65 lines
1.9 KiB
JavaScript
65 lines
1.9 KiB
JavaScript
"use strict";
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Object.defineProperty(exports, "__esModule", { value: true });
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exports.precisionAdd = exports.getLogPositiveMin = exports.log = exports.calBase = void 0;
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var util_1 = require("@antv/util");
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// 求以a为次幂，结果为b的基数，如 x^^a = b;求x
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// 虽然数学上 b 不支持负数，但是这里需要支持 负数
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function calBase(a, b) {
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var e = Math.E;
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var value;
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if (b >= 0) {
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value = Math.pow(e, Math.log(b) / a); // 使用换底公式求底
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}
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else {
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value = Math.pow(e, Math.log(-b) / a) * -1; // 使用换底公式求底
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}
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return value;
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}
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exports.calBase = calBase;
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function log(a, b) {
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if (a === 1) {
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return 1;
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}
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return Math.log(b) / Math.log(a);
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}
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exports.log = log;
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function getLogPositiveMin(values, base, max) {
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if (util_1.isNil(max)) {
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max = Math.max.apply(null, values);
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}
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var positiveMin = max;
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util_1.each(values, function (value) {
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if (value > 0 && value < positiveMin) {
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positiveMin = value;
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}
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});
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if (positiveMin === max) {
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positiveMin = max / base;
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}
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if (positiveMin > 1) {
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positiveMin = 1;
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}
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return positiveMin;
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}
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exports.getLogPositiveMin = getLogPositiveMin;
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function digitLength(num) {
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// Get digit length of e
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var eSplit = num.toString().split(/[eE]/);
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var len = (eSplit[0].split('.')[1] || '').length - +(eSplit[1] || 0);
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return len > 0 ? len : 0;
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}
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/**
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* 高精度加法，解决 0.1 + 0.2 !== 0.3 的经典问题
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*
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* @param num1 加数
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* @param num2 被加数
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* @return {number} 返回值
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*/
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function precisionAdd(num1, num2) {
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var num1Digits = digitLength(num1);
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var num2Digits = digitLength(num2);
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var baseNum = Math.pow(10, Math.max(num1Digits, num2Digits));
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return (num1 * baseNum + num2 * baseNum) / baseNum;
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}
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exports.precisionAdd = precisionAdd;
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//# sourceMappingURL=math.js.map
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