How to get a certain number of unique random numbers in a certain range?

Question

I have a function that takes user input to tell it how many random numbers to output, and the range to make the random numbers between (say 1-90). The problem is that it will give repeat numbers, but I want only unique numbers. Does anyone know how I can change the code to achieve this?

function random() {
    let randomNums = [];

// Get how many random numbers to output
    let numBox = document.querySelector('.numBox').value;

// Get the highest number range should go to (ex. 1-70)
    let highestNumber = document.querySelector('.highestNumber').value;

// Loop to generate random numbers and push to randomNums array
    for (let i = 0; i < numBox; i++) {
        let num = Math.floor(Math.random() * highestNumber) + 1;
        randomNums.push(` ${num}`)  
    }     

// Sort numbers from lowest to highest
    randomNums.sort(function(a, b) {return a - b;});

// Output numbers
    document.querySelector('.randomOutput').innerHTML = randomNums;
}

You can keep track of previously included value, [`Unique number`](https://stackoverflow.com/a/55624470/9624435) something like this can be done — Code Maniac, Oct 07 '19 at 16:10

Bilal Siddiqui · Answer 1 · 2019-10-07T16:13:08.223

1

Just change your loop to be a while-loop and check if array doesn't have that value already:

let i = 0;
while(i < numBox) {
    let num = Math.floor(Math.random() * highestNumber) + 1;
    if (randomNums.indexOf(num) == -1) {
        randomNums.push(num);
        i++;
    }
}

edited Oct 07 '19 at 16:13

answered Oct 07 '19 at 16:06

Bilal Siddiqui

3,579
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Sorry, no it still gives repeating numbers – fitz2882 Oct 07 '19 at 16:09
coreected, there was a missing bracket. – Bilal Siddiqui Oct 07 '19 at 16:10
I'm checking by generating 5 numbers between 1 and 10 and still getting some repeated numbers – fitz2882 Oct 07 '19 at 16:11
Ahhh, there was a space in push data ` ${num}` . Now you may try again – Bilal Siddiqui Oct 07 '19 at 16:13
if you are using ES7, you can use array.includes() for checking if an item exist in the array or not instead of array.indexOf() method. – Akash Oct 07 '19 at 16:14
@Akash, yeah but you know IE.. ;) – Bilal Siddiqui Oct 07 '19 at 16:15
With a while loop containing randomness, there's a chance, however small it may be, to create an infinite (or at least quote long-lasting) loop. Of course, you may decide to choose to ignore this probabilty. ;-) – Constantin Groß Oct 07 '19 at 16:16
@BilalSiddiqui use a `Set` instead of array, so the read access time will be `O(1)` instead of `O(n)` – Code Maniac Oct 07 '19 at 16:16
@BilalSiddiqui yeah, if need to support IE then fallback is needed OR better use ES5. I agree. – Akash Oct 07 '19 at 16:18
Also if IE is concern you use `Object` for the same purpose. – Code Maniac Oct 07 '19 at 16:21

score 0 · Answer 2 · answered Oct 07 '19 at 16:18

you can simply generate and verify if it exist until get the necessary random unique numbers

while (randomNums.length < numBox) {
    let num = Math.floor(Math.random() * highestNumber) + 1;
    if (randomNums.indexOf(num) === -1) randomNums.push(num);
}

score 0 · Answer 3 · answered Oct 07 '19 at 16:23

Here's an approach avoiding a while() loop.

Populate an array with the numbers from 1 to the maximum value
Randomise the array (I've borrowed the Fisher-Yates algorithm from this answer)
Pick the first N items from the array

function random() {
  let randomNums = [];

  // Get how many random numbers to output
  let numBox = document.querySelector('.numBox').value;

  // Get the highest number range should go to (ex. 1-70)
  let highestNumber = document.querySelector('.highestNumber').value;

  // Loop to generate random numbers and push to randomNums array
  for (let i = 1; i <= highestNumber; i++) {
    randomNums.push(i)
  }

  randomNums = shuffle(randomNums).slice(0, numBox);

  // Sort numbers from lowest to highest
  randomNums.sort(function(a, b) {
    return a - b;
  });

  // Output numbers
  document.querySelector('.randomOutput').innerHTML = randomNums;
}

function shuffle(a) {
  var j, x, i;
  for (i = a.length - 1; i > 0; i--) {
    j = Math.floor(Math.random() * (i + 1));
    x = a[i];
    a[i] = a[j];
    a[j] = x;
  }
  return a;
}

random();

<input class="numBox" value="4"> out of <input class="highestNumber" value="48">
<div class="randomOutput"></div>
<button onclick="random()">randomize</button>

score 0 · Accepted Answer · answered Oct 07 '19 at 16:28

This is the code that fixed it, if anyone else has a similar issue:

function random() {
    let randomNums = [];
    let included = {};
    let numBox = document.querySelector('.numBox').value;
    let highestNumber = document.querySelector('.highestNumber').value;

    for (let i =0; i<numBox; i++) {
        let temp = true;
        while(temp) {
            let num = Math.floor(Math.random() * highestNumber) + 1;
            if (included[num] === undefined) {
                temp = false
                randomNums.push(` ${num}`)
                included[num] = num
               }
        }
    }
    randomNums.sort(function(a, b) {return a - b;});
    document.querySelector('.randomOutput').innerHTML = randomNums;
}

score 0 · Answer 5 · answered Oct 07 '19 at 16:44

0

as @CodeManiac mentioned, using "Set" in the @BilalSiddiqui solution.

let randomNums = new Set();
while(randomNums.size < 5) {
    let num = Math.floor(Math.random() * 10) + 1;
        randomNums.add(num);
}
console.log(randomNums)

answered Oct 07 '19 at 16:44

Akash

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How to get a certain number of unique random numbers in a certain range?

5 Answers5