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 }

进一步改进@Yash的回答，支持负数:

function nFormatter(num) {
    isNegative = false
    if (num < 0) {
        isNegative = true
    }
    num = Math.abs(num)
    if (num >= 1000000000) {
        formattedNumber = (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
    } else if (num >= 1000000) {
        formattedNumber =  (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
    } else  if (num >= 1000) {
        formattedNumber =  (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
    } else {
        formattedNumber = num;
    }   
    if(isNegative) { formattedNumber = '-' + formattedNumber }
    return formattedNumber;
}

nFormatter(-120000)
"-120K"
nFormatter(120000)
"120K"

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)

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
}

function AmountConveter(amount) {
  return Math.abs(amount) > 999
    ? Math.sign(amount) * (Math.abs(amount) / 1000).toFixed(1) + "k"
    : Math.sign(amount) * Math.abs(amount);
}

console.log(AmountConveter(1200)); // 1.2k
console.log(AmountConveter(-1200)); // -1.2k
console.log(AmountConveter(900)); // 900
console.log(AmountConveter(-900)); // -900

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);