbols[tier]; const scale = 10 ** (tier * 3); // scale the number const scaled = number / scale; // format number and add suffix return scaled.toLocaleString(undefined, { minimumFractionDigits: minDigits, maximumFractionDigits: maxDigits, }) + suffix; } ); const SI_SYMBOLS = ["", "k", "M", "G", "T", "P", "E"]; const SHORT_SYMBOLS = ["", "K", "M", "B", "T", "Q"]; const LONG_SYMBOLS = ["", " thousand", " million", " billion", " trillion", " quadrillion"]; const abbreviateNumberSI = abbreviateNumberFactory(SI_SYMBOLS); const abbreviateNumberShort = abbreviateNumberFactory(SHORT_SYMBOLS); const abbreviateNumberLong = abbreviateNumberFactory(LONG_SYMBOLS); const tests = [1e5, -9e7, [1009999.999, 2], [245345235.34513, 1, 1], [-72773144123, 3] ]; const functions = { abbreviateNumberSI, abbreviateNumberShort, abbreviateNumberLong, }; tests.forEach((test) => { const testValue = Array.isArray(test) ? test : [test]; Object.entries(functions).forEach(([key, func]) => { console.log(`${key}(${testValue.join(', ')}) = ${func(...testValue)}`); }); });

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"

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;
}