我想最多四舍五入两位小数,但只有在必要时。
输入:
10
1.7777777
9.1
输出:
10
1.78
9.1
如何在JavaScript中执行此操作?
我想最多四舍五入两位小数,但只有在必要时。
输入:
10
1.7777777
9.1
输出:
10
1.78
9.1
如何在JavaScript中执行此操作?
当前回答
我的解决方案将输入视为字符串,并使用n位数字的“数学舍入”算法:取n位数字,如果数字n+1等于或大于5,则加1。它还允许指定负数,例如,将123.45舍入为-1的数字为120。它也适用于科学符号(例如1.2e-3)。我没有测量它的速度,我认为这不是最好的表现。
function safeRound( numInput, numPrecision ) {
const strNumber = numInput.toString().replace( 'E', 'e' );
const bSign = '+-'.indexOf( strNumber[ 0 ] ) !== -1;
const strSign = bSign ? strNumber[ 0 ] : '';
const numSign = strSign !== '-' ? +1 : -1;
const ixExponent = ( ixFound => ixFound !== -1 ? ixFound : strNumber.length )( strNumber.indexOf( 'e' ) );
const strExponent = strNumber.substr( ixExponent + 1 );
const numExponent = ixExponent !== strNumber.length ? Number.parseInt( strExponent ) : 0;
const ixDecimal = ( ixFound => ixFound !== -1 ? ixFound : ixExponent )( strNumber.indexOf( '.' ) );
const strInteger = strNumber.substring( !bSign ? 0 : 1, ixDecimal );
const strFraction = strNumber.substring( ixDecimal + 1, ixExponent );
const numPrecisionAdjusted = numPrecision + numExponent;
const strIntegerKeep = strInteger.substring( 0, strInteger.length + Math.min( 0, numPrecisionAdjusted ) ) + '0'.repeat( -Math.min( 0, numPrecisionAdjusted ) );
const strFractionKeep = strFraction.substring( 0, Math.max( 0, numPrecisionAdjusted ) );
const strRoundedDown = strSign + ( strIntegerKeep === '' ? '0' : strIntegerKeep ) + ( strFractionKeep === '' ? '' : '.' + strFractionKeep ) + ( strExponent === '' ? '' : 'e' + strExponent );
const chRoundUp = 0 <= numPrecisionAdjusted ? strFraction.substr( numPrecisionAdjusted, 1 ) : ( '0' + strInteger ).substr( numPrecisionAdjusted, 1 );
const bRoundUp = '5' <= chRoundUp && chRoundUp <= '9';
const numRoundUp = bRoundUp ? numSign * Math.pow( 10, -numPrecision ) : 0;
return Number.parseFloat( strRoundedDown ) + numRoundUp;
}
function safeRoundTest( numInput, numPrecision, strExpected ) {
const strActual = safeRound( numInput, numPrecision ).toString();
const bPassed = strActual === strExpected;
console.log( 'numInput', numInput, 'numPrecision', numPrecision, 'strExpected', strExpected, 'strActual', strActual, 'bPassed', bPassed );
return bPassed ? 0 : 1;
}
function safeRoundTests() {
let numFailed = 0;
numFailed += safeRoundTest( 0, 0, '0' );
numFailed += safeRoundTest( '0', 0, '0' );
numFailed += safeRoundTest( '0.1', 0, '0' );
numFailed += safeRoundTest( '+0.1', 0, '0' );
numFailed += safeRoundTest( '-0.1', 0, '0' );
numFailed += safeRoundTest( '0.1', 1, '0.1' );
numFailed += safeRoundTest( '+0.1', 1, '0.1' );
numFailed += safeRoundTest( '-0.1', 1, '-0.1' );
numFailed += safeRoundTest( '0.9', 0, '1' );
numFailed += safeRoundTest( '+0.9', 0, '1' );
numFailed += safeRoundTest( '-0.9', 0, '-1' );
numFailed += safeRoundTest( '0.9', 1, '0.9' );
numFailed += safeRoundTest( '+0.9', 1, '0.9' );
numFailed += safeRoundTest( '-0.9', 1, '-0.9' );
numFailed += safeRoundTest( '0.5', 0, '1' );
numFailed += safeRoundTest( '+0.5', 0, '1' );
numFailed += safeRoundTest( '-0.5', 0, '-1' );
numFailed += safeRoundTest( '0.4999', 0, '0' );
numFailed += safeRoundTest( '+0.4999', 0, '0' );
numFailed += safeRoundTest( '-0.4999', 0, '0' );
numFailed += safeRoundTest( '1.005', 2, '1.01' );
numFailed += safeRoundTest( '1.00499999999', 2, '1' );
numFailed += safeRoundTest( '012.3456', -4, '0' );
numFailed += safeRoundTest( '012.3456', -3, '0' );
numFailed += safeRoundTest( '012.3456', -2, '0' );
numFailed += safeRoundTest( '012.3456', -1, '10' );
numFailed += safeRoundTest( '012.3456', 0, '12' );
numFailed += safeRoundTest( '012.3456', 1, '12.3' );
numFailed += safeRoundTest( '012.3456', 2, '12.35' );
numFailed += safeRoundTest( '012.3456', 3, '12.346' );
numFailed += safeRoundTest( '012.3456', 4, '12.3456' );
numFailed += safeRoundTest( '012.3456', 5, '12.3456' );
numFailed += safeRoundTest( '12.', 0, '12' );
numFailed += safeRoundTest( '.12', 2, '0.12' );
numFailed += safeRoundTest( '0e0', 0, '0' );
numFailed += safeRoundTest( '1.2e3', 0, '1200' );
numFailed += safeRoundTest( '1.2e+3', 0, '1200' );
numFailed += safeRoundTest( '1.2e-3', 0, '0' );
numFailed += safeRoundTest( '1.2e-3', 3, '0.001' );
numFailed += safeRoundTest( '1.2e-3', 4, '0.0012' );
numFailed += safeRoundTest( '1.2e-3', 5, '0.0012' );
numFailed += safeRoundTest( '+12.', 0, '12' );
numFailed += safeRoundTest( '+.12', 2, '0.12' );
numFailed += safeRoundTest( '+0e0', 0, '0' );
numFailed += safeRoundTest( '+1.2e3', 0, '1200' );
numFailed += safeRoundTest( '+1.2e+3', 0, '1200' );
numFailed += safeRoundTest( '+1.2e-3', 0, '0' );
numFailed += safeRoundTest( '+1.2e-3', 3, '0.001' );
numFailed += safeRoundTest( '+1.2e-3', 4, '0.0012' );
numFailed += safeRoundTest( '+1.2e-3', 5, '0.0012' );
numFailed += safeRoundTest( '-12.', 0, '-12' );
numFailed += safeRoundTest( '-.12', 2, '-0.12' );
numFailed += safeRoundTest( '-0e0', 0, '0' );
numFailed += safeRoundTest( '-1.2e3', 0, '-1200' );
numFailed += safeRoundTest( '-1.2e+3', 0, '-1200' );
numFailed += safeRoundTest( '-1.2e-3', 0, '0' );
numFailed += safeRoundTest( '-1.2e-3', 3, '-0.001' );
numFailed += safeRoundTest( '-1.2e-3', 4, '-0.0012' );
numFailed += safeRoundTest( '-1.2e-3', 5, '-0.0012' );
numFailed += safeRoundTest( '9876.543e210', 0, '9.876543e+213' );
numFailed += safeRoundTest( '9876.543e210', -210, '9.877e+213' );
numFailed += safeRoundTest( '9876.543e210', -209, '9.8765e+213' );
numFailed += safeRoundTest( '9876.543e+210', 0, '9.876543e+213' );
numFailed += safeRoundTest( '9876.543e+210', -210, '9.877e+213' );
numFailed += safeRoundTest( '9876.543e+210', -209, '9.8765e+213' );
numFailed += safeRoundTest( '9876.543e-210', 213, '9.876543e-207' );
numFailed += safeRoundTest( '9876.543e-210', 210, '9.877e-207' );
numFailed += safeRoundTest( '9876.543e-210', 211, '9.8765e-207' );
console.log( 'numFailed', numFailed );
}
safeRoundTests();
其他回答
有两种方法可以做到这一点。对于像我这样的人,Lodash的变体
function round(number, precision) {
var pair = (number + 'e').split('e')
var value = Math.round(pair[0] + 'e' + (+pair[1] + precision))
pair = (value + 'e').split('e')
return +(pair[0] + 'e' + (+pair[1] - precision))
}
用法:
round(0.015, 2) // 0.02
round(1.005, 2) // 1.01
如果您的项目使用jQuery或Lodash,您也可以在库中找到适当的舍入方法。
请参阅AmrAli的答案,以了解此解决方案的所有不同调整的更全面的运行和性能细分。
var DecimalPrecision=(函数){if(数字.EPSILON===未定义){Number.EPSILON=数学功率(2,-52);}if(Number.isInteger==未定义){Number.isInteger=函数(值){返回值类型==“number”&&isFinite(值)&&数学下限(值)==值;};}this.isRound=函数(n,p){设l=n.toString().split('.')[1].length;返回(p>=l);}this.round=函数(n,p=2){if(Number.isInteger(n)|| this.isRound(n,p))返回n;设r=0.5*Number.EPSILON*n;设o=1;而(p-->0)o*=10;如果(n<0)o*=-1;返回数学舍入((n+r)*o)/o;}this.ceil=函数(n,p=2){if(Number.isInteger(n)|| this.isRound(n,p))返回n;设r=0.5*Number.EPSILON*n;设o=1;而(p-->0)o*=10;返回Math.ceil((n+r)*o)/o;}this.flor=函数(n,p=2){if(Number.isInteger(n)|| this.isRound(n,p))返回n;设r=0.5*Number.EPSILON*n;设o=1;而(p-->0)o*=10;返回数学楼层((n+r)*o)/o;}返回此;})();console.log(DecimalPrecision.round(1.005));console.log(DecimalPrecision.ceil(1.005));console.log(DecimalPrecision.floor(1.005));console.log(DecimalPrecision.round(1.0049999));console.log(DecimalPrecision.ceil(1.0049999));console.log(DecimalPrecision.floor(1.0049999));console.log(DecimalPrecision.round(2.175495134384,7));console.log(DecimalPrecision.round(2.1753543549,8));console.log(DecimalPrecision.round(2.1755465135333,4));console.log(DecimalPrecision.ceil(17,4));console.log(DecimalPrecision.ceil(17.1,4));console.log(DecimalPrecision.ceil(17.1,15));
简单的通用舍入函数如下:
步骤如下:
使用Math.pow(10,位)将数字乘以(10乘以小数位数的幂)。使用Math.Round将结果舍入为整数。将结果除以(10乘以小数位数的幂)Math.pow(10,位)。
例子:
数字为:1.2375四舍五入至小数点后三位
1.2375 * (10^3) ==> 1.2375 * 1000 = 1237.5舍入为整数==>1238将1238除以(10^3)==>1238/1000=1.238
(注:10^3表示数学功率(10,3))。
函数编号RoundDecimal(v,n){return Math.round((v+Number.EPSILON)*Math.pow(10,n))/Math.pow(1,n)}//-------测试--------console.log(numberRoundDecimal(-0.0246411603862896567,3))//-0.025console.log(numberRoundDecimal(0.9993360575508052,3))//0.999console.log(numberRoundDecimal(1.0020739645577939,3))//1.002console.log(numberRoundDecimal(0.975,0))//1console.log(numberRoundDecimal(0.975,1))//1console.log(numberRoundDecimal(0.975,2))//0.98console.log(numberRoundDecimal(1.005,2))//1.01
避免舍入到任意位数的二进制问题的适当方法是:
function roundToDigits(number, digits) {
return Number(Math.round(Number(number + 'e' + digits)) + 'e-' + digits);
}
修复toFixed()函数的一种方法是:
Number.prototype.toFixed = (prototype => {
const toFixed = prototype.toFixed;
// noinspection JSVoidFunctionReturnValueUsed
return function (fractionDigits) {
if (!fractionDigits) {
return toFixed.call(this);
} else {
// Avoid binary rounding issues
fractionDigits = Math.floor(fractionDigits);
const n = Number(Math.round(Number(+this + 'e' + fractionDigits)) + 'e-' + fractionDigits);
return toFixed.call(n, fractionDigits);
}
};
})(Number.prototype);
我仍然不认为有人给他答案,告诉他如何在需要时只进行舍入。我认为最简单的方法是检查数字中是否有小数,如下所示:
var num = 3.21;
if ( (num+"").indexOf('.') >= 0 ) { //at least assert to string first...
// whatever code you decide to use to round
}