How to round up last digit in two decimal point using JavaScript - javascript

I want to round up last digit of two decimal point if last digit more then 5.
For example,
2.58 to 2.60
Besides that, I also want to round down last digit of two decimal point if last digit less then 5.
For example, 6.54 to 6.50
How to do this by using javascript

You can achieve this by Math.round and then dividing the result by 10 (which results in an extra 0 in a floating point number:
let number = 2.58;
let rounded = Math.round(number * 10, 2) / 10; // 2.60
let number2 = 2.54;
let rounded2 = Math.round(number2 * 10, 2) / 10; // 2.50

Use the toFixed() method.
var number = 2.58;
var rounded = number.toFixed(1);
console.log(parseFloat(rounded).toFixed(2));

Related

how to get exact two digits after decimal point in jquery or javascript

I want the exact 2 digits after the decimal point. I tried the toFixed(2) function but it returns rounded off 2 digits. Here is my Code it returns 18.70 but I want 18.69extract number
rim_weight = (23* 36 *70);
rim_weight = rim_weight/3100; // retuns 18.69677419354839
rim_weight = rim_weight.toFixed(2); // returns 18.70
If you just want to truncate the number to two decimal places, first divide by 31, convert to an integral value using Math.floor and then divide by 100:
rim_weight = (23 * 36 * 70);
rim_weight = Math.floor(rim_weight / 31);
rim_weight = rim_weight / 100;
console.log(rim_weight);
just convert it to string then match the number up to the second place

Rounding Column Footer in Datatables [duplicate]

How will i round margin_total to 3 decimal places?
margin_total = margin_total + parseFloat(marginObj.value);
document.getElementById('margin_total').value = margin_total;
Use num.toFixed(d) to convert a number into the String representation of that number in base 10 with d digits after the decimal point, in this case,
margin_total.toFixed(3);
The toFixed() method converts a number into a string, keeping a specified number of decimals. A string representation of input that does not use exponential notation and has exactly digits digits after the decimal place. The number is rounded if necessary, and the fractional part is padded with zeros if necessary so that it has the specified length.
function myFunction() {
var num = 5.56789;
var n = num.toFixed(3)
document.getElementById("demo").innerHTML = n;
}
<p>Click the button to display the fixed number.</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>
const roundedDown = Math.round(6.426475 * 1000) / 1000
console.log(roundedDown)
const roundedUp = Math.round(6.426575 * 1000) / 1000
console.log(roundedUp)
The above code rounds to 3 decimal places (dp)
To round to 2 dp use 100 instead of 1000 (all occurrences)
To round to 4 dp use 10000 instead of 1000
You get the idea!
const num = 70.012345678900
console.log(parseFloat(num.toFixed(3)));
// expected output: 70.012

Javascript round / floor / toFixed on decimals

I am having an issue with how javascript is dividing and rounding the number.
I have two float , 0.11 and 0.12
I want to calculate the mid of these two numbers and round it to the nearest highest value with 2 decimal price.
For example, if I do this on Calculator
0.11+0.12 / 2 = 0.115, and I need to round it to 0.12 as it is mid or above mid.
If I do this with Javascript, I am not getting an accurate number
Example,
var high = parseFloat(0.12);
var low = parseFloat(0.11);
var mid = (high + low) / 2;
document.getElementById("demo1").innerHTML = mid;
document.getElementById("demo2").innerHTML = mid.toFixed(2);
var another = mid.toFixed(3);
document.getElementById("demo3").innerHTML =another;
var last = Math.floor(another)
document.getElementById("demo4").innerHTML =last;
http://jsfiddle.net/gzqwbp6c/9/
Any input would be appreciated.
As the 0.11499999999999999 shows, the result is very slightly less than 0.115. This is because 0.11 and 0.12 cannot be represented with perfect accuracy using floating-point-numbers.
When you don't want to deal with floating-point-error, it's often easier to work with integers directly. Small integers are represented exactly by floating point numbers.
You can multiply by 100 before, and round, to ensure your numbers are integers, and only divide after you get your final result:
var a = Math.round(100 * parseFloat("0.12")) // 12
var b = Math.round(100 * parseFloat("0.11")) // 11
var mid = (a + b) / 2 // 11.5.
// 0.5 can be represented exactly in floating point for small numbers.
var midRound = (Math.round(mid) / 100).toFixed(2) // "0.12"
Need to multiply (workout on int part, i.e. find mid, and divide to reconvert to origin):
function myMid(high,low, precision){
var precision=2
var convFactor = Math.pow(10,precision);
return
(Math.round((low*convFactor+high*convFactor)/2)/convFactor).toFixed(precision);
}
Float is not precise, you cant rely on that, you'll have unexpected results.
everything *100 to prevent inaccuracies
.toFixed() does the rounding
var a = 0.11;
var b = 0.12;
c = parseFloat((((a*100) + (b*100))/200).toFixed(2));
console.log(c);

Opposite of toString(36)?

var a = (123.456).toString(36) //"3f.gez4w97ry0a18ymf6qadcxr"
Now, how do I revert back to the original number using that string?
Note: parseInt(number,36) only works for integers.
You could try parsing the integer and float parts separately with parseInt, as parseFloat does not support a radix:
function parseFloatInBase(n, radix) {
var nums = n.split(".")
// get the part before the decimal point
var iPart = parseInt(nums[0], radix)
// get the part after the decimal point
var fPart = parseInt(nums[1], radix) / Math.pow(radix, nums[1].length)
return iPart + fPart
}
// this will log 123.456:
console.log(parseFloatInBase("3f.gez4w97ry0a18ymf6qadcxr", 36))
I am dividing by radix ^ numLength because I am basically moving the decimal point over numLength spaces. You would do this just like in math class, because as you know dividing by 10 moves the decimal over one space, because most math is in base 10. Example:
123456 / 10 / 10 / 10 = 123.456
This is equivalent to
123456 / (10 * 10 * 10) = 123.456
And therefore
123456 / (10 ^ 3) = 123.456

How to perform an integer division, and separately get the remainder, in JavaScript?

In JavaScript, how do I get:
The whole number of times a given integer goes into another?
The remainder?
For some number y and some divisor x compute the quotient (quotient)[1] and remainder (remainder) as:
const quotient = Math.floor(y/x);
const remainder = y % x;
Example:
const quotient = Math.floor(13/3); // => 4 => the times 3 fits into 13
const remainder = 13 % 3; // => 1
[1] The integer number resulting from the division of one number by another
I'm no expert in bitwise operators, but here's another way to get the whole number:
var num = ~~(a / b);
This will work properly for negative numbers as well, while Math.floor() will round in the wrong direction.
This seems correct as well:
var num = (a / b) >> 0;
I did some speed tests on Firefox.
-100/3 // -33.33..., 0.3663 millisec
Math.floor(-100/3) // -34, 0.5016 millisec
~~(-100/3) // -33, 0.3619 millisec
(-100/3>>0) // -33, 0.3632 millisec
(-100/3|0) // -33, 0.3856 millisec
(-100-(-100%3))/3 // -33, 0.3591 millisec
/* a=-100, b=3 */
a/b // -33.33..., 0.4863 millisec
Math.floor(a/b) // -34, 0.6019 millisec
~~(a/b) // -33, 0.5148 millisec
(a/b>>0) // -33, 0.5048 millisec
(a/b|0) // -33, 0.5078 millisec
(a-(a%b))/b // -33, 0.6649 millisec
The above is based on 10 million trials for each.
Conclusion: Use (a/b>>0) (or (~~(a/b)) or (a/b|0)) to achieve about 20% gain in efficiency. Also keep in mind that they are all inconsistent with Math.floor, when a/b<0 && a%b!=0.
ES6 introduces the new Math.trunc method. This allows to fix #MarkElliot's answer to make it work for negative numbers too:
var div = Math.trunc(y/x);
var rem = y % x;
Note that Math methods have the advantage over bitwise operators that they work with numbers over 231.
I normally use:
const quotient = (a - a % b) / b;
const remainder = a % b;
It's probably not the most elegant, but it works.
var remainder = x % y;
return (x - remainder) / y;
You can use the function parseInt to get a truncated result.
parseInt(a/b)
To get a remainder, use mod operator:
a%b
parseInt have some pitfalls with strings, to avoid use radix parameter with base 10
parseInt("09", 10)
In some cases the string representation of the number can be a scientific notation, in this case, parseInt will produce a wrong result.
parseInt(100000000000000000000000000000000, 10) // 1e+32
This call will produce 1 as result.
Math.floor(operation) returns the rounded down value of the operation.
Example of 1st question:
const x = 5;
const y = 10.4;
const z = Math.floor(x + y);
console.log(z);
Example of 2nd question:
const x = 14;
const y = 5;
const z = Math.floor(x % y);
console.log(x);
JavaScript calculates right the floor of negative numbers and the remainder of non-integer numbers, following the mathematical definitions for them.
FLOOR is defined as "the largest integer number smaller than the parameter", thus:
positive numbers: FLOOR(X)=integer part of X;
negative numbers: FLOOR(X)=integer part of X minus 1 (because it must be SMALLER than the parameter, i.e., more negative!)
REMAINDER is defined as the "left over" of a division (Euclidean arithmetic). When the dividend is not an integer, the quotient is usually also not an integer, i.e., there is no remainder, but if the quotient is forced to be an integer (and that's what happens when someone tries to get the remainder or modulus of a floating-point number), there will be a non-integer "left over", obviously.
JavaScript does calculate everything as expected, so the programmer must be careful to ask the proper questions (and people should be careful to answer what is asked!) Yarin's first question was NOT "what is the integer division of X by Y", but, instead, "the WHOLE number of times a given integer GOES INTO another". For positive numbers, the answer is the same for both, but not for negative numbers, because the integer division (dividend by divisor) will be -1 smaller than the times a number (divisor) "goes into" another (dividend). In other words, FLOOR will return the correct answer for an integer division of a negative number, but Yarin didn't ask that!
gammax answered correctly, that code works as asked by Yarin. On the other hand, Samuel is wrong, he didn't do the maths, I guess, or he would have seen that it does work (also, he didn't say what was the divisor of his example, but I hope it was 3):
Remainder = X % Y = -100 % 3 = -1
GoesInto = (X - Remainder) / Y = (-100 - -1) / 3 = -99 / 3 = -33
By the way, I tested the code on Firefox 27.0.1, it worked as expected, with positive and negative numbers and also with non-integer values, both for dividend and divisor. Example:
-100.34 / 3.57: GoesInto = -28, Remainder = -0.3800000000000079
Yes, I noticed, there is a precision problem there, but I didn't had time to check it (I don't know if it's a problem with Firefox, Windows 7 or with my CPU's FPU). For Yarin's question, though, which only involves integers, the gammax's code works perfectly.
const idivmod = (a, b) => [a/b |0, a%b];
there is also a proposal working on it
Modulus and Additional Integer Math
Alex Moore-Niemi's comment as an answer:
For Rubyists here from Google in search of divmod, you can implement it as such:
function divmod(x, y) {
var div = Math.trunc(x/y);
var rem = x % y;
return [div, rem];
}
Result:
// [2, 33]
If you need to calculate the remainder for very large integers, which the JS runtime cannot represent as such (any integer greater than 2^32 is represented as a float and so it loses precision), you need to do some trick.
This is especially important for checking many case of check digits which are present in many instances of our daily life (bank account numbers, credit cards, ...)
First of all you need your number as a string (otherwise you have already lost precision and the remainder does not make sense).
str = '123456789123456789123456789'
You now need to split your string in smaller parts, small enough so the concatenation of any remainder and a piece of string can fit in 9 digits.
digits = 9 - String(divisor).length
Prepare a regular expression to split the string
splitter = new RegExp(`.{1,${digits}}(?=(.{${digits}})+$)`, 'g')
For instance, if digits is 7, the regexp is
/.{1,7}(?=(.{7})+$)/g
It matches a nonempty substring of maximum length 7, which is followed ((?=...) is a positive lookahead) by a number of characters that is multiple of 7. The 'g' is to make the expression run through all string, not stopping at first match.
Now convert each part to integer, and calculate the remainders by reduce (adding back the previous remainder - or 0 - multiplied by the correct power of 10):
reducer = (rem, piece) => (rem * Math.pow(10, digits) + piece) % divisor
This will work because of the "subtraction" remainder algorithm:
n mod d = (n - kd) mod d
which allows to replace any 'initial part' of the decimal representation of a number with its remainder, without affecting the final remainder.
The final code would look like:
function remainder(num, div) {
const digits = 9 - String(div).length;
const splitter = new RegExp(`.{1,${digits}}(?=(.{${digits}})+$)`, 'g');
const mult = Math.pow(10, digits);
const reducer = (rem, piece) => (rem * mult + piece) % div;
return str.match(splitter).map(Number).reduce(reducer, 0);
}
If you are just dividing with powers of two, you can use bitwise operators:
export function divideBy2(num) {
return [num >> 1, num & 1];
}
export function divideBy4(num) {
return [num >> 2, num & 3];
}
export function divideBy8(num) {
return [num >> 3, num & 7];
}
(The first is the quotient, the second the remainder)
function integerDivison(dividend, divisor){
this.Division = dividend/divisor;
this.Quotient = Math.floor(dividend/divisor);
this.Remainder = dividend%divisor;
this.calculate = ()=>{
return {Value:this.Division,Quotient:this.Quotient,Remainder:this.Remainder};
}
}
var divide = new integerDivison(5,2);
console.log(divide.Quotient) //to get Quotient of two value
console.log(divide.division) //to get Floating division of two value
console.log(divide.Remainder) //to get Remainder of two value
console.log(divide.calculate()) //to get object containing all the values
You can use ternary to decide how to handle positive and negative integer values as well.
var myInt = (y > 0) ? Math.floor(y/x) : Math.floor(y/x) + 1
If the number is a positive, all is fine. If the number is a negative, it will add 1 because of how Math.floor handles negatives.
This will always truncate towards zero.
Not sure if it is too late, but here it goes:
function intdiv(dividend, divisor) {
divisor = divisor - divisor % 1;
if (divisor == 0) throw new Error("division by zero");
dividend = dividend - dividend % 1;
var rem = dividend % divisor;
return {
remainder: rem,
quotient: (dividend - rem) / divisor
};
}
Calculating number of pages may be done in one step:
Math.ceil(x/y)
Here is a way to do this. (Personally I would not do it this way, but thought it was a fun way to do it for an example) The ways mentioned above are definitely better as this calls multiple functions and is therefore slower as well as takes up more room in your bundle.
function intDivide(numerator, denominator) {
return parseInt((numerator/denominator).toString().split(".")[0]);
}
let x = intDivide(4,5);
let y = intDivide(5,5);
let z = intDivide(6,5);
console.log(x);
console.log(y);
console.log(z);

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