I need numbers to have only 2 decimals (as in money), and I was using this:
Number(parseFloat(Math.trunc(amount_to_truncate * 100) / 100));
But I can no longer support the Math library.
How can I achieve this without the Math library AND withou rounding the decimals?
You can use toFixed
Number(amount_to_truncate.toFixed(2))
If you are sure that your input always will be lower or equal than 21474836.47 ((2^31 - 1) / 100) (32bit) then:
if you need as string (to make sure result will have 2 decimals)
((amount_to_truncate * 100|0)/100).toFixed(2)
Otherwise
((amount_to_truncate * 100|0)/100)
Else: See Nina Schols's answer
console.log((((15.555 * 100)|0)/100)) // will not round: 15.55
console.log((((15 * 100)|0)/100).toFixed(2)) // will not round: 15.55
Make it simple
const trunc = (n, decimalPlaces) => {
const decimals = decimalPlaces ? decimalPlaces : 2;
const asString = n.toString();
const pos = asString.indexOf('.') != -1 ? asString.indexOf('.') + decimals + 1 : asString.length;
return parseFloat(n.toString().substring(0, pos));
};
console.log(trunc(3.14159265359));
console.log(trunc(11.1111111));
console.log(trunc(3));
console.log(trunc(11));
console.log(trunc(3.1));
console.log(trunc(11.1));
console.log(trunc(3.14));
console.log(trunc(11.11));
console.log(trunc(3.141));
console.log(trunc(11.111));
The only thing I see wrong with toFixed is that it rounds the precision which OP specifically states they don't want to do. Truncate is more equivalent to floor for positive numbers and ceil for negative than round or toFixed. On the MDN page for the Math.trunc there is a polyfill replacement function that would do what OP is expecting.
Math.trunc = Math.trunc || function(x) {
return x - x % 1;
}
If you just used that, then the code wouldn't have to change.
You could use parseInt for a non rounded number.
console.log(parseInt(15.555 * 100, 10) / 100); // 15.55 no rounding
console.log((15.555 * 100 | 0) / 100); // 15.55 no rounding, 32 bit only
console.log((15.555).toFixed(2)); // 15.56 rounding
Try using toFixed:
number.toFixed(2)
Truncate does also a rounding, so your statement: "I need numbers to have only 2 decimals ... without rounding the decimals" seems to me a little bit convoluted and would lead to a long discussion.
Beside this, when dealing with money, the problem isn't Math but how you are using it. I suggest you read the Floating-point cheat sheet for JavaScript - otherwise you will fail even with a simple calculation like 1.40 - 1.00.
The solution to your question is to use a well-tested library for arbitrary-precision decimals like bignumber.js or decimals.js (just as an example).
EDIT:
If you absolutely need a snippet, this is how i did it some time ago:
function round2(d) { return Number(((d+'e'+2)|0)+'e-'+2); }
You could parseInt to truncate, then divide by 100 and parseFloat.
var num = 123.4567;
num=parseInt(num*100);
num=parseFloat(num/100);
alert(num);
See fiddle
Edit: in order to deal with javascript math craziness, you can use .toFixed and an additional digit of multiplication/division:
var num = 123.4567;
num = (num*1000).toFixed();
num = parseInt(num/10);
num = parseFloat(num/100);
alert(num);
Updated fiddle
This was a lot easier than I thought:
const trunc = (number, precision) => {
let index = number.toString().indexOf(".");
let subStr;
// in case of no decimal
if (index === -1) {
subStr = number.toString();
}
// in case of 0 precision
else if (precision === 0) {
subStr = number.toString().substring(0, index);
}
// all else
else {
subStr = number.toString().substring(0, index + 1 + precision);
}
return parseFloat(subStr);
};
let x = trunc(99.12, 1);
console.log("x", x);
You can try this
function trunc(value){
return (!!value && typeof value == "number")? value - value%1 : 0;
}
console.log(trunc(1.4));
console.log(trunc(111.9));
console.log(trunc(0.4));
console.log(trunc("1.4"));
I would like to trim a number into 3 decimal places then round it in 2 decimal places.
For example:
1.234567
Trim it to 1.234
Then round it = 1.23
Another example:
1.389999
Trim it to 1.389
Then round it: 1.39
I tried using toFixed() function but it automatically round it.
Thanks in advance.
Use the Math.floor method to trim it, then the Math.round method to round it:
var n = 1.23456;
n = Math.round(Math.floor(n * 1000) / 10) / 100;
You can multiply the number by a power of 10, use the appropriate math methods, and divide by the same factor.
For rounding, there is Math.round:
function myRound(num, decimals) {
var factor = Math.pow(10, decimals);
return Math.round(num * factor) / factor;
}
For truncating, ECMAScript 6 introduces Math.trunc. For old browsers it can be polyfilled or, assuming the number will be positive, you can use Math.floor.
function myTruncate(num, decimals) {
var factor = Math.pow(10, decimals);
return Math.trunc(num * factor) / factor;
}
Use them like
myTruncate(1.234567, 3); // 1.234
myTruncate(1.389999, 3); // 1.389
myRound(1.234567, 2); // 1.23
myRound(1.389999, 2); // 1.39
Question
Does anyone know of a way to round a float to the nearest 0.05 in JavaScript?
Example
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.50
2.56 | 2.60
Current Code
var _ceil = Math.ceil;
Math.ceil = function(number, decimals){
if (arguments.length == 1)
return _ceil(number);
multiplier = Math.pow(10, decimals);
return _ceil(number * multiplier) / multiplier;
}
Then elsewhere...
return (Math.ceil((amount - 0.05), 1) + 0.05).toFixed(2);
Which is resulting in...
BEFORE | AFTER
2.51 | 2.55
2.50 | 2.55
2.56 | 2.65
Multiply by 20, then divide by 20:
(Math.ceil(number*20)/20).toFixed(2)
Rob's answer with my addition:
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
Otherwise it always rounds up to the nearest 0.05.
** UPDATE **
Sorry has been pointed out this is not what the orig poster wanted.
I would go for the standard of actually dividing by the number you're factoring it to, and rounding that and multiplying it back again after. That seems to be a proper working method which you can use with any number and maintain the mental image of what you are trying to achieve.
var val = 26.14,
factor = 0.05;
val = Math.round(val / factor) * factor;
This will work for tens, hundreds or any number. If you are specifically rounding to the higher number then use Math.ceil instead of Math.round.
Another method specifically for rounding just to 1 or more decimal places (rather than half a place) is the following:
Number(Number(1.5454545).toFixed(1));
It creates a fixed number string and then turns it into a real Number.
I would write a function that does it for you by
move the decimal over two places (multiply by 100)
then mod (%) that inflatedNumber by 5 and get the remainder
subtract the remainder from 5 so that you know what the 'gap'(ceilGap) is between your number and the next closest .05
finally, divide your inflatedNumber by 100 so that it goes back to your original float, and voila, your num will be rounded up to the nearest .05.
function calcNearestPointZeroFive(num){
var inflatedNumber = num*100,
remainder = inflatedNumber % 5;
ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}
If you want to leave numbers like 5.50 untouched you can always add this checker:
if (remainder===0){
return num
} else {
var ceilGap = 5 - remainder
return (inflatedNumber + ceilGap)/100
}
You need to put -1 to round half down and after that multiply by -1 like the example down bellow.
<script type="text/javascript">
function roundNumber(number, precision, isDown) {
var factor = Math.pow(10, precision);
var tempNumber = number * factor;
var roundedTempNumber = 0;
if (isDown) {
tempNumber = -tempNumber;
roundedTempNumber = Math.round(tempNumber) * -1;
} else {
roundedTempNumber = Math.round(tempNumber);
}
return roundedTempNumber / factor;
}
</script>
<div class="col-sm-12">
<p>Round number 1.25 down: <script>document.write(roundNumber(1.25, 1, true));</script>
</p>
<p>Round number 1.25 up: <script>document.write(roundNumber(1.25, 1, false));</script></p>
</div>
I ended up using this function in my project, successfully:
roundToNearestFiveCents( number: any ) {
return parseFloat((Math.round(number / 0.05) * 0.05).toFixed(2));
}
Might be of use to someone wanting to simply round to the nearest 5 cents on their monetary results, keeps the result a number, so if you perform addition on it further it won't result in string concatenation; also doesn't unnecessarily round up as a few of the other answers pointed out. Also limits it to two decimals, which is customary with finance.
My solution and test:
let round = function(number, precision = 2, rounding = 0.05) {
let multiply = 1 / rounding;
return parseFloat((Math.round(number * multiply) / multiply)).toFixed(precision);
};
https://jsfiddle.net/maciejSzewczyk/7r1tvhdk/40/
Even though the OP is not explicit about banker rounding, rounding up to the nearest $0.05 (5 cents) should be compatible with banker rounding. What suggested by Arth is more accurate than the accepted answer by Rob W.
(Math.ceil(number*20 - 0.5)/20).toFixed(2)
With banker rounding, you need a basic banker rounding function as suggested at Gaussian/banker's rounding in JavaScript, and I rewrite in TypeScript:
static bankerRound(num: number, decimalPlaces?: number) {
const d = decimalPlaces || 0;
const m = Math.pow(10, d);
const n = +(d ? num * m : num).toFixed(8);
const i = Math.floor(n), f = n - i;
const e = 1e-8;
const r = (f > 0.5 - e && f < 0.5 + e) ?
((i % 2 === 0) ? i : i + 1) : Math.round(n);
return d ? r / m : r;
}
static roundTo5cents(num: number) {
const r = bankerRound(Math.ceil(num * 20 - 0.5) / 20, 2);
return r;
}
The correctness of this algorithm could be verified through MBS Online, e.g. http://www9.health.gov.au/mbs/ready_reckoner.cfm?item_num=60
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(input, digits){
var rounder = Math.pow(10, digits);
return (Math.round(input * rounder) / rounder).toFixed(digits);
}
lodash has a round method:
_.round(4.006);
// => 4
_.round(4.006, 2);
// => 4.01
_.round(4060, -2);
// => 4100
Docs.
Source.
You can simply do the following:
let n = 1.25
let result = Number(n).toFixed(1)
// output string: 1.3
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
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.
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).
Using toPrecision method:
var a = 1.2345
a.toPrecision(2)
// result "1.2"
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;
}
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.
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;
}
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}}
:)
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 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
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;
}
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