JavaScript generate random number except some values

Question

I'm generating random numbers from 1 to 20 by calling generateRandom(). How can I exclude some values, say 8 and 15?

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

var test = generateRandom(1, 20)

Answer 1

To generate random number between 1 and 20 excluding some given numbers, you can simply do this:

function generateRandom(min, max, exclude) {
  let random;
  while (!random) {
    const x = Math.floor(Math.random() * (max - min + 1)) + min;
    if (exclude.indexOf(x) === -1) random = x;
  }
  return random;
}
const test = generateRandom(1, 20, [8, 15]);

Answer 2

    /**
    * Returns a random integer between min (inclusive) and max (inclusive).
    * Pass all values as an array, as 3rd argument which values shouldn't be generated by the function.
    * The value is no lower than min (or the next integer greater than min
    * if min isn't an integer) and no greater than max (or the next integer
    * lower than max if max isn't an integer).
    * Using Math.round() will give you a non-uniform distribution!
    */
    
function getRandomInt(min, max) {
   const minimum = Math.ceil(min);
   const maximum = Math.floor(max);

   return Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;
}


function getRandomIntExcludingExistingNumbers(min, max, excludeArrayNumbers) {
  let randomNumber;

  if(!Array.isArray(excludeArrayNumbers)) {
    randomNumber = getRandomInt(min, max);
    return randomNumber;
  }

  do {
    randomNumber = getRandomInt(min, max);
  } while ((excludeArrayNumbers || []).includes(randomNumber));

  return randomNumber;
}

const randomNumber = getRandomIntExcludingExistingNumbers(1, 10, [1, 2, 4, 5, 9]);

// It will return random integer between 1 to 10 excluding 1,2,4,5,9

Explanation:

getRandomInt function generates random numbers between min and max values.

I am utilizing that function to make "getRandomIntExcludingExistingNumbers" function to avoid specific values.

we will simply call getRandomInt(min, max) values. Then in do while loop we will check if randomly generated values belongs to any of the values which shouldn't be generated.

If it is unique integer outside exclude values then we will return the value.

If our value is from the excluded values, then from do -- while loop, we will once again call getRandomInt to generate new values.

Answer 3

You could take an offset for random values greater or equal than zerow ith a sorted (ascending) array and return a sum with adjusted random value.

const
    getRandomWithoutZero = (lower, upper, gaps) => () => {
        const r = Math.floor(Math.random() * (upper - lower + 1 - gaps.length) + lower);
        return gaps.reduce((s, g) => s + (s >= g), r);
    },
    random = getRandomWithoutZero(-9, 9, [-3, 0, 4]),
    count = {};

for (let i = 0; i < 1.6e6; i++) {
    const r = random();
    count[r] = (count[r] || 0) + 1;
}

console.log(count);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

This is a simple and neat idea, I am a electromechanical engineer and I am just learning JS. This is going to print a random numeber between 1 and 100. Except 8 and 15

 var r;   // this is the random integer.
 var val; //this will serve as validator for the random integer.

 val=0; 
 while(val==0)    {
 r=Math.round(Math.random()*100)+1;
 if(r!=8 && r!=15){val=1;}    //a valid number will be any number different from 8 and 15
                              //then validator will change and go out from the loop.
                }   
document.write(r);

Answer 5

You can simply do like this

function generatedRandExclude(showed,max) {
     let randNo = -1;
     while(showed.length < max) {
        	randNo = Math.floor(Math.random() * Math.floor(max));	
         	if(!showed.includes(randNo)) {
            	showed.push(randNo);
             	break;
          }
     }
     return randNo;
}

let showed = [];
function run() {
    console.log(generatedRandExclude(showed,6));
}
run();
run();
run();
run();

generatedRandExclude generate random number excluded using array showed.

Answer 6

I've read through all these answers and they differ a lot in philosophy, so I thought I might add my very own 2 bits, despite of this question having an answer, because I do think there is a better and more elegant way of approaching this problem.

We can make a function that takes min, max and blacklist as parameters and outputs a random result without using recursion (and with close to 0 if statements):

const blrand = function(min, max, blacklist) { if(!blacklist) blacklist = [] let rand = (min, max) => Math.floor(Math.random() * (max - min + 1)) + min; let retv = 0; while(blacklist.indexOf(retv = rand(min,max)) > -1) { } return retv; }

usage: let randomNumber = blrand(0, 20, [8, 15]);

Answer 7

Here is a slightly modified answer that is similar to all the others but it allows your to pass a single or an array of failing numbers

function generateRandom(min, max, failOn) {
    failOn = Array.isArray(failOn) ? failOn : [failOn]
    var num = Math.floor(Math.random() * (max - min + 1)) + min;
    return failOn.includes(num) ? generateRandom(min, max, failOn) : num;
}

Answer 8

I have answered a similar question for Java: Generate random numbers except certain values. I just copy and paste the answer as follows.

Actually, we do not need to use contains(random) with a while loop.

To simplify the question, let's see what happens if we only have one excluding value. We can split the result to 2 parts. Then the number of possible values is range-1. If the random number is less than the excluded value, just return it. Otherwise, we could add 1.

For multiple excluding values, We can split the result set into size+1 parts, where size means the number of excluding values. Then the number of possible values is range-size. Then we sort excluding values in ascending order. If random number is less than the excluding value minus i, then we just return the random number add i, where i is the index of the the excluding value.

public int generateRandomNumberWithExcepts(int start, int end, List<Integer> excepts) {
    int size = excepts.size();
    int range = end - start + 1 - size;
    int randNum = random.nextInt(range) + start;
    excepts.sort(null); // sort excluding values in ascending order
    int i=0;
    for(int except : excepts) {
        if(randNum < except-i){
            return randNum + i;
        }
        i++;
    }
    return randNum + i;
}

Answer 9

Right now I'm using this and it works without causing browser issues with infinities loops, also tested in mobile devices (using Ionic/Cordova):

function getRandomIndex(usedIndexs, maxIndex) {
    var result = 0;
    var min = 0;
    var max = maxIndex - 1;
    var index = Math.floor(Math.random()*(max-min+1)+min);

    while(usedIndexs.indexOf(index) > -1) {
        if (index < max) {
            index++;
        } else {
          index = 0;
        }
    }

    return index;
}

Answer 10

Here is a really stupidly overcomplicated solution...

        <script>
    var excludedNumbers = [];
    excludedNumbers.push(8);
    excludedNumbers.push(15);
    excludedNumbers.push(10);

    var array = generateExclude(excludedNumbers, 1, 20);

    function generateExclude(excludedNumbers, min, max){
        var newNumbers = [];
        for(var i = min; i <= max; i++) {
            for(var j = 0; j < excludedNumbers.length; j++) {
                var checker = $.inArray(i, excludedNumbers)
                if(checker > -1){

                }else{
                    if($.inArray(i, newNumbers)<= -1){
                        newNumbers.push(i); 
                    }

                }
            };
        };
        return newNumbers;
    }
    function generateRandomNumbers(items){
        var num = items[Math.floor(Math.random()*items.length)];;
        return num;
    }
    console.log(generateRandomNumbers(array))
    </script>

Answer 11

You can build an array dynamically. Depending on where you are getting the excluded numbers. Something like:

var excluded = [8, 15];
var random = [];
for(var i = min; i <= max; i++) {
  if(excluded.indexOf(i) !== -1) {
    random.push(i);
  }
}

Then use the tips found in the answer for this post: How can I generate a random number within a range but exclude some?. Should get you to where you want to go.

Answer 12

it should be or instead of and

function generateRandom(min, max) {
    var num = Math.floor(Math.random() * (max - min + 1)) + min;
    return (num === 8 || num === 15) ? generateRandom(min, max) : num;
}

var test = generateRandom(1, 20)

Answer 13

I think it should be like this, if you want good distribution on all numbers. and, for this solution, it is required to higher max than 15 and lower min that 8

function generateRandom(min, max) {
    var v =  Math.floor(Math.random() * (max - min + 1 - 2)) + min;
    if (v == 8) return max-1;
    else if (v == 15) return max-2;
    else return v;
}

var test = generateRandom(1, 20)

Answer 14

You could make use of a recursive function

function generateRandom(min, max, num1, num2) {
    var rtn = Math.floor(Math.random() * (max - min + 1)) + min;
    return rtn == num1 || rtn == num2 ? generateRandom(min, max, num1, num2) : rtn;
}

Answer 15

One way, which will maintain the generator's statistical properties, is to generate a number in [1, 18]. Then apply, in this order:

1. If the number is 8 or more, add 1.

2. If the number is 15 or more, add 1.

I'd be reluctant to reject and re-sample as that can cause correlation plains to appear in linear congruential generators.