How do you round to 1 decimal place in Javascript?

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(input, digits){
  var rounded = Math.pow(10, digits);
  return (Math.round(input * rounded) / rounded).toFixed(digits);
}

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lodash has a round method:

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

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

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

Docs.

Source.

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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.

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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

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You can simply do the following:

let n = 1.25
let result = Number(n).toFixed(1)
// output string: 1.3

---

Why not just

let myNumber = 213.27321;
+myNumber.toFixed(1); // => 213.3

toFixed: returns a string representing the given number using fixed-point notation. 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.

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Using toPrecision method:

var a = 1.2345
a.toPrecision(2)

// result "1.2"

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

Naive implementation

Using the following function with halfway numbers, you will get either the upper rounded value as expected, or the lower rounded value sometimes depending on the input.

This inconsistency in rounding may introduce hard to detect bugs in the client code.

function naiveRound(num, decimalPlaces) { var p = Math.pow(10, decimalPlaces); return Math.round(num * p) / p; } console.log( naiveRound(1.245, 2) ); // 1.25 correct (rounded as expected) console.log( naiveRound(1.255, 2) ); // 1.25 incorrect (should be 1.26)

Better implementations

By converting the number to a string in the exponential notation, positive numbers are rounded as expected. But, be aware that negative numbers round differently than positive numbers.

In fact, it performs what is basically equivalent to "round half up" as the rule, you will see that round(-1.005, 2) evaluates to -1 even though round(1.005, 2) evaluates to 1.01. The lodash _.round method uses this technique.

/** * Round half up ('round half towards positive infinity') * Uses exponential notation to avoid floating-point issues. * Negative numbers round differently than positive numbers. */ function round(num, decimalPlaces) { num = Math.round(num + "e" + decimalPlaces); return Number(num + "e" + -decimalPlaces); } // test rounding of half console.log( round(0.5, 0) ); // 1 console.log( round(-0.5, 0) ); // 0 // testing edge cases console.log( round(1.005, 2) ); // 1.01 console.log( round(2.175, 2) ); // 2.18 console.log( round(5.015, 2) ); // 5.02 console.log( round(-1.005, 2) ); // -1 console.log( round(-2.175, 2) ); // -2.17 console.log( round(-5.015, 2) ); // -5.01

If you want the usual behavior when rounding negative numbers, you would need to convert negative numbers to positive before calling Math.round(), and then convert them back to negative numbers before returning.

// Round half away from zero
function round(num, decimalPlaces) {
    num = Math.round(Math.abs(num) + "e" + decimalPlaces) * Math.sign(num);
    return Number(num + "e" + -decimalPlaces);
}

There is a different purely mathematical technique to perform round-to-nearest (using "round half away from zero"), in which epsilon correction is applied before calling the rounding function.

Simply, we add the smallest possible float value (= 1.0 ulp; unit in the last place) to the number before rounding. This moves to the next representable value after the number, away from zero.

/** * Round half away from zero ('commercial' rounding) * Uses correction to offset floating-point inaccuracies. * Works symmetrically for positive and negative numbers. */ function round(num, decimalPlaces) { var p = Math.pow(10, decimalPlaces); var e = Number.EPSILON * num * p; return Math.round((num * p) + e) / p; } // test rounding of half console.log( round(0.5, 0) ); // 1 console.log( round(-0.5, 0) ); // -1 // testing edge cases console.log( round(1.005, 2) ); // 1.01 console.log( round(2.175, 2) ); // 2.18 console.log( round(5.015, 2) ); // 5.02 console.log( round(-1.005, 2) ); // -1.01 console.log( round(-2.175, 2) ); // -2.18 console.log( round(-5.015, 2) ); // -5.02

This is needed to offset the implicit round-off error that may occur during encoding of decimal numbers, particularly those having "5" in the last decimal position, like 1.005, 2.675 and 16.235. Actually, 1.005 in decimal system is encoded to 1.0049999999999999 in 64-bit binary float; while, 1234567.005 in decimal system is encoded to 1234567.0049999998882413 in 64-bit binary float.

It is worth noting that the maximum binary round-off error is dependent upon (1) the magnitude of the number and (2) the relative machine epsilon (2^-52).

---

var num = 34.7654;

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

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

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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.

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ES 6 Version of Accepted Answer:

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

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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.

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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) {
    const precisionWithPow10 = Math.pow(10, precision);
    return Math.round(n * precisionWithPow10) / precisionWithPow10;
}

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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}}

:)

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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

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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)

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

For example, 1455581777.8-145558160.4 gives you 1310023617.3999999.

So only use num.toFixed(1)

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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

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

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

Thanks

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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

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If your source code is typescript you could use a function like this:

public static ToFixedRounded(decimalNumber: number, fractionDigits: number): number {
    var rounded = Math.pow(10, fractionDigits);
    return (Math.round(decimalNumber * rounded) / rounded).toFixed(fractionDigits) as unknown as number;
}

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const solds = 136780000000; 
const number = (solds >= 1000000000 && solds < 1000000000000) ? { divisor: 1000000000, postfix: "B" }: (solds >= 1000000 && solds < 1000000000) ? { divisor: 1000000, postfix: "M" }: (solds >= 1000 && solds < 1000000) ? { divisor: 1000, postfix: "K" }: { divisor: 1, postfix: null }; 
const floor = Math.floor(solds / number.divisor).toLocaleString(); 
const firstDecimalIndex = solds.toLocaleString().charAt(floor.length+1); 
const final =firstDecimalIndex.match("0")? floor + number.postfix: floor + "." + firstDecimalIndex + number.postfix; 
console.log(final);

136780000000 --> 136.7B

1367800 --> 1.3M

1342 --> 1.3K

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