最简单最简单的方法是

new Intl.NumberFormat('en-IN', { 
    notation: "compact",
    compactDisplay: "short",
    style: 'currency',
    currency: 'INR'
}).format(1000).replace("T", "K")

这对任何数字都适用。包括L、Cr等。

注意:不能在safari中工作。

/**
 * Shorten number to thousands, millions, billions, etc.
 * http://en.wikipedia.org/wiki/Metric_prefix
 *
 * @param {number} num Number to shorten.
 * @param {number} [digits=0] The number of digits to appear after the decimal point.
 * @returns {string|number}
 *
 * @example
 * // returns '12.5k'
 * shortenLargeNumber(12543, 1)
 *
 * @example
 * // returns '-13k'
 * shortenLargeNumber(-12567)
 *
 * @example
 * // returns '51M'
 * shortenLargeNumber(51000000)
 *
 * @example
 * // returns 651
 * shortenLargeNumber(651)
 *
 * @example
 * // returns 0.12345
 * shortenLargeNumber(0.12345)
 */
function shortenLargeNumber(num, digits) {
    var units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
        decimal;

    for(var i=units.length-1; i>=0; i--) {
        decimal = Math.pow(1000, i+1);

        if(num <= -decimal || num >= decimal) {
            return +(num / decimal).toFixed(digits) + units[i];
        }
    }

    return num;
}

谢谢@Cos的评论，我删除了Math。round10依赖。

一个简短的替代方案:

function nFormatter(num) { const format = [ { 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: 1, symbol: '' }, ]; const formatIndex = format.findIndex((data) => num >= data.value); console.log(formatIndex) return (num / format[formatIndex === -1? 6: formatIndex].value).toFixed(2) + format[formatIndex === -1?6: formatIndex].symbol; }

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"

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 }

改进@tfmontague的答案，进一步格式化小数点。33.0k到33k

largeNumberFormatter(value: number): any {
   let result: any = value;

   if (value >= 1e3 && value < 1e6) { result = (value / 1e3).toFixed(1).replace(/\.0$/, '') + 'K'; }
   if (value >= 1e6 && value < 1e9) { result = (value / 1e6).toFixed(1).replace(/\.0$/, '') + 'M'; }
   if (value >= 1e9) { result = (value / 1e9).toFixed(1).replace(/\.0$/, '') + 'T'; }

   return result;
}