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
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);
其他回答
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$/, '') + 'K';
} else {
formattedNumber = num;
}
if(isNegative) { formattedNumber = '-' + formattedNumber }
return formattedNumber;
}
nFormatter(-120000)
"-120K"
nFormatter(120000)
"120K"
您可以使用模仿Python高级字符串格式化PEP3101的d3格式包:
var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"
简单通用的方法
可以将COUNT_FORMATS配置对象设置为您想要的长度或长度，这取决于您测试的值范围。
// Configuration const COUNT_FORMATS = [ { // 0 - 999 letter: '', limit: 1e3 }, { // 1,000 - 999,999 letter: 'K', limit: 1e6 }, { // 1,000,000 - 999,999,999 letter: 'M', limit: 1e9 }, { // 1,000,000,000 - 999,999,999,999 letter: 'B', limit: 1e12 }, { // 1,000,000,000,000 - 999,999,999,999,999 letter: 'T', limit: 1e15 } ]; // Format Method: function formatCount(value) { const format = COUNT_FORMATS.find(format => (value < format.limit)); value = (1000 * value / format.limit); value = Math.round(value * 10) / 10; // keep one decimal number, only if needed return (value + format.letter); } // Test: const test = [274, 1683, 56512, 523491, 9523489, 5729532709, 9421032489032]; test.forEach(value => console.log(`${ value } >>> ${ formatCount(value) }`));
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
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
