Generate 10 unique random integers

Question

I'm working on a loop that adds 10 unique random integers from 1-10 to an array, so the expected result would be 2,1,4,6,10,7,9,8,5,3 I manage to generate random numbers and only add the unique ones, my problem is, its not complete 10 items, I think the problem is upon checking I it doesn't generate again.

Hope you help me.

Thanks.

for (let i = 0; i < 10; i++) {
  var arrNum = [];
  setTimeout(function() {
    var r = Math.floor(Math.random() * 10) + 1;
    while (arrNum.includes(r) == false) {
      arrNum.push(r);
    }
    if (i == 9) {
      $('ul').append('<li>' + arrNum + '</li>');
    }
  }, 500);
}

ul li{
  list-style-type: none;
  display: inline-block;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul></ul>

Possible duplicate of [Generate unique random numbers between 1 and 100](https://stackoverflow.com/questions/2380019/generate-unique-random-numbers-between-1-and-100). See [this answer](https://stackoverflow.com/a/50652249/2019247) specifically. — 31piy, Aug 20 '18 at 05:20
Or, one could ask: "How to shuffle an array?" (really, ask SO this question) — user2864740, Aug 20 '18 at 05:23
@user2864740 oh you mean [this?](https://stackoverflow.com/questions/2450954/how-to-randomize-shuffle-a-javascript-array). too complicated right? — user123, Aug 20 '18 at 05:27

Jonas Wilms · Accepted Answer · 2018-08-20T06:05:26.623

2

  while (arrNum.includes(r) == false) {
   arrNum.push(r);
 }

That adds the number as long as it doesnt exist, so it will only run once, and if the number exists already it doesnt generate a new one, you want:

   const result = [];
   for(let i = 1; i <= 10; i++) {
       let random;
       do {
         random = 1 + Math.floor(Math.random() * 10);
      } while(result.includes(random));
      result.push(random);
  }

But its probably way more easy to just shuffle an array with the numbers 1 to 10.

edited Aug 20 '18 at 06:05

answered Aug 20 '18 at 05:28

Jonas Wilms

132,000
20
149
151

thanks this for sir, the approach is different that's why I choose this method. – user123 Aug 20 '18 at 05:38
@Mark While this does show valid logic and focuses on the original code/issue, a do-while is a *non-scaling* approach. For 10 items, it won't matter. – user2864740 Aug 20 '18 at 16:30
... and to continue on @user2864740 it might even happen that this one runs forever ... (very unlikely though) – Jonas Wilms Aug 20 '18 at 17:40

bitsobits · Answer 2 · 2018-08-20T06:21:58.680

1

Your code has a mistake. The i is not the ideal way to check for 10 occurrences as it is incremented even if variable r does not satisfy your condition - arrNum.includes(r) == false.

The check should be made by array length.

Replace if (i == 9) with if(arrNum.length = 10) in your code.

edited Aug 20 '18 at 06:21

answered Aug 20 '18 at 05:28

bitsobits

104
11

score 1 · Answer 3 · answered Aug 20 '18 at 05:30

Create an array with numbers from 1 to N (first line). Replace N with 10 in your case. Then sort it randomly.

const numbers = [ ...Array(N).keys() ].map(o => o + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers); // [ 6, 4, 5, 2, 1, 9, 3, 7, 8, 10 ]

Generate 10 unique random integers

3 Answers3