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Given any number between 0 and 1, such as 0.84729347293923, is there a simple way to make it into 84729347293923 without string or regex manipulation? I can think of using a loop, which probably is no worse than using a string because it is O(n) with n being the number of digits. But is there a better way?
function getRandom() {
let r = Math.random();
while (Math.floor(r) !== r) r *= 10;
return r;
}
for (let i = 0; i < 10; i++)
console.log(getRandom());
Integers mod 1 = 0, non integers mod 1 != 0.
while ((r*=10) % 1);
Ok, just want to refactor my code (i realized that was bad so this is what i discovered to correctly get the value as you requested).
NOTE: As the question says that "given any number between 0 and 1", this solution only works for values between 0 and 1:
window.onload = ()=>{
function getLen(num){
let currentNumb = num;
let integratedArray = [];
let realLen = 0;
/*While the number is not an integer, we will multiply the copy of the original
*value by ten, and when the loop detects that the number is already an integer
*the while simply breaks, in this process we are storing each transformations
*of the number in an array called integratedArray*/
while(!(Number.isInteger(currentNumb))){
currentNumb *= 10;
integratedArray.push(currentNumb);
}
/*We iterate over the array and compare each value of the array with an operation
*in which the resultant value should be exactly the same as the actual item of the
*array, in the case that both are equal we assign the var realLen to i, and
*in case that the values were not the same, we simply breaks the loop, if the
*values are not the same, this indicates that we found the "trash numbers", so
*we simply skip them.*/
for(let i = 0; i < integratedArray.length; i++){
if(Math.floor(integratedArray[i]) === Math.floor(num * Math.pow(10, i + 1))){
realLen = i;
}else{
break;
}
}
return realLen;
}
//Get the float value of a number between 0 and 1 as an integer.
function getShiftedNumber(num){
//First we need the length to get the float part of the number as an integer
const len = getLen(num);
/*Once we have the length of the number we simply multiply the number by
*(10) ^ numberLength, this eliminates the comma (,), or point (.), and
*automatically transforms the number to an integer in this case a large integer*/
return num * (Math.pow(10, len));
}
console.log(getShiftedNumber(0.84729347293923));
}
So the explanation is the next:
Because we want to convert this number without using any string, regex or any another thing, first we need to get the length of the number, this is a bit hard to do without using string conversions... so i did the function getLen for this purpose.
In the function getLen, we have 3 variables:
currentNumb: This var is a copy of the original value (the original number), this value help us to found the length of the number and we can do some transforms to this value whitout changing the original reference of the number.
We need to multiply this value any times is needed to transform the number to an integer and then multiplyng this value by ten to ten.
with the help of a while (this method makes the number a false integer).
NOTE: I saw "False integer" because when i was making the tests i realized that in the number is being adding more digits than normal... (Very very strange), so this stupid but important thing makes neccesary the filter of these "trash numbers", so later we proccess them.
integratedArray: This array stores the values of the result of the first while operations, so the last number stored in this array is an integer, but this number is one of the "fake integers", so with this array we need to iterate later to compare what of those stored values are different to the original value multiplied by (10 * i + 1), so here is the hint:
In this case the first 12 values of this array are exactly the same with the operation of Math.floor(num * Math.pow(10, i + 1))), but in the 13th value of the array these values are not the same so... yes!, there are those "trash numbers" that we were searching for.
realLen: This is the variable where we will store the real length of the number converting the float part of this number in an integer.
Some binary search approach:
Its useless if avarage length < 8;
It contains floating point issues.
But hey it is O(log n) with tons of wasted side computations - i guess if one counts them its event worse than just plain multiplication.
I prefer #chiliNUT answer. One line stamp.
function floatToIntBinarySearch(number){
const max_safe_int_length = 16;
const powers = [
1,
10,
100,
1000,
10000,
100000,
1000000,
10000000,
100000000,
1000000000,
10000000000,
100000000000,
1000000000000,
10000000000000,
100000000000000,
1000000000000000,
10000000000000000
]
let currentLength = 16
let step = 16
let _number = number * powers[currentLength]
while(_number % 1 != 0 || (_number % 10 | 0) == 0){
step /= 2
if( (_number % 10 | 0) == 0 && !(_number % 1 != 0)){
currentLength = currentLength - step;
} else {
currentLength = step + currentLength;
}
if(currentLength < 1 || currentLength > max_safe_int_length * 2) throw Error("length is weird: " + currentLength)
_number = number * powers[currentLength]
console.log(currentLength, _number)
if(Number.isNaN(_number)) throw Error("isNaN: " + ((number + "").length - 2) + " maybe greater than 16?")
}
return number * powers[currentLength]
}
let randomPower = 10 ** (Math.random() * 10 | 0)
let test = (Math.random() * randomPower | 0) / randomPower
console.log(test)
console.log(floatToIntBinarySearch(test))
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"));
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generating random numbers in Javascript in a specific range?
How can i get a random value between, for example, from -99 to 99, excluding 0?
var num = Math.floor(Math.random()*99) + 1; // this will get a number between 1 and 99;
num *= Math.round(Math.random()) ? 1 : -1; // this will add minus sign in 50% of cases
Altogether
var ranNum = Math.ceil(Math.random() * 99) * (Math.round(Math.random()) ? 1 : -1)
This returns what you want
function getNonZeroRandomNumber(){
var random = Math.floor(Math.random()*199) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Here's a functional fiddle
EDIT
To contribute for future readers with a little debate happened in the comments which the user #MarkDickinson made a indeed relevant contribution to my first code posted, I've decided to make another fiddle with a fast comparison between using Math.floor() and Math.round() functions to return the value the op wanted.
First Scenario: Using var random = Math.round(Math.random()*198) - 99; (My first suggestion)
function getNonZeroRandomNumberWithMathRound(){
var random = Math.round(Math.random()*198) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Second scenario: Using var random=Math.floor(Math.random()*199) - 99; (Mark suggestion)
function getNonZeroRandomNumberWithMathFloor(){
var random = Math.floor(Math.random()*199) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Methodology
Since it's a short debate I've chosen fiddle.net to do the comparison.
The test consists of running the above functions 100.000 times and then retrieving how much times the extreme numbers 99 and -99 would appear against a other number, let's say 33 and -33.
The test will then give a simple output consisting of the percentage of appearances from 99 and -99 and the percentage of appearances of 33 and -33.
It'll be used the Webkit implementation from Safari 6.0.2 to the give the output from this answer but anyone can test with your favourite browser late on fiddle.net
Result from first scenario:
Percentage of normal ocurrences:0.97%
Percentage of extreme ocurrences:0.52%
Percentage of extreme ocurrences relative to normal ocurrences:53.4% // Half the chances indeed
Result from second scenario:
Percentage of normal ocurrences:1.052%
Percentage of extreme ocurrences:0.974%
Percentage of extreme ocurrences relative to normal ocurrences:92% //Closer of a fair result with a minimal standard deviation
The result can be seen here: http://jsfiddle.net/brunovieira/LrXqh/
Here's a generalized solution that will let you set the boundaries, and opt in/out of including the 0.
var pos = 99,
neg = 99,
includeZero = false,
result;
do result = Math.ceil(Math.random() * (pos + neg)) - neg;
while (includeZero === false && result === 0);
The pos and neg values are inclusive.
This way there's no requirement that the positive and negative ranges be balanced.
Or if you're worried about the rerun due to a single excluded value, you can just make the initial range less by one, and add 1 to any result greater than or equal to 0.
var pos = 5,
neg = 5,
result;
result = Math.floor(Math.random() * (pos + neg)) - neg;
result = result < 0 ? result : result + 1;
That last line could be shorter if you prefer:
result += (result >= 0)
Is there a math function in JavaScript that converts numbers to positive value?
You could use this...
Math.abs(x)
Math​.abs() | MDN
What about x *= -1? I like its simplicity.
I know this is a bit late, but for people struggling with this, you can use the following functions:
Turn any number positive
let x = 54;
let y = -54;
let resultx = Math.abs(x); // 54
let resulty = Math.abs(y); // 54
Turn any number negative
let x = 54;
let y = -54;
let resultx = -Math.abs(x); // -54
let resulty = -Math.abs(y); // -54
Invert any number
let x = 54;
let y = -54;
let resultx = -(x); // -54
let resulty = -(y); // 54
Math.abs(x) or if you are certain the value is negative before the conversion just prepend a regular minus sign: x = -x.
The minus sign (-) can convert positive numbers to negative numbers and negative numbers to positive numbers. x=-y is visual sugar for x=(y*-1).
var y = -100;
var x =- y;
unsigned_value = Math.abs(signed_value);
var posNum = (num < 0) ? num * -1 : num; // if num is negative multiple by negative one ...
I find this solution easy to understand.
If you'd like to write interesting code that nobody else can ever update, try this:
~--x
Multiplying by (-1) is the fastest way to convert negative number to positive. But you have to be careful not to convert my mistake a positive number to negative! So additional check is needed...
Then Math.abs, Math.floor and parseInt is the slowest.
https://jsperf.com/test-parseint-and-math-floor-and-mathabs/1
Negative to positive
var X = -10 ;
var number = Math.abs(X); //result 10
Positive to negative
var X = 10 ;
var number = (X)*(-1); //result -10
I did something like this myself.
num<0?num*=-1:'';
It checks if the number is negative and if it is, multiply with -1
This does return a value, its up to you if you capture it. In case you want to assign it to something, you should probably do something like:
var out = num<0?num*=-1:num; //I think someone already mentioned this variant.
But it really depends what your goal is. For me it was simple, make it positive if negative, else do nothing. Hence the '' in the code.
In this case i used tertiary operator cause I wanted to, it could very well be:
if(num<0)num*=-1;
I saw the bitwise solution here and wanted to comment on that one too.
~--num; //Drawback for this is that num original value will be reduced by 1
This soultion is very fancy in my opinion, we could rewrite it like this:
~(num = num-1);
In simple terms, we take the negative number, take one away from it and then bitwise invert it. If we had bitwise inverted it normally we would get a value 1 too small.
You can also do this:
~num+1; //Wont change the actual num value, merely returns the new value
That will do the same but will invert first and then add 1 to the positive number.
Although you CANT do this:
~num++; //Wont display the right value.
That will not work cause of precedence, postfix operators such as num++ would be evaluated before ~ and the reason prefix ++num wouldnt work even though it is on the same precedence as bitwise NOT(~), is cause it is evaluated from right to left. I did try to swap them around but it seems that prefix is a little finicky compared to bitwise NOT.
The +1 will work because '+' has a higher precedence and will be evaluated later.
I found that solution to be rather fun and decided to expand on it as it was just thrown in there and post people looking at it were probably ignoring it. Although yes, it wont work with floats.
My hopes are that this post hasn't moved away from the original question. :/
My minimal approach
For converting negative number to positive & vice-versa
var num = -24;
num -= num*2;
console.log(num)
// result = 24
If you want the number to always be positive no matter what you can do this.
function toPositive(n){
if(n < 0){
n = n * -1;
}
return n;
}
var a = toPositive(2); // 2
var b = toPositive(-2); // 2
You could also try this, but i don't recommended it:
function makePositive(n){
return Number((n*-n).toString().replace('-',''));
}
var a = makePositive(2); // 2
var b = makePositive(-2); // 2
The problem with this is that you could be changing the number to negative, then converting to string and removing the - from the string, then converting back to int. Which I would guess would take more processing then just using the other function.
I have tested this in php and the first function is faster, but sometimes JS does some crazy things, so I can't say for sure.
You can use ~ operator that logically converts the number to negative and adds 1 to the negative:
var x = 3;
x = (~x + 1);
console.log(x)
// result = -3
I know another way to do it. This technique works negative to positive & Vice Versa
var x = -24;
var result = x * -1;
Vice Versa:
var x = 58;
var result = x * -1;
LIVE CODE:
// NEGATIVE TO POSITIVE: ******************************************
var x = -24;
var result = x * -1;
console.log(result);
// VICE VERSA: ****************************************************
var x = 58;
var result = x * -1;
console.log(result);
// FLOATING POINTS: ***********************************************
var x = 42.8;
var result = x * -1;
console.log(result);
// FLOATING POINTS VICE VERSA: ************************************
var x = -76.8;
var result = x * -1;
console.log(result);
For a functional programming Ramda has a nice method for this. The same method works going from positive to negative and vice versa.
https://ramdajs.com/docs/#negate
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);