I am trying to truncate decimal numbers to decimal places. Something like this:
5.467 -> 5.46
985.943 -> 985.94
toFixed(2) does just about the right thing but it rounds off the value. I don't need the value rounded off. Hope this is possible in javascript.
Dogbert's answer is good, but if your code might have to deal with negative numbers, Math.floor by itself may give unexpected results.
E.g. Math.floor(4.3) = 4, but Math.floor(-4.3) = -5
Use a helper function like this one instead to get consistent results:
truncateDecimals = function (number) {
return Math[number < 0 ? 'ceil' : 'floor'](number);
};
// Applied to Dogbert's answer:
var a = 5.467;
var truncated = truncateDecimals(a * 100) / 100; // = 5.46
Here's a more convenient version of this function:
truncateDecimals = function (number, digits) {
var multiplier = Math.pow(10, digits),
adjustedNum = number * multiplier,
truncatedNum = Math[adjustedNum < 0 ? 'ceil' : 'floor'](adjustedNum);
return truncatedNum / multiplier;
};
// Usage:
var a = 5.467;
var truncated = truncateDecimals(a, 2); // = 5.46
// Negative digits:
var b = 4235.24;
var truncated = truncateDecimals(b, -2); // = 4200
If that isn't desired behaviour, insert a call to Math.abs on the first line:
var multiplier = Math.pow(10, Math.abs(digits)),
EDIT: shendz correctly points out that using this solution with a = 17.56 will incorrectly produce 17.55. For more about why this happens, read What Every Computer Scientist Should Know About Floating-Point Arithmetic. Unfortunately, writing a solution that eliminates all sources of floating-point error is pretty tricky with javascript. In another language you'd use integers or maybe a Decimal type, but with javascript...
This solution should be 100% accurate, but it will also be slower:
function truncateDecimals (num, digits) {
var numS = num.toString(),
decPos = numS.indexOf('.'),
substrLength = decPos == -1 ? numS.length : 1 + decPos + digits,
trimmedResult = numS.substr(0, substrLength),
finalResult = isNaN(trimmedResult) ? 0 : trimmedResult;
return parseFloat(finalResult);
}
For those who need speed but also want to avoid floating-point errors, try something like BigDecimal.js. You can find other javascript BigDecimal libraries in this SO question: "Is there a good Javascript BigDecimal library?" and here's a good blog post about math libraries for Javascript
upd:
So, after all it turned out, rounding bugs will always haunt you, no matter how hard you try to compensate them. Hence the problem should be attacked by representing numbers exactly in decimal notation.
Number.prototype.toFixedDown = function(digits) {
var re = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)"),
m = this.toString().match(re);
return m ? parseFloat(m[1]) : this.valueOf();
};
[ 5.467.toFixedDown(2),
985.943.toFixedDown(2),
17.56.toFixedDown(2),
(0).toFixedDown(1),
1.11.toFixedDown(1) + 22];
// [5.46, 985.94, 17.56, 0, 23.1]
Old error-prone solution based on compilation of others':
Number.prototype.toFixedDown = function(digits) {
var n = this - Math.pow(10, -digits)/2;
n += n / Math.pow(2, 53); // added 1360765523: 17.56.toFixedDown(2) === "17.56"
return n.toFixed(digits);
}
var a = 5.467;
var truncated = Math.floor(a * 100) / 100; // = 5.46
You can fix the rounding by subtracting 0.5 for toFixed, e.g.
(f - 0.005).toFixed(2)
Nice one-line solution:
function truncate (num, places) {
return Math.trunc(num * Math.pow(10, places)) / Math.pow(10, places);
}
Then call it with:
truncate(3.5636232, 2); // returns 3.56
truncate(5.4332312, 3); // returns 5.433
truncate(25.463214, 4); // returns 25.4632
Consider taking advantage of the double tilde: ~~.
Take in the number. Multiply by significant digits after the decimal so that you can truncate to zero places with ~~. Divide that multiplier back out. Profit.
function truncator(numToTruncate, intDecimalPlaces) {
var numPower = Math.pow(10, intDecimalPlaces); // "numPowerConverter" might be better
return ~~(numToTruncate * numPower)/numPower;
}
I'm trying to resist wrapping the ~~ call in parens; order of operations should make that work correctly, I believe.
alert(truncator(5.1231231, 1)); // is 5.1
alert(truncator(-5.73, 1)); // is -5.7
alert(truncator(-5.73, 0)); // is -5
JSFiddle link.
EDIT: Looking back over, I've unintentionally also handled cases to round off left of the decimal as well.
alert(truncator(4343.123, -2)); // gives 4300.
The logic's a little wacky looking for that usage, and may benefit from a quick refactor. But it still works. Better lucky than good.
I thought I'd throw in an answer using | since it is simple and works well.
truncate = function(number, places) {
var shift = Math.pow(10, places);
return ((number * shift) | 0) / shift;
};
Truncate using bitwise operators:
~~0.5 === 0
~~(-0.5) === 0
~~14.32794823 === 14
~~(-439.93) === -439
#Dogbert's answer can be improved with Math.trunc, which truncates instead of rounding.
There is a difference between rounding and truncating. Truncating is
clearly the behaviour this question is seeking. If I call
truncate(-3.14) and receive -4 back, I would definitely call that
undesirable. – #NickKnowlson
var a = 5.467;
var truncated = Math.trunc(a * 100) / 100; // = 5.46
var a = -5.467;
var truncated = Math.trunc(a * 100) / 100; // = -5.46
I wrote an answer using a shorter method. Here is what I came up with
function truncate(value, precision) {
var step = Math.pow(10, precision || 0);
var temp = Math.trunc(step * value);
return temp / step;
}
The method can be used like so
truncate(132456.25456789, 5)); // Output: 132456.25456
truncate(132456.25456789, 3)); // Output: 132456.254
truncate(132456.25456789, 1)); // Output: 132456.2
truncate(132456.25456789)); // Output: 132456
Or, if you want a shorter syntax, here you go
function truncate(v, p) {
var s = Math.pow(10, p || 0);
return Math.trunc(s * v) / s;
}
I think this function could be a simple solution:
function trunc(decimal,n=2){
let x = decimal + ''; // string
return x.lastIndexOf('.')>=0?parseFloat(x.substr(0,x.lastIndexOf('.')+(n+1))):decimal; // You can use indexOf() instead of lastIndexOf()
}
console.log(trunc(-241.31234,2));
console.log(trunc(241.312,5));
console.log(trunc(-241.233));
console.log(trunc(241.2,0));
console.log(trunc(241));
Number.prototype.trim = function(decimals) {
var s = this.toString();
var d = s.split(".");
d[1] = d[1].substring(0, decimals);
return parseFloat(d.join("."));
}
console.log((5.676).trim(2)); //logs 5.67
I'm a bit confused as to why there are so many different answers to such a fundamentally simple question; there are only two approaches which I saw which seemed to be worth looking at. I did a quick benchmark to see the speed difference using https://jsbench.me/.
This is the solution which is currently (9/26/2020) flagged as the answer:
function truncate(n, digits) {
var re = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)"),
m = n.toString().match(re);
return m ? parseFloat(m[1]) : n.valueOf();
};
[ truncate(5.467,2),
truncate(985.943,2),
truncate(17.56,2),
truncate(0, 1),
truncate(1.11, 1) + 22];
However, this is doing string and regex stuff, which is usually not very efficient, and there is a Math.trunc function which does exactly what the OP wants just with no decimals. Therefore, you can easily use that plus a little extra arithmetic to get the same thing.
Here is another solution I found on this thread, which is the one I would use:
function truncate(n, digits) {
var step = Math.pow(10, digits || 0);
var temp = Math.trunc(step * n);
return temp / step;
}
[ truncate(5.467,2),
truncate(985.943,2),
truncate(17.56,2),
truncate(0, 1),
truncate(1.11, 1) + 22];
The first method is "99.92% slower" than the second, so the second is definitely the one I would recommend using.
Okay, back to finding other ways to avoid work...
I found a problem: considering the next situation: 2.1 or 1.2 or -6.4
What if you want always 3 decimals or two or wharever, so, you have to complete the leading zeros to the right
// 3 decimals numbers
0.5 => 0.500
// 6 decimals
0.1 => 0.10000
// 4 decimales
-2.1 => -2.1000
// truncate to 3 decimals
3.11568 => 3.115
This is the fixed function of Nick Knowlson
function truncateDecimals (num, digits)
{
var numS = num.toString();
var decPos = numS.indexOf('.');
var substrLength = decPos == -1 ? numS.length : 1 + decPos + digits;
var trimmedResult = numS.substr(0, substrLength);
var finalResult = isNaN(trimmedResult) ? 0 : trimmedResult;
// adds leading zeros to the right
if (decPos != -1){
var s = trimmedResult+"";
decPos = s.indexOf('.');
var decLength = s.length - decPos;
while (decLength <= digits){
s = s + "0";
decPos = s.indexOf('.');
decLength = s.length - decPos;
substrLength = decPos == -1 ? s.length : 1 + decPos + digits;
};
finalResult = s;
}
return finalResult;
};
https://jsfiddle.net/huttn155/7/
function toFixed(number, digits) {
var reg_ex = new RegExp("(\\d+\\.\\d{" + digits + "})(\\d)")
var array = number.toString().match(reg_ex);
return array ? parseFloat(array[1]) : number.valueOf()
}
var test = 10.123456789
var __fixed = toFixed(test, 6)
console.log(__fixed)
// => 10.123456
The answer by #kirilloid seems to be the correct answer, however, the main code needs to be updated. His solution doesn't take care of negative numbers (which someone did mention in the comment section but has not been updated in the main code).
Updating that to a complete final tested solution:
Number.prototype.toFixedDown = function(digits) {
var re = new RegExp("([-]*\\d+\\.\\d{" + digits + "})(\\d)"),
m = this.toString().match(re);
return m ? parseFloat(m[1]) : this.valueOf();
};
Sample Usage:
var x = 3.1415629;
Logger.log(x.toFixedDown(2)); //or use whatever you use to log
Fiddle: JS Number Round down
PS: Not enough repo to comment on that solution.
Here my take on the subject:
convert.truncate = function(value, decimals) {
decimals = (decimals === undefined ? 0 : decimals);
return parseFloat((value-(0.5/Math.pow(10, decimals))).toFixed(decimals),10);
};
It's just a slightly more elaborate version of
(f - 0.005).toFixed(2)
Here is simple but working function to truncate number upto 2 decimal places.
function truncateNumber(num) {
var num1 = "";
var num2 = "";
var num1 = num.split('.')[0];
num2 = num.split('.')[1];
var decimalNum = num2.substring(0, 2);
var strNum = num1 +"."+ decimalNum;
var finalNum = parseFloat(strNum);
return finalNum;
}
The resulting type remains a number...
/* Return the truncation of n wrt base */
var trunc = function(n, base) {
n = (n / base) | 0;
return base * n;
};
var t = trunc(5.467, 0.01);
Lodash has a few Math utility methods that can round, floor, and ceil a number to a given decimal precision. This leaves off trailing zeroes.
They take an interesting approach, using the exponent of a number. Apparently this avoids rounding issues.
(Note: func is Math.round or ceil or floor in the code below)
// Shift with exponential notation to avoid floating-point issues.
var pair = (toString(number) + 'e').split('e'),
value = func(pair[0] + 'e' + (+pair[1] + precision));
pair = (toString(value) + 'e').split('e');
return +(pair[0] + 'e' + (+pair[1] - precision));
Link to the source code
const TO_FIXED_MAX = 100;
function truncate(number, decimalsPrecison) {
// make it a string with precision 1e-100
number = number.toFixed(TO_FIXED_MAX);
// chop off uneccessary digits
const dotIndex = number.indexOf('.');
number = number.substring(0, dotIndex + decimalsPrecison + 1);
// back to a number data type (app specific)
return Number.parseFloat(number);
}
// example
truncate(0.00000001999, 8);
0.00000001
works with:
negative numbers
very small numbers (Number.EPSILON precision)
The one that is mark as the solution is the better solution I been found until today, but has a serious problem with 0 (for example, 0.toFixedDown(2) gives -0.01). So I suggest to use this:
Number.prototype.toFixedDown = function(digits) {
if(this == 0) {
return 0;
}
var n = this - Math.pow(10, -digits)/2;
n += n / Math.pow(2, 53); // added 1360765523: 17.56.toFixedDown(2) === "17.56"
return n.toFixed(digits);
}
Here is what I use:
var t = 1;
for (var i = 0; i < decimalPrecision; i++)
t = t * 10;
var f = parseFloat(value);
return (Math.floor(f * t)) / t;
You can work with strings.
It Checks if '.' exists, and then removes part of string.
truncate (7.88, 1) --> 7.8
truncate (7.889, 2) --> 7.89
truncate (-7.88, 1 ) --> -7.88
function truncate(number, decimals) {
const tmp = number + '';
if (tmp.indexOf('.') > -1) {
return +tmp.substr(0 , tmp.indexOf('.') + decimals+1 );
} else {
return +number
}
}
function trunc(num, dec) {
const pow = 10 ** dec
return Math.trunc(num * pow) / pow
}
// ex.
trunc(4.9634, 1) // 4.9
trunc(4.9634, 2) // 4.96
trunc(-4.9634, 1) // -4.9
You can use toFixed(2) to convert your float to a string with 2 decimal points. Then you can wrap that in floatParse() to convert that string back to a float to make it usable for calculations or db storage.
const truncatedNumber = floatParse(num.toFixed(2))
I am not sure of the potential drawbacks of this answer like increased processing time but I tested edge cases from other comments like .29 which returns .29 (not .28 like other solutions). It also handles negative numbers.
just to point out a simple solution that worked for me
convert it to string and then regex it...
var number = 123.45678;
var number_s = '' + number;
var number_truncated_s = number_s.match(/\d*\.\d{4}/)[0]
var number_truncated = parseFloat(number_truncated_s)
It can be abbreviated to
var number_truncated = parseFloat(('' + 123.4568908).match(/\d*\.\d{4}/)[0])
Here is an ES6 code which does what you want
const truncateTo = (unRouned, nrOfDecimals = 2) => {
const parts = String(unRouned).split(".");
if (parts.length !== 2) {
// without any decimal part
return unRouned;
}
const newDecimals = parts[1].slice(0, nrOfDecimals),
newString = `${parts[0]}.${newDecimals}`;
return Number(newString);
};
// your examples
console.log(truncateTo(5.467)); // ---> 5.46
console.log(truncateTo(985.943)); // ---> 985.94
// other examples
console.log(truncateTo(5)); // ---> 5
console.log(truncateTo(-5)); // ---> -5
console.log(truncateTo(-985.943)); // ---> -985.94
Suppose you want to truncate number x till n digits.
Math.trunc(x * pow(10,n))/pow(10,n);
Number.prototype.truncate = function(places) {
var shift = Math.pow(10, places);
return Math.trunc(this * shift) / shift;
};
I have this line of code which rounds my numbers to two decimal places. But I get numbers like this: 10.8, 2.4, etc. These are not my idea of two decimal places so how I can improve the following?
Math.round(price*Math.pow(10,2))/Math.pow(10,2);
I want numbers like 10.80, 2.40, etc. Use of jQuery is fine with me.
To format a number using fixed-point notation, you can simply use the toFixed method:
(10.8).toFixed(2); // "10.80"
var num = 2.4;
alert(num.toFixed(2)); // "2.40"
Note that toFixed() returns a string.
IMPORTANT: Note that toFixed does not round 90% of the time, it will return the rounded value, but for many cases, it doesn't work.
For instance:
2.005.toFixed(2) === "2.00"
UPDATE:
Nowadays, you can use the Intl.NumberFormat constructor. It's part of the ECMAScript Internationalization API Specification (ECMA402). It has pretty good browser support, including even IE11, and it is fully supported in Node.js.
const formatter = new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(formatter.format(2.005)); // "2.01"
console.log(formatter.format(1.345)); // "1.35"
You can alternatively use the toLocaleString method, which internally will use the Intl API:
const format = (num, decimals) => num.toLocaleString('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
});
console.log(format(2.005)); // "2.01"
console.log(format(1.345)); // "1.35"
This API also provides you a wide variety of options to format, like thousand separators, currency symbols, etc.
This is an old topic but still top-ranked Google results and the solutions offered share the same floating point decimals issue. Here is the (very generic) function I use, thanks to MDN:
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
As we can see, we don't get these issues:
round(1.275, 2); // Returns 1.28
round(1.27499, 2); // Returns 1.27
This genericity also provides some cool stuff:
round(1234.5678, -2); // Returns 1200
round(1.2345678e+2, 2); // Returns 123.46
round("123.45"); // Returns 123
Now, to answer the OP's question, one has to type:
round(10.8034, 2).toFixed(2); // Returns "10.80"
round(10.8, 2).toFixed(2); // Returns "10.80"
Or, for a more concise, less generic function:
function round2Fixed(value) {
value = +value;
if (isNaN(value))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + 2) : 2)));
// Shift back
value = value.toString().split('e');
return (+(value[0] + 'e' + (value[1] ? (+value[1] - 2) : -2))).toFixed(2);
}
You can call it with:
round2Fixed(10.8034); // Returns "10.80"
round2Fixed(10.8); // Returns "10.80"
Various examples and tests (thanks to #t-j-crowder!):
function round(value, exp) {
if (typeof exp === 'undefined' || +exp === 0)
return Math.round(value);
value = +value;
exp = +exp;
if (isNaN(value) || !(typeof exp === 'number' && exp % 1 === 0))
return NaN;
// Shift
value = value.toString().split('e');
value = Math.round(+(value[0] + 'e' + (value[1] ? (+value[1] + exp) : exp)));
// Shift back
value = value.toString().split('e');
return +(value[0] + 'e' + (value[1] ? (+value[1] - exp) : -exp));
}
function naive(value, exp) {
if (!exp) {
return Math.round(value);
}
var pow = Math.pow(10, exp);
return Math.round(value * pow) / pow;
}
function test(val, places) {
subtest(val, places);
val = typeof val === "string" ? "-" + val : -val;
subtest(val, places);
}
function subtest(val, places) {
var placesOrZero = places || 0;
var naiveResult = naive(val, places);
var roundResult = round(val, places);
if (placesOrZero >= 0) {
naiveResult = naiveResult.toFixed(placesOrZero);
roundResult = roundResult.toFixed(placesOrZero);
} else {
naiveResult = naiveResult.toString();
roundResult = roundResult.toString();
}
$("<tr>")
.append($("<td>").text(JSON.stringify(val)))
.append($("<td>").text(placesOrZero))
.append($("<td>").text(naiveResult))
.append($("<td>").text(roundResult))
.appendTo("#results");
}
test(0.565, 2);
test(0.575, 2);
test(0.585, 2);
test(1.275, 2);
test(1.27499, 2);
test(1234.5678, -2);
test(1.2345678e+2, 2);
test("123.45");
test(10.8034, 2);
test(10.8, 2);
test(1.005, 2);
test(1.0005, 2);
table {
border-collapse: collapse;
}
table, td, th {
border: 1px solid #ddd;
}
td, th {
padding: 4px;
}
th {
font-weight: normal;
font-family: sans-serif;
}
td {
font-family: monospace;
}
<table>
<thead>
<tr>
<th>Input</th>
<th>Places</th>
<th>Naive</th>
<th>Thorough</th>
</tr>
</thead>
<tbody id="results">
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
I usually add this to my personal library, and after some suggestions and using the #TIMINeutron solution too, and making it adaptable for decimal length then, this one fits best:
function precise_round(num, decimals) {
var t = Math.pow(10, decimals);
return (Math.round((num * t) + (decimals>0?1:0)*(Math.sign(num) * (10 / Math.pow(100, decimals)))) / t).toFixed(decimals);
}
will work for the exceptions reported.
FAST AND EASY
parseFloat(number.toFixed(2))
Example
let number = 2.55435930
let roundedString = number.toFixed(2) // "2.55"
let twoDecimalsNumber = parseFloat(roundedString) // 2.55
let directly = parseFloat(number.toFixed(2)) // 2.55
One way to be 100% sure that you get a number with 2 decimals:
(Math.round(num*100)/100).toFixed(2)
If this causes rounding errors, you can use the following as James has explained in his comment:
(Math.round((num * 1000)/10)/100).toFixed(2)
I don't know why can't I add a comment to a previous answer (maybe I'm hopelessly blind, I don't know), but I came up with a solution using #Miguel's answer:
function precise_round(num,decimals) {
return Math.round(num*Math.pow(10, decimals)) / Math.pow(10, decimals);
}
And its two comments (from #bighostkim and #Imre):
Problem with precise_round(1.275,2) not returning 1.28
Problem with precise_round(6,2) not returning 6.00 (as he wanted).
My final solution is as follows:
function precise_round(num,decimals) {
var sign = num >= 0 ? 1 : -1;
return (Math.round((num*Math.pow(10,decimals)) + (sign*0.001)) / Math.pow(10,decimals)).toFixed(decimals);
}
As you can see I had to add a little bit of "correction" (it's not what it is, but since Math.round is lossy - you can check it on jsfiddle.net - this is the only way I knew how to "fix" it). It adds 0.001 to the already padded number, so it is adding a 1 three 0s to the right of the decimal value. So it should be safe to use.
After that I added .toFixed(decimal) to always output the number in the correct format (with the right amount of decimals).
So that's pretty much it. Use it well ;)
EDIT: added functionality to the "correction" of negative numbers.
toFixed(n) provides n length after the decimal point; toPrecision(x)
provides x total length.
Use this method below
// Example: toPrecision(4) when the number has 7 digits (3 before, 4 after)
// It will round to the tenths place
num = 500.2349;
result = num.toPrecision(4); // result will equal 500.2
AND if you want the number to be fixed use
result = num.toFixed(2);
I didn't find an accurate solution for this problem, so I created my own:
function inprecise_round(value, decPlaces) {
return Math.round(value*Math.pow(10,decPlaces))/Math.pow(10,decPlaces);
}
function precise_round(value, decPlaces){
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val-parseInt(val))*10)/10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if(fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
Examples:
function inprecise_round(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
function precise_round(value, decPlaces) {
var val = value * Math.pow(10, decPlaces);
var fraction = (Math.round((val - parseInt(val)) * 10) / 10);
//this line is for consistency with .NET Decimal.Round behavior
// -342.055 => -342.06
if (fraction == -0.5) fraction = -0.6;
val = Math.round(parseInt(val) + fraction) / Math.pow(10, decPlaces);
return val;
}
// This may produce different results depending on the browser environment
console.log("342.055.toFixed(2) :", 342.055.toFixed(2)); // 342.06 on Chrome & IE10
console.log("inprecise_round(342.055, 2):", inprecise_round(342.055, 2)); // 342.05
console.log("precise_round(342.055, 2) :", precise_round(342.055, 2)); // 342.06
console.log("precise_round(-342.055, 2) :", precise_round(-342.055, 2)); // -342.06
console.log("inprecise_round(0.565, 2) :", inprecise_round(0.565, 2)); // 0.56
console.log("precise_round(0.565, 2) :", precise_round(0.565, 2)); // 0.57
Here's a simple one
function roundFloat(num,dec){
var d = 1;
for (var i=0; i<dec; i++){
d += "0";
}
return Math.round(num * d) / d;
}
Use like alert(roundFloat(1.79209243929,4));
Jsfiddle
Round down
function round_down(value, decPlaces) {
return Math.floor(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round up
function round_up(value, decPlaces) {
return Math.ceil(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Round nearest
function round_nearest(value, decPlaces) {
return Math.round(value * Math.pow(10, decPlaces)) / Math.pow(10, decPlaces);
}
Merged https://stackoverflow.com/a/7641824/1889449 and
https://www.kirupa.com/html5/rounding_numbers_in_javascript.htm Thanks
them.
Building on top of Christian C. Salvadó's answer, doing the following will output a Number type, and also seems to be dealing with rounding well:
const roundNumberToTwoDecimalPlaces = (num) => Number(new Intl.NumberFormat('en-US', {
minimumFractionDigits: 2,
maximumFractionDigits: 2,
}).format(num));
roundNumberToTwoDecimalPlaces(1.344); // => 1.34
roundNumberToTwoDecimalPlaces(1.345); // => 1.35
The difference between the above and what has already been mentioned is that you don't need the .format() chaining when you're using it[, and that it outputs a Number type].
#heridev and I created a small function in jQuery.
You can try next:
HTML
<input type="text" name="one" class="two-digits"><br>
<input type="text" name="two" class="two-digits">
jQuery
// apply the two-digits behaviour to elements with 'two-digits' as their class
$( function() {
$('.two-digits').keyup(function(){
if($(this).val().indexOf('.')!=-1){
if($(this).val().split(".")[1].length > 2){
if( isNaN( parseFloat( this.value ) ) ) return;
this.value = parseFloat(this.value).toFixed(2);
}
}
return this; //for chaining
});
});
DEMO ONLINE:
http://jsfiddle.net/c4Wqn/
The trouble with floating point values is that they are trying to represent an infinite amount of (continuous) values with a fixed amount of bits. So naturally, there must be some loss in play, and you're going to be bitten with some values.
When a computer stores 1.275 as a floating point value, it won't actually remember whether it was 1.275 or 1.27499999999999993, or even 1.27500000000000002. These values should give different results after rounding to two decimals, but they won't, since for computer they look exactly the same after storing as floating point values, and there's no way to restore the lost data. Any further calculations will only accumulate such imprecision.
So, if precision matters, you have to avoid floating point values from the start. The simplest options are to
use a devoted library
use strings for storing and passing around the values (accompanied by string operations)
use integers (e.g. you could be passing around the amount of hundredths of your actual value, e.g. amount in cents instead of amount in dollars)
For example, when using integers to store the number of hundredths, the function for finding the actual value is quite simple:
function descale(num, decimals) {
var hasMinus = num < 0;
var numString = Math.abs(num).toString();
var precedingZeroes = '';
for (var i = numString.length; i <= decimals; i++) {
precedingZeroes += '0';
}
numString = precedingZeroes + numString;
return (hasMinus ? '-' : '')
+ numString.substr(0, numString.length-decimals)
+ '.'
+ numString.substr(numString.length-decimals);
}
alert(descale(127, 2));
With strings, you'll need rounding, but it's still manageable:
function precise_round(num, decimals) {
var parts = num.split('.');
var hasMinus = parts.length > 0 && parts[0].length > 0 && parts[0].charAt(0) == '-';
var integralPart = parts.length == 0 ? '0' : (hasMinus ? parts[0].substr(1) : parts[0]);
var decimalPart = parts.length > 1 ? parts[1] : '';
if (decimalPart.length > decimals) {
var roundOffNumber = decimalPart.charAt(decimals);
decimalPart = decimalPart.substr(0, decimals);
if ('56789'.indexOf(roundOffNumber) > -1) {
var numbers = integralPart + decimalPart;
var i = numbers.length;
var trailingZeroes = '';
var justOneAndTrailingZeroes = true;
do {
i--;
var roundedNumber = '1234567890'.charAt(parseInt(numbers.charAt(i)));
if (roundedNumber === '0') {
trailingZeroes += '0';
} else {
numbers = numbers.substr(0, i) + roundedNumber + trailingZeroes;
justOneAndTrailingZeroes = false;
break;
}
} while (i > 0);
if (justOneAndTrailingZeroes) {
numbers = '1' + trailingZeroes;
}
integralPart = numbers.substr(0, numbers.length - decimals);
decimalPart = numbers.substr(numbers.length - decimals);
}
} else {
for (var i = decimalPart.length; i < decimals; i++) {
decimalPart += '0';
}
}
return (hasMinus ? '-' : '') + integralPart + (decimals > 0 ? '.' + decimalPart : '');
}
alert(precise_round('1.275', 2));
alert(precise_round('1.27499999999999993', 2));
Note that this function rounds to nearest, ties away from zero, while IEEE 754 recommends rounding to nearest, ties to even as the default behavior for floating point operations. Such modifications are left as an exercise for the reader :)
Round your decimal value, then use toFixed(x) for your expected digit(s).
function parseDecimalRoundAndFixed(num,dec){
var d = Math.pow(10,dec);
return (Math.round(num * d) / d).toFixed(dec);
}
Call
parseDecimalRoundAndFixed(10.800243929,4) => 10.80
parseDecimalRoundAndFixed(10.807243929,2) => 10.81
Number(Math.round(1.005+'e2')+'e-2'); // 1.01
This worked for me: Rounding Decimals in JavaScript
With these examples you will still get an error when trying to round the number 1.005 the solution is to either use a library like Math.js or this function:
function round(value: number, decimals: number) {
return Number(Math.round(value + 'e' + decimals) + 'e-' + decimals);
}
Here is my 1-line solution: Number((yourNumericValueHere).toFixed(2));
Here's what happens:
1) First, you apply .toFixed(2) onto the number that you want to round off the decimal places of. Note that this will convert the value to a string from number. So if you are using Typescript, it will throw an error like this:
"Type 'string' is not assignable to type 'number'"
2) To get back the numeric value or to convert the string to numeric value, simply apply the Number() function on that so-called 'string' value.
For clarification, look at the example below:
EXAMPLE:
I have an amount that has upto 5 digits in the decimal places and I would like to shorten it to upto 2 decimal places. I do it like so:
var price = 0.26453;
var priceRounded = Number((price).toFixed(2));
console.log('Original Price: ' + price);
console.log('Price Rounded: ' + priceRounded);
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).
Put the following in some global scope:
Number.prototype.getDecimals = function ( decDigCount ) {
return this.toFixed(decDigCount);
}
and then try:
var a = 56.23232323;
a.getDecimals(2); // will return 56.23
Update
Note that toFixed() can only work for the number of decimals between 0-20 i.e. a.getDecimals(25) may generate a javascript error, so to accomodate that you may add some additional check i.e.
Number.prototype.getDecimals = function ( decDigCount ) {
return ( decDigCount > 20 ) ? this : this.toFixed(decDigCount);
}
Number(((Math.random() * 100) + 1).toFixed(2))
this will return a random number from 1 to 100 rounded to 2 decimal places.
Using this response by reference: https://stackoverflow.com/a/21029698/454827
I build a function to get dynamic numbers of decimals:
function toDec(num, dec)
{
if(typeof dec=='undefined' || dec<0)
dec = 2;
var tmp = dec + 1;
for(var i=1; i<=tmp; i++)
num = num * 10;
num = num / 10;
num = Math.round(num);
for(var i=1; i<=dec; i++)
num = num / 10;
num = num.toFixed(dec);
return num;
}
here working example: https://jsfiddle.net/wpxLduLc/
parse = function (data) {
data = Math.round(data*Math.pow(10,2))/Math.pow(10,2);
if (data != null) {
var lastone = data.toString().split('').pop();
if (lastone != '.') {
data = parseFloat(data);
}
}
return data;
};
$('#result').html(parse(200)); // output 200
$('#result1').html(parse(200.1)); // output 200.1
$('#result2').html(parse(200.10)); // output 200.1
$('#result3').html(parse(200.109)); // output 200.11
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.0/jquery.min.js"></script>
<div id="result"></div>
<div id="result1"></div>
<div id="result2"></div>
<div id="result3"></div>
I got some ideas from this post a few months back, but none of the answers here, nor answers from other posts/blogs could handle all the scenarios (e.g. negative numbers and some "lucky numbers" our tester found). In the end, our tester did not find any problem with this method below. Pasting a snippet of my code:
fixPrecision: function (value) {
var me = this,
nan = isNaN(value),
precision = me.decimalPrecision;
if (nan || !value) {
return nan ? '' : value;
} else if (!me.allowDecimals || precision <= 0) {
precision = 0;
}
//[1]
//return parseFloat(Ext.Number.toFixed(parseFloat(value), precision));
precision = precision || 0;
var negMultiplier = value < 0 ? -1 : 1;
//[2]
var numWithExp = parseFloat(value + "e" + precision);
var roundedNum = parseFloat(Math.round(Math.abs(numWithExp)) + 'e-' + precision) * negMultiplier;
return parseFloat(roundedNum.toFixed(precision));
},
I also have code comments (sorry i forgot all the details already)...I'm posting my answer here for future reference:
9.995 * 100 = 999.4999999999999
Whereas 9.995e2 = 999.5
This discrepancy causes Math.round(9.995 * 100) = 999 instead of 1000.
Use e notation instead of multiplying /dividing by Math.Pow(10,precision).
I'm fix the problem the modifier.
Support 2 decimal only.
$(function(){
//input number only.
convertNumberFloatZero(22); // output : 22.00
convertNumberFloatZero(22.5); // output : 22.50
convertNumberFloatZero(22.55); // output : 22.55
convertNumberFloatZero(22.556); // output : 22.56
convertNumberFloatZero(22.555); // output : 22.55
convertNumberFloatZero(22.5541); // output : 22.54
convertNumberFloatZero(22222.5541); // output : 22,222.54
function convertNumberFloatZero(number){
if(!$.isNumeric(number)){
return 'NaN';
}
var numberFloat = number.toFixed(3);
var splitNumber = numberFloat.split(".");
var cNumberFloat = number.toFixed(2);
var cNsplitNumber = cNumberFloat.split(".");
var lastChar = splitNumber[1].substr(splitNumber[1].length - 1);
if(lastChar > 0 && lastChar < 5){
cNsplitNumber[1]--;
}
return Number(splitNumber[0]).toLocaleString('en').concat('.').concat(cNsplitNumber[1]);
};
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
(Math.round((10.2)*100)/100).toFixed(2)
That should yield: 10.20
(Math.round((.05)*100)/100).toFixed(2)
That should yield: 0.05
(Math.round((4.04)*100)/100).toFixed(2)
That should yield: 4.04
etc.
/*Due to all told stuff. You may do 2 things for different purposes:
When showing/printing stuff use this in your alert/innerHtml= contents:
YourRebelNumber.toFixed(2)*/
var aNumber=9242.16;
var YourRebelNumber=aNumber-9000;
alert(YourRebelNumber);
alert(YourRebelNumber.toFixed(2));
/*and when comparing use:
Number(YourRebelNumber.toFixed(2))*/
if(YourRebelNumber==242.16)alert("Not Rounded");
if(Number(YourRebelNumber.toFixed(2))==242.16)alert("Rounded");
/*Number will behave as you want in that moment. After that, it'll return to its defiance.
*/
This is very simple and works just as well as any of the others:
function parseNumber(val, decimalPlaces) {
if (decimalPlaces == null) decimalPlaces = 0
var ret = Number(val).toFixed(decimalPlaces)
return Number(ret)
}
Since toFixed() can only be called on numbers, and unfortunately returns a string, this does all the parsing for you in both directions. You can pass a string or a number, and you get a number back every time! Calling parseNumber(1.49) will give you 1, and parseNumber(1.49,2) will give you 1.50. Just like the best of 'em!
You could also use the .toPrecision() method and some custom code, and always round up to the nth decimal digit regardless the length of int part.
function glbfrmt (number, decimals, seperator) {
return typeof number !== 'number' ? number : number.toPrecision( number.toString().split(seperator)[0].length + decimals);
}
You could also make it a plugin for a better use.
Here's a TypeScript implementation of https://stackoverflow.com/a/21323330/916734. It also dries things up with functions, and allows for a optional digit offset.
export function round(rawValue: number | string, precision = 0, fractionDigitOffset = 0): number | string {
const value = Number(rawValue);
if (isNaN(value)) return rawValue;
precision = Number(precision);
if (precision % 1 !== 0) return NaN;
let [ stringValue, exponent ] = scientificNotationToParts(value);
let shiftExponent = exponentForPrecision(exponent, precision, Shift.Right);
const enlargedValue = toScientificNotation(stringValue, shiftExponent);
const roundedValue = Math.round(enlargedValue);
[ stringValue, exponent ] = scientificNotationToParts(roundedValue);
const precisionWithOffset = precision + fractionDigitOffset;
shiftExponent = exponentForPrecision(exponent, precisionWithOffset, Shift.Left);
return toScientificNotation(stringValue, shiftExponent);
}
enum Shift {
Left = -1,
Right = 1,
}
function scientificNotationToParts(value: number): Array<string> {
const [ stringValue, exponent ] = value.toString().split('e');
return [ stringValue, exponent ];
}
function exponentForPrecision(exponent: string, precision: number, shift: Shift): number {
precision = shift * precision;
return exponent ? (Number(exponent) + precision) : precision;
}
function toScientificNotation(value: string, exponent: number): number {
return Number(`${value}e${exponent}`);
}
fun Any.twoDecimalPlaces(numInDouble: Double): String {
return "%.2f".format(numInDouble)
}