10 different number in array

Question

I need 10 different random numbers in an array.

I am using the following function:

    for (k=0;k<10;k++) {
      x[k] = Math.floor((Math.random()*10));
      for (l=0;l<k;l++){
        if (x[l]==x[k]){fail=true;}
      }
      if (fail){k=k-1;}
    }

THX for answer.

Well, what's wrong with the code you have now? Does it do nothing? Make an error? Crash your computer? Make flying monkeys fall from the sky? Give you free waffles? Please clarify. — tckmn, Jun 01 '13 at 13:09

score 3 · Accepted Answer · edited May 23 '17 at 11:49

3

Make sure all your variables are declared:

var x = [], k, l, fail;

Make sure you reset "fail" before the inner loop:

  fail = false;
  for (l = 0; l < k; l++)
    // ...

Now that said, if you want 10 numbers in an array, and the array is 10 elements long, you're better off filling the array with the numbers from 1 to 10 (or 0 to 9; whatever) and then shuffling the array. If you do things your way, then the performance of the process is unpredictable. When you get to the last couple of elements, it may take two or three tries, or it may take hundreds; you never know what the random number sequence may do.

edited May 23 '17 at 11:49

Community

1
1

answered Jun 01 '13 at 13:14

Pointy

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+1 for being able to fix this code that barely makes sense :P – tckmn Jun 01 '13 at 13:19
y this was stupid bad ... but i make more bads XD still doesnt work. – maty Jun 01 '13 at 13:21
@maty if you make the changes I described it works just fine. – Pointy Jun 01 '13 at 13:26

score 3 · Answer 2 · answered Jun 01 '13 at 13:16

It would be better to fill the array with numbers from 1 to 10, and then shuffle it, like this:

var myArray = [1,2,3,4,5,6,7,8,9,10];

//shuffle array
//code from http://stackoverflow.com/a/2450976/1223693
var i = myArray.length, j, temp;
if ( i === 0 ) return false;
while ( --i ) {
   j = Math.floor( Math.random() * ( i + 1 ) );
   temp = myArray[i];
   myArray[i] = myArray[j]; 
   myArray[j] = temp;
}

//done
alert(myArray);

Your current method is very slow because

it has an inner loop
it may generate wrong numbers many many times before it generates a correct number

score 1 · Answer 3 · answered Jun 01 '13 at 13:19

You really want to have a array with 10 elements with random numbers between 0 and 9 and none of it is allowed to be the same? So you want to have the numbers 0-9 in 10 Elements shuffled around?

Then just init them and shuffle them:

//+ Jonas Raoni Soares Silva
//@ http://jsfromhell.com/array/shuffle [v1.0]
function shuffle(o){ //v1.0
    for(var j, x, i = o.length; i; j = parseInt(Math.random() * i), x = o[--i], o[i] = o[j], o[j] = x);
    return o;
};


// init the array:
var a = new Array();
for (var i=0;i<10;i++){
  a.push(i);
} 
// shuffle it.
a = shuffle(a);

score 1 · Answer 4 · answered Jun 01 '13 at 13:41

1

could be simpler-

var nums= [0, 1, 2, 3, 4, 5, 6, 7, 8, 9].sort(
function(){return Math.random()-.5});

answered Jun 01 '13 at 13:41

kennebec

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See [Is it correct to use JavaScript Array.sort() method for shuffling?](http://stackoverflow.com/q/962802/1048572) - no, it is not. – Bergi Jun 01 '13 at 13:48
@bergi: sometimes a simple method is all you need- what is not appropriate for cryptography may be enough for a slide show, for example – kennebec Jun 01 '13 at 14:27

10 different number in array

4 Answers4