How do you round to 1 decimal place in Javascript?


Can you round a number in javascript to 1 character after the decimal point (properly rounded)?

I tried the *10, round, /10 but it leaves two decimals at the end of the int.

Math.round(num * 10) / 10 works, here is an example...

var number = 12.3456789
var rounded = Math.round(number * 10) / 10
// rounded is 12.3

if you want it to have one decimal place, even when that would be a 0, then add...

var fixed = rounded.toFixed(1)
// fixed is always to 1 d.p.
// NOTE: .toFixed() returns a string!

// To convert back to number format
parseFloat(number.toFixed(2))
// 12.34
// but that will not retain any trailing zeros

// So, just make sure it is the last step before output,
// and use a number format during calculations!

EDIT: Add round with precision function...

Using this principle, for reference, here is a handy little round function that takes precision...

function round(value, precision) {
    var multiplier = Math.pow(10, precision || 0);
    return Math.round(value * multiplier) / multiplier;
}

... usage ...

round(12345.6789, 2) // 12345.68
round(12345.6789, 1) // 12345.7

... defaults to round to nearest whole number (precision 0) ...

round(12345.6789) // 12346

... and can be used to round to nearest 10 or 100 etc...

round(12345.6789, -1) // 12350
round(12345.6789, -2) // 12300

... and correct handling of negative numbers ...

round(-123.45, 1) // -123.4
round(123.45, 1) // 123.5

... and can be combined with toFixed to format consistently as string ...

round(456.7, 2).toFixed(2) // "456.70"

var number = 123.456;

console.log(number.toFixed(1)); // should round to 123.5

If you use Math.round(5.01) you will get 5 instead of 5.0.

If you use toFixed you run into rounding issues.

If you want the best of both worlds combine the two:

(Math.round(5.01 * 10) / 10).toFixed(1)

You might want to create a function for this:

function roundedToFixed(_float, _digits){
  var rounded = Math.pow(10, _digits);
  return (Math.round(_float * rounded) / rounded).toFixed(_digits);
}

lodash has a round method:

_.round(4.006);
// => 4

_.round(4.006, 2);
// => 4.01

_.round(4060, -2);
// => 4100

Docs.

Source.


I vote for toFixed(), but, for the record, here's another way that uses bit shifting to cast the number to an int. So, it always rounds towards zero (down for positive numbers, up for negatives).

var rounded = ((num * 10) << 0) * 0.1;

But hey, since there are no function calls, it's wicked fast. :)

And here's one that uses string matching:

var rounded = (num + '').replace(/(^.*?\d+)(\.\d)?.*/, '$1$2');

I don't recommend using the string variant, just sayin.


Try with this:

var original=28.453

// 1.- round "original" to two decimals
var result = Math.round (original * 100) / 100  //returns 28.45

// 2.- round "original" to 1 decimal
var result = Math.round (original * 10) / 10  //returns 28.5

// 3.- round 8.111111 to 3 decimals
var result = Math.round (8.111111 * 1000) / 1000  //returns 8.111

less complicated and easier to implement...

with this, you can create a function to do:

function RoundAndFix (n, d) {
    var m = Math.pow (10, d);
    return Math.round (n * m) / m;
}

function RoundAndFix (n, d) {
    var m = Math.pow (10, d);
    return Math.round (n * m) / m;
}
console.log (RoundAndFix(8.111111, 3));

EDIT: see this How to round using ROUND HALF UP. Rounding mode that most of us were taught in grade school


x = number, n = decimal-places:

function round(x, n) {
    return Math.round(x * Math.pow(10, n)) / Math.pow(10, n)
}

var num = 34.7654;

num = Math.round(num * 10) / 10;

console.log(num); // Logs: 34.8

To complete the Best Answer:

var round = function ( number, precision )
{
    precision = precision || 0;
    return parseFloat( parseFloat( number ).toFixed( precision ) );
}

The input parameter number may "not" always be a number, in this case .toFixed does not exist.


ES 6 Version of Accepted Answer:

function round(value, precision) {
    const multiplier = 10 ** (precision || 0);
    return Math.round(value * multiplier) / multiplier;
}

Why not just

let myNumber = 213.27321;
+myNumber.toFixed(1); // => 213.3
  1. toFixed: returns a string representing the given number using fixed-point notation.
  2. Unary plus (+): The unary plus operator precedes its operand and evaluates to its operand but attempts to convert it into a number, if it isn't already.

If your method does not work, plz post your code.

However,you could accomplish the rounding off task as:

var value = Math.round(234.567*100)/100

Will give you 234.56

Similarly

 var value = Math.round(234.567*10)/10

Will give 234.5

In this way you can use a variable in the place of the constant as used above.


Using toPrecision method:

var a = 1.2345
a.toPrecision(2)

// result "1.2"

Little Angular filter if anyone wants it:

angular.module('filters').filter('decimalPlace', function() {
    return function(num, precision) {
        var multiplier = Math.pow(10, precision || 0);
        return Math.round(num * multiplier) / multiplier;
    };
});

use if via:

{{model.value| decimalPlace}}
{{model.value| decimalPlace:1}}
{{model.value| decimalPlace:2}}

:)


In general, rounding is done by scaling: round(num / p) * p

Using the exponential notation handles rounding of +ve numbers, correctly. However, this method fails to round -ve edge cases correctly.

function round(num, precision = 2) {
	var scaled = Math.round(num + "e" + precision);
	return Number(scaled + "e" + -precision);
}

// testing some edge cases
console.log( round(1.005, 2) );  // 1.01 correct
console.log( round(2.175, 2) );  // 2.18 correct
console.log( round(5.015, 2) );  // 5.02 correct

console.log( round(-1.005, 2) );  // -1    wrong
console.log( round(-2.175, 2) );  // -2.17 wrong
console.log( round(-5.015, 2) );  // -5.01 wrong

Here, also is one function I wrote to do arithmetic rounding, you can test it yourself.

/**
 * MidpointRounding away from zero ('arithmetic' rounding)
 * Uses a half-epsilon for correction. (This offsets IEEE-754
 * half-to-even rounding that was applied at the edge cases).
 */

function RoundCorrect(num, precision = 2) {
	// half epsilon to correct edge cases.
	var c = 0.5 * Number.EPSILON * num;
//	var p = Math.pow(10, precision); //slow
	var p = 1; while (precision--> 0) p *= 10;
	if (num < 0)
		p *= -1;
	return Math.round((num + c) * p) / p;
}

// testing some edge cases
console.log(RoundCorrect(1.005, 2));  // 1.01 correct
console.log(RoundCorrect(2.175, 2));  // 2.18 correct
console.log(RoundCorrect(5.015, 2));  // 5.02 correct

console.log(RoundCorrect(-1.005, 2));  // -1.01 correct
console.log(RoundCorrect(-2.175, 2));  // -2.18 correct
console.log(RoundCorrect(-5.015, 2));  // -5.02 correct


This seems to work reliably across anything I throw at it:

function round(val, multiplesOf) {
  var s = 1 / multiplesOf;
  var res = Math.ceil(val*s)/s;
  res = res < val ? res + multiplesOf: res;
  var afterZero = multiplesOf.toString().split(".")[1];
  return parseFloat(res.toFixed(afterZero ? afterZero.length : 0));
}

It rounds up, so you may need to modify it according to use case. This should work:

console.log(round(10.01, 1)); //outputs 11
console.log(round(10.01, 0.1)); //outputs 10.1

If you care about proper rounding up then:

function roundNumericStrings(str , numOfDecPlacesRequired){ 
     var roundFactor = Math.pow(10, numOfDecPlacesRequired);  
     return (Math.round(parseFloat(str)*roundFactor)/roundFactor).toString();  }

Else if you don't then you already have a reply from previous posts

str.slice(0, -1)

Math.round( num * 10) / 10 doesn't work.

For example, 1455581777.8-145558160.4 gives you 1310023617.3999999.

So only use num.toFixed(1)


I made one that returns number type and also places decimals only if are needed (no 0 padding).

Examples:

roundWithMaxPrecision(11.234, 2); //11.23
roundWithMaxPrecision(11.234, 1); //11.2
roundWithMaxPrecision(11.234, 4); //11.23
roundWithMaxPrecision(11.234, -1); //10

roundWithMaxPrecision(4.2, 2); //4.2
roundWithMaxPrecision(4.88, 1); //4.9

The code:

function roundWithMaxPrecision (n, precision) {
    return Math.round(n * Math.pow(10, precision)) / Math.pow(10, precision);
}

I found a way to avoid the precision problems:

function badRound (num, precision) {
    const x = 10 ** precision;
    return Math.round(num * x) / x
}
// badRound(1.005, 2) --> 1

function round (num, precision) {
    const x = 10 ** (precision + 1);
    const y = 10 ** precision;
    return Math.round(Math.round(num * x) / 10) / y
}
// round(1.005, 2) --> 1.01

Math.round( mul/count * 10 ) / 10

Math.round(Math.sqrt(sqD/y) * 10 ) / 10

Thanks


function rnd(v,n=2) {
    return Math.round((v+Number.EPSILON)*Math.pow(10,n))/Math.pow(10,n)
}

this one catch the corner cases well