Generate random number between two numbers in JavaScript

JavascriptRandom

Javascript Problem Overview


Is there a way to generate a random number in a specified range with JavaScript ?

For example: a specified range from 1 to 6 were the random number could be either 1, 2, 3, 4, 5, or 6.

Javascript Solutions


Solution 1 - Javascript

function randomIntFromInterval(min, max) { // min and max included 
  return Math.floor(Math.random() * (max - min + 1) + min)
}

const rndInt = randomIntFromInterval(1, 6)
console.log(rndInt)

What it does "extra" is it allows random intervals that do not start with 1. So you can get a random number from 10 to 15 for example. Flexibility.

Solution 2 - Javascript

Important
The following code works only if the minimum value is 1. It does not work for minimum values other than 1.

If you wanted to get a random integer between 1 (and only 1) and 6, you would calculate:

    const rndInt = Math.floor(Math.random() * 6) + 1
    console.log(rndInt)

Where:

  • 1 is the start number
  • 6 is the number of possible results (1 + start (6) - end (1))

Solution 3 - Javascript

Math.random()

Returns an integer random number between min (included) and max (included):

function randomInteger(min, max) {
  return Math.floor(Math.random() * (max - min + 1)) + min;
}

Or any random number between min (included) and max (not included):

function randomNumber(min, max) {
  return Math.random() * (max - min) + min;
}

Useful examples (integers):

// 0 -> 10
Math.floor(Math.random() * 11);

// 1 -> 10
Math.floor(Math.random() * 10) + 1;

// 5 -> 20
Math.floor(Math.random() * 16) + 5;

// -10 -> (-2)
Math.floor(Math.random() * 9) - 10;

** And always nice to be reminded (Mozilla):

> Math.random() does not provide cryptographically secure random > numbers. Do not use them for anything related to security. Use the Web > Crypto API instead, and more precisely the > window.crypto.getRandomValues() method.

Solution 4 - Javascript

Other solutions:

  • (Math.random() * 6 | 0) + 1
  • ~~(Math.random() * 6) + 1

Try online

Solution 5 - Javascript

TL;DR

function generateRandomInteger(min, max) {
  return Math.floor(min + Math.random()*(max - min + 1))
}

To get the random number generateRandomInteger(-20, 20);

EXPLANATION BELOW

integer - A number which is not a fraction; a whole number

We need to get a random number , say X between min and max. X, min and max are all integers

i.e min <= X <= max

If we subtract min from the equation, this is equivalent to

0 <= (X - min) <= (max - min)

Now, lets multiply this with a random number r which is

0 <= (X - min) * r <= (max - min) * r

Now, lets add back min to the equation

min <= min + (X - min) * r <= min + (max - min) * r

For, any given X, the above equation satisfies only when r has range of [0,1] For any other values of r the above equation is unsatisfied.

Learn more about ranges [x,y] or (x,y) here

Our next step is to find a function which always results in a value which has a range of [0,1]

Now, the range of r i.e [0,1] is very similar to Math.random() function in Javascript. Isn't it? > The Math.random() function returns a floating-point, pseudo-random > number in the range [0, 1); that is, from 0 (inclusive) up to but not > including 1 (exclusive)

Random Function using Math.random() 0 <= r < 1

Notice that in Math.random() left bound is inclusive and the right bound is exclusive. This means min + (max - min) * r will evaluate to having a range from [min, max)

To include our right bound i.e [min,max] we increase the right bound by 1 and floor the result.

function generateRandomInteger(min, max) {
  return Math.floor(min + Math.random()*(max - min + 1))
}

To get the random number

generateRandomInteger(-20, 20);

Solution 6 - Javascript

Or, in Underscore

_.random(min, max)

Solution 7 - Javascript

var x = 6; // can be any number
var rand = Math.floor(Math.random()*x) + 1;

Solution 8 - Javascript

jsfiddle: https://jsfiddle.net/cyGwf/477/

Random Integer: to get a random integer between min and max, use the following code

function getRandomInteger(min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  return Math.floor(Math.random() * (max - min)) + min;
}

Random Floating Point Number: to get a random floating point number between min and max, use the following code

function getRandomFloat(min, max) {
  return Math.random() * (max - min) + min;
}

Reference: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/random

Solution 9 - Javascript

Math is not my strong point, but I've been working on a project where I needed to generate a lot of random numbers between both positive and negative.

function randomBetween(min, max) {
    if (min < 0) {
    	return min + Math.random() * (Math.abs(min)+max);
    }else {
	    return min + Math.random() * max;
    }
}

E.g

randomBetween(-10,15)//or..
randomBetween(10,20)//or...
randomBetween(-200,-100)

Of course, you can also add some validation to make sure you don't do this with anything other than numbers. Also make sure that min is always less than or equal to max.

Solution 10 - Javascript

ES6 / Arrow functions version based on Francis' code (i.e. the top answer):

const randomIntFromInterval = (min, max) => Math.floor(Math.random() * (max - min + 1) + min);

Solution 11 - Javascript

Get a random integer between 0 and 400

let rand = Math.round(Math.random() * 400)

document.write(rand)

Get a random integer between 200 and 1500

let range = {min: 200, max: 1500}
let delta = range.max - range.min

const rand = Math.round(range.min + Math.random() * delta)

document.write(rand)

Using functions

function randBetween(min, max){
  let delta = max - min
  return Math.round(min + Math.random() * delta)
}

document.write(randBetween(10, 15));

// JavaScript ES6 arrow function

const randBetween = (min, max) => {
  let delta = max - min
  return Math.round(min + Math.random() * delta)
}

document.write(randBetween(10, 20))

Solution 12 - Javascript

I wrote more flexible function which can give you random number but not only integer.

function rand(min,max,interval)
{
    if (typeof(interval)==='undefined') interval = 1;
    var r = Math.floor(Math.random()*(max-min+interval)/interval);
    return r*interval+min;
}

var a = rand(0,10); //can be 0, 1, 2 (...) 9, 10
var b = rand(4,6,0.1); //can be 4.0, 4.1, 4.2 (...) 5.9, 6.0

Fixed version.

Solution 13 - Javascript

Example

Return a random number between 1 and 10:

Math.floor((Math.random() * 10) + 1);

The result could be: 3

Try yourself: here

--

or using lodash / undescore:

_.random(min, max)

Docs:

Solution 14 - Javascript

I was searching random number generator written in TypeScript and I have written this after reading all of the answers, hope It would work for TypeScript coders.

    Rand(min: number, max: number): number {
        return (Math.random() * (max - min + 1) | 0) + min;
    }   

Solution 15 - Javascript

The top rated solution is not mathematically correct as same as comments under it -> Math.floor(Math.random() * 6) + 1.

Task: generate random number between 1 and 6.

Math.random() returns floating point number between 0 and 1 (like 0.344717274374 or 0.99341293123 for example), which we will use as a percentage, so Math.floor(Math.random() * 6) + 1 returns some percentage of 6 (max: 5, min: 0) and adds 1. The author got lucky that lower bound was 1., because percentage floor will "maximumly" return 5 which is less than 6 by 1, and that 1 will be added by lower bound 1.

The problems occurs when lower bound is greater than 1. For instance, Task: generate random between 2 and 6.

(following author's logic) Math.floor(Math.random() * 6) + 2, it is obviously seen that if we get 5 here -> Math.random() * 6 and then add 2, the outcome will be 7 which goes beyond the desired boundary of 6.

Another example, Task: generate random between 10 and 12.

(following author's logic) Math.floor(Math.random() * 12) + 10, (sorry for repeating) it is obvious that we are getting 0%-99% percent of number "12", which will go way beyond desired boundary of 12.

So, the correct logic is to take the difference between lower bound and upper bound add 1, and only then floor it (to substract 1, because Math.random() returns 0 - 0.99, so no way to get full upper bound, thats why we adding 1 to upper bound to get maximumly 99% of (upper bound + 1) and then we floor it to get rid of excess). Once we got the floored percentage of (difference + 1), we can add lower boundary to get the desired randomed number between 2 numbers.

The logic formula for that will be: Math.floor(Math.random() * ((up_boundary - low_boundary) + 1)) + 10.

P.s.: Even comments under the top-rated answer were incorrect, since people forgot to add 1 to the difference, meaning that they will never get the up boundary (yes it might be a case if they dont want to get it at all, but the requirenment was to include the upper boundary).

Solution 16 - Javascript

Inspite of many answers and almost same result. I would like to add my answer and explain its working. Because it is important to understand its working rather than copy pasting one line code. Generating random numbers is nothing but simple maths.

CODE:

function getR(lower, upper) {

  var percent = (Math.random() * 100);
  // this will return number between 0-99 because Math.random returns decimal number from 0-0.9929292 something like that
  //now you have a percentage, use it find out the number between your INTERVAL :upper-lower 
  var num = ((percent * (upper - lower) / 100));
  //num will now have a number that falls in your INTERVAL simple maths
  num += lower;
  //add lower to make it fall in your INTERVAL
  //but num is still in decimal
  //use Math.floor>downward to its nearest integer you won't get upper value ever
  //use Math.ceil>upward to its nearest integer upper value is possible
  //Math.round>to its nearest integer 2.4>2 2.5>3	both lower and upper value possible
  console.log(Math.floor(num), Math.ceil(num), Math.round(num));
}

Solution 17 - Javascript

Math.random() is fast and suitable for many purposes, but it's not appropriate if you need cryptographically-secure values (it's not secure), or if you need integers from a completely uniform unbiased distribution (the multiplication approach used in others answers produces certain values slightly more often than others).

In such cases, we can use crypto.getRandomValues() to generate secure integers, and reject any generated values that we can't map uniformly into the target range. This will be slower, but it shouldn't be significant unless you're generating extremely large numbers of values.

To clarify the biased distribution concern, consider the case where we want to generate a value between 1 and 5, but we have a random number generator that produces values between 1 and 16 (a 4-bit value). We want to have the same number of generated values mapping to each output value, but 16 does not evenly divide by 5: it leaves a remainder of 1. So we need to reject 1 of the possible generated values, and only continue when we get one of the 15 lesser values that can be uniformly mapped into our target range. Our behaviour could look like this pseudocode:

Generate a 4-bit integer in the range 1-16.
If we generated  1,  6, or 11 then output 1.
If we generated  2,  7, or 12 then output 2.
If we generated  3,  8, or 13 then output 3.
If we generated  4,  9, or 14 then output 4.
If we generated  5, 10, or 15 then output 5.
If we generated 16 then reject it and try again.

The following code uses similar logic, but generates a 32-bit integer instead, because that's the largest common integer size that can be represented by JavaScript's standard number type. (This could be modified to use BigInts if you need a larger range.) Regardless of the chosen range, the fraction of generated values that are rejected will always be less than 0.5, so the expected number of rejections will always be less than 1.0 and usually close to 0.0; you don't need to worry about it looping forever.

const randomInteger = (min, max) => {
  const range = max - min;
  const maxGeneratedValue = 0xFFFFFFFF;
  const possibleResultValues = range + 1;
  const possibleGeneratedValues = maxGeneratedValue + 1;
  const remainder = possibleGeneratedValues % possibleResultValues;
  const maxUnbiased = maxGeneratedValue - remainder;

  if (!Number.isInteger(min) || !Number.isInteger(max) ||
       max > Number.MAX_SAFE_INTEGER || min < Number.MIN_SAFE_INTEGER) {
    throw new Error('Arguments must be safe integers.');
  } else if (range > maxGeneratedValue) {
    throw new Error(`Range of ${range} (from ${min} to ${max}) > ${maxGeneratedValue}.`);
  } else if (max < min) {
    throw new Error(`max (${max}) must be >= min (${min}).`);
  } else if (min === max) {
    return min;
  } 

  let generated;
  do {
    generated = crypto.getRandomValues(new Uint32Array(1))[0];
  } while (generated > maxUnbiased);

  return min + (generated % possibleResultValues);
};

console.log(randomInteger(-8, 8));          // -2
console.log(randomInteger(0, 0));           // 0
console.log(randomInteger(0, 0xFFFFFFFF));  // 944450079
console.log(randomInteger(-1, 0xFFFFFFFF));
// Error: Range of 4294967296 covering -1 to 4294967295 is > 4294967295.
console.log(new Array(12).fill().map(n => randomInteger(8, 12)));
// [11, 8, 8, 11, 10, 8, 8, 12, 12, 12, 9, 9]

Solution 18 - Javascript

to return 1-6 like a dice basically,

return Math.round(Math.random() * 5 + 1);

Solution 19 - Javascript

Adding float with fixed precision version based on the int version in @Francisc's answer:

function randomFloatFromInterval (min, max, fractionDigits) {
  const fractionMultiplier = Math.pow(10, fractionDigits)
  return Math.round(
    (Math.random() * (max - min) + min) * fractionMultiplier,
  ) / fractionMultiplier
}

so:

randomFloatFromInterval(1,3,4) // => 2.2679, 1.509, 1.8863, 2.9741, ...

and for int answer

randomFloatFromInterval(1,3,0) // => 1, 2, 3

Solution 20 - Javascript

This function can generate a random integer number between (and including) min and max numbers:

function randomNumber(min, max) {
  if (min > max) {
	let temp = max;
	max = min;
	min = temp;
  }

  if (min <= 0) {
	return Math.floor(Math.random() * (max + Math.abs(min) + 1)) + min;
  } else {
	return Math.floor(Math.random() * (max - min + 1)) + min;
  }
}

Example:

randomNumber(-2, 3); // can be -2, -1, 0, 1, 2 and 3
randomNumber(-5, -2); // can be -5, -4, -3 and -2
randomNumber(0, 4); // can be 0, 1, 2, 3 and 4
randomNumber(4, 0); // can be 0, 1, 2, 3 and 4

Solution 21 - Javascript

Using random function, which can be reused.

function randomNum(min, max) {
  return Math.floor(Math.random() * (max - min + 1)) + min;
}
randomNum(1, 6);

Solution 22 - Javascript

This should work:

const getRandomNum = (min, max) => Math.floor(Math.random() * (max - min + 1)) + min

Solution 23 - Javascript

Crypto-strong random integer number in range [a,b] (assumption: a < b )

let rand= (a,b)=> a+(b-a+1)*crypto.getRandomValues(new Uint32Array(1))[0]/2**32|0

console.log( rand(1,6) );

Solution 24 - Javascript

This is about nine years late, but randojs.com makes this a simple one-liner:

rando(1, 6)

You just need to add this to the head of your html document, and you can do pretty much whatever you want with randomness easily. Random values from arrays, random jquery elements, random properties from objects, and even preventing repetitions if needed.

<script src="https://randojs.com/1.0.0.js"></script>

Solution 25 - Javascript

Try using:

function random(min, max) {
   return Math.round((Math.random() *( Math.abs(max - min))) + min);
}
console.log(random(1, 6));

Solution 26 - Javascript

If the starting number is 1, as in your example (1-6), you can use Math.ceil() method instead of Math.floor().

Math.ceil(Math.random() * 6)

instead of

Math.floor(Math.random() * 6) + 1

Let's not forget other useful Math methods.

Solution 27 - Javascript

I discovered a great new way to do this using ES6 default parameters. It is very nifty since it allows either one argument or two arguments. Here it is:

function random(n, b = 0) {
    return Math.random() * (b-n) + n;
}

Solution 28 - Javascript

This works for me and produces values like Python's random.randint standard library function:


function randint(min, max) {
   return Math.round((Math.random() * Math.abs(max - min)) + min);
}
    
console.log("Random integer: " + randint(-5, 5));

Solution 29 - Javascript

for big number.

var min_num = 900;
var max_num = 1000;
while(true){
	
    let num_random = Math.random()* max_num;
	console.log('input : '+num_random);
    if(num_random >= min_num){
        console.log(Math.floor(num_random));
       break; 
    } else {
        console.log(':::'+Math.floor(num_random));
    }
}

Solution 30 - Javascript

For Postman: pm.environment.set("random_number", _.random(11111111111, 99999999999))

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionMirgorodView Question on Stackoverflow
Solution 1 - JavascriptFranciscView Answer on Stackoverflow
Solution 2 - Javascriptkhr055View Answer on Stackoverflow
Solution 3 - JavascriptLior ElromView Answer on Stackoverflow
Solution 4 - JavascriptVishalView Answer on Stackoverflow
Solution 5 - JavascriptFaiz Mohamed HaneefView Answer on Stackoverflow
Solution 6 - JavascriptvladiimView Answer on Stackoverflow
Solution 7 - Javascriptryebr3adView Answer on Stackoverflow
Solution 8 - JavascriptRazan PaulView Answer on Stackoverflow
Solution 9 - JavascriptPetter ThowsenView Answer on Stackoverflow
Solution 10 - JavascriptThe OninView Answer on Stackoverflow
Solution 11 - JavascriptGassView Answer on Stackoverflow
Solution 12 - JavascriptElChupacabraView Answer on Stackoverflow
Solution 13 - JavascriptSebastián LaraView Answer on Stackoverflow
Solution 14 - JavascriptErdi İzgiView Answer on Stackoverflow
Solution 15 - JavascriptM22View Answer on Stackoverflow
Solution 16 - JavascriptArun SharmaView Answer on Stackoverflow
Solution 17 - JavascriptJeremyView Answer on Stackoverflow
Solution 18 - JavascriptHusky931View Answer on Stackoverflow
Solution 19 - JavascriptyonatanmnView Answer on Stackoverflow
Solution 20 - JavascriptRayron VictorView Answer on Stackoverflow
Solution 21 - JavascriptOkiemute GoldView Answer on Stackoverflow
Solution 22 - JavascriptSabbir AhmedView Answer on Stackoverflow
Solution 23 - JavascriptKamil KiełczewskiView Answer on Stackoverflow
Solution 24 - JavascriptAaron PlocharczykView Answer on Stackoverflow
Solution 25 - JavascriptAyushKatiyarView Answer on Stackoverflow
Solution 26 - JavascriptmaxxxView Answer on Stackoverflow
Solution 27 - Javascriptmaburdi94View Answer on Stackoverflow
Solution 28 - JavascriptgnrfanView Answer on Stackoverflow
Solution 29 - JavascriptRutheView Answer on Stackoverflow
Solution 30 - JavascriptMarcoView Answer on Stackoverflow