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
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);
您应该使用:
Math.round( num * 100 + Number.EPSILON ) / 100
似乎没有人知道数字EPSILON。
此外,值得注意的是,这并不像某些人所说的那样是JavaScript的怪异之处。
这就是浮点数在计算机中的工作方式。与99%的编程语言一样,JavaScript没有自制的浮点数;它依赖于CPU/FPU。计算机使用二进制,在二进制中,没有像0.1这样的数字,而只是二进制的近似值。为什么?出于同样的原因,1/3不能用十进制写:它的值是0.33333333……无穷大为三。
这里是Number.EPSILON。这个数字是1和双精度浮点数字中存在的下一个数字之间的差值。就是这样:在1和1+number.EPSILON之间没有数字。
编辑:
正如评论中所问的,让我们澄清一件事:添加Number.EPSILON仅当要舍入的值是算术运算的结果时才相关,因为它可以吞下一些浮点误差增量。
当值来自直接来源(例如:文字、用户输入或传感器)时,它不起作用。
编辑(2019):
像@maganap和一些人指出的那样,最好在相乘之前加上Number.EPSILON:
Math.round( ( num + Number.EPSILON ) * 100 ) / 100
编辑(2019年12月):
最近,我使用了一个类似于此的函数来比较epsilon感知的数字:
const ESPILON_RATE = 1 + Number.EPSILON ;
const ESPILON_ZERO = Number.MIN_VALUE ;
function epsilonEquals( a , b ) {
if ( Number.isNaN( a ) || Number.isNaN( b ) ) {
return false ;
}
if ( a === 0 || b === 0 ) {
return a <= b + EPSILON_ZERO && b <= a + EPSILON_ZERO ;
}
return a <= b * EPSILON_RATE && b <= a * EPSILON_RATE ;
}
我的用例是我多年来开发的断言+数据验证库。
事实上,在代码中,我使用的是ESPILON_RATE=1+4*数字.EPSILON和EPSILON_ZERO=4*数字.MIN_VALUE(四倍于EPSILON),因为我想要一个足够宽松的等式检查器来累积浮点错误。
到目前为止,它看起来很适合我。我希望这会有所帮助。
使用Math.rround():
Math.round(num * 100) / 100
或者更具体地说,为了确保1.005这样的数字正确,请使用Number.EPSILON:
Math.round((num + Number.EPSILON) * 100) / 100
这可能会帮助您:
var result = Math.round(input*100)/100;
有关更多信息,您可以查看Math.round(num)vs num.toFixed(0)和浏览器不一致性
如果值是文本类型:
parseFloat("123.456").toFixed(2);
如果值是数字:
var numb = 123.23454;
numb = numb.toFixed(2);
有一个缺点,像1.5这样的值将给出“1.50”作为输出。@minitech建议的修复方法:
var numb = 1.5;
numb = +numb.toFixed(2);
// Note the plus sign that drops any "extra" zeroes at the end.
// It changes the result (which is a string) into a number again (think "0 + foo"),
// which means that it uses only as many digits as necessary.
Math.round似乎是一个更好的解决方案。但事实并非如此!在某些情况下,它不会正确舍入:
Math.round(1.005 * 100)/100 // Returns 1 instead of expected 1.01!
toFixed()在某些情况下也不会正确舍入(在Chrome v.55.0.2883.87中测试)!
示例:
parseFloat("1.555").toFixed(2); // Returns 1.55 instead of 1.56.
parseFloat("1.5550").toFixed(2); // Returns 1.55 instead of 1.56.
// However, it will return correct result if you round 1.5551.
parseFloat("1.5551").toFixed(2); // Returns 1.56 as expected.
1.3555.toFixed(3) // Returns 1.355 instead of expected 1.356.
// However, it will return correct result if you round 1.35551.
1.35551.toFixed(2); // Returns 1.36 as expected.
我想,这是因为1.555实际上就像是幕后的浮球1.55499994。
解决方案1是使用具有所需舍入算法的脚本,例如:
function roundNumber(num, scale) {
if(!("" + num).includes("e")) {
return +(Math.round(num + "e+" + scale) + "e-" + scale);
} else {
var arr = ("" + num).split("e");
var sig = ""
if(+arr[1] + scale > 0) {
sig = "+";
}
return +(Math.round(+arr[0] + "e" + sig + (+arr[1] + scale)) + "e-" + scale);
}
}
它也在Plunker。
注意:这并不是每个人都能通用的解决方案。有几种不同的舍入算法。您的实现可能不同,这取决于您的需求。请参见舍入。
解决方案2是避免前端计算,并从后端服务器提取舍入值。
另一种可能的解决方案,也不是防弹的。
Math.round((num + Number.EPSILON) * 100) / 100
在某些情况下,当您舍入像1.3549999999999998这样的数字时,它将返回错误的结果。它应该是1.35,但结果是1.36。