我需要显示一个货币值的格式1K等于一千，或1.1K, 1.2K, 1.9K等，如果它不是一个偶数千，否则如果低于一千，显示正常500,100,250等，使用JavaScript格式化的数字?
当前回答
通过消除@martin-sznapka解决方案中的循环，您将减少40%的执行时间。
function formatNum(num,digits) {
let units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'];
let floor = Math.floor(Math.abs(num).toString().length / 3);
let value=+(num / Math.pow(1000, floor))
return value.toFixed(value > 1?digits:2) + units[floor - 1];
}
速度测试(200000随机样本)从这个线程不同的解决方案
Execution time: formatNum 418 ms
Execution time: kFormatter 438 ms it just use "k" no "M".."T"
Execution time: beautify 593 ms doesnt support - negatives
Execution time: shortenLargeNumber 682 ms
Execution time: Intl.NumberFormat 13197ms
其他回答
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);
}
通过消除@martin-sznapka解决方案中的循环，您将减少40%的执行时间。
function formatNum(num,digits) {
let units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'];
let floor = Math.floor(Math.abs(num).toString().length / 3);
let value=+(num / Math.pow(1000, floor))
return value.toFixed(value > 1?digits:2) + units[floor - 1];
}
速度测试(200000随机样本)从这个线程不同的解决方案
Execution time: formatNum 418 ms
Execution time: kFormatter 438 ms it just use "k" no "M".."T"
Execution time: beautify 593 ms doesnt support - negatives
Execution time: shortenLargeNumber 682 ms
Execution time: Intl.NumberFormat 13197ms
这里有一个简单的解决方案，可以避免所有的if语句(借助Math的力量)。
var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];
function abbreviateNumber(number){
// what tier? (determines SI symbol)
var tier = Math.log10(Math.abs(number)) / 3 | 0;
// if zero, we don't need a suffix
if(tier == 0) return number;
// get suffix and determine scale
var suffix = SI_SYMBOL[tier];
var scale = Math.pow(10, tier * 3);
// scale the number
var scaled = number / scale;
// format number and add suffix
return scaled.toFixed(1) + suffix;
}
奖金模因
SI代表什么?
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; }
不满足任何张贴的解决方案，所以这是我的版本:
Supports positive and negative numbers Supports negative exponents Rounds up to next exponent if possible Performs bounds checking (doesn't error out for very large/small numbers) Strips off trailing zeros/spaces Supports a precision parameter function abbreviateNumber(number,digits=2) { var expK = Math.floor(Math.log10(Math.abs(number)) / 3); var scaled = number / Math.pow(1000, expK); if(Math.abs(scaled.toFixed(digits))>=1000) { // Check for rounding to next exponent scaled /= 1000; expK += 1; } var SI_SYMBOLS = "apμm kMGTPE"; var BASE0_OFFSET = SI_SYMBOLS.indexOf(' '); if (expK + BASE0_OFFSET>=SI_SYMBOLS.length) { // Bound check expK = SI_SYMBOLS.length-1 - BASE0_OFFSET; scaled = number / Math.pow(1000, expK); } else if (expK + BASE0_OFFSET < 0) return 0; // Too small return scaled.toFixed(digits).replace(/(\.|(\..*?))0+$/,'$2') + SI_SYMBOLS[expK+BASE0_OFFSET].trim(); } ////////////////// const tests = [ [0.0000000000001,2], [0.00000000001,2], [0.000000001,2], [0.000001,2], [0.001,2], [0.0016,2], [-0.0016,2], [0.01,2], [1,2], [999.99,2], [999.99,1], [-999.99,1], [999999,2], [999999999999,2], [999999999999999999,2], [99999999999999999999,2], ]; for (var i = 0; i < tests.length; i++) { console.log(abbreviateNumber(tests[i][0], tests[i][1]) ); }
