1

I am using the following code which generates random number between 0 to Totalfriends, I would like to get the random numbers but they should not be repeated. Any idea how?

This is the code I am using

FB.getLoginStatus(function(response) {
    var profilePicsDiv = document.getElementById('profile_pics');
FB.api({ method: 'friends.get' }, function(result) {

     // var result =resultF.data;
   // console.log(result);
   var user_ids="" ;
   var totalFriends = result.length;
   // console.log(totalFriends);
   var numFriends = result ? Math.min(25, result.length) : 0;
  // console.log(numFriends);
   if (numFriends > 0) {
      for (var i=0; i<numFriends; i++) {
        var randNo = Math.floor(Math.random() * (totalFriends + 1))
        user_ids+= (',' + result[randNo]);
         console.log(user_ids);

          }
        }
        profilePicsDiv.innerHTML = user_ids;
      });
});
Delimitry
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Yahoo
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  • You should be using `Math.floor(Math.random() * totalFriends)`, else you'll sometimes go past the end of the array – Eric May 25 '12 at 14:26

4 Answers4

3

Here's a function that will take n random elements from array, and return them, based off a fisher-yates shuffle. Note that it will modify the array argument.

function randomFrom(array, n) {
    var at = 0;
    var tmp, current, top = array.length;

    if(top) while(--top && at++ < n) {
        current = Math.floor(Math.random() * (top - 1));
        tmp = array[current];
        array[current] = array[top];
        array[top] = tmp;
    }

    return array.slice(-n);
}

Assuming your code works how I think it does, you already have an array of userids:

var random10 = randomFrom(friendIds, 10);
Eric
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  • I have Update the complete code , I am actually getting a list of ID's from Those list of ID's i am select random 25 . – Yahoo May 25 '12 at 14:50
  • So should i add this `randomFrom` function in the page and then call it inplace of `randNo` ? – Yahoo May 25 '12 at 14:51
2
  1. create an array (e.g. yourarray) of numbers in range [1..totalfriends]
  2. shuffle the array (e.g. using a javascript implementation of Fisher-Yates algorithm)
  3. inside the for (from 0 to yourarray.length - 1) make a pop() from the array (or just get the n-th element) so you will get everytime a different number

Doing so you you will avoid to get duplicated numbers

Fabrizio Calderan
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0

I would perform random iterations, create an array with all your numbers in, such as:

var friendIndexes = [];

for (var i=0; i<numFriends; i++)
{
   friendIndexes.push(i);
}

Then once you have an array of all the numbers, I would perform some number of iterations, maybe 1,000, where you generate two random numbers, and swap the values in those indexes.

for (var s = 0; s<1000; s++)
{
    var rnd1 = Math.floor(Math.random() * (numFriends + 1);
    var rnd2 = Math.floor(Math.random() * (numFriends + 1);

    // Swap the two values (remember to use a temp variable)
    var tmp = friendIndexes[rnd1];
    friendIndexes[rnd1] = friendIndexes[rnd2];
    friendIndexes[rnd2] = tmp;
}

You're essentially shuffling them, and the result is going to give you the numbers in a random order.

NibblyPig
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  • The idea is good, but the [Fisher-Yates shuffle](http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle) is a much more efficient way to shuffle an array. – Matthew Crumley May 25 '12 at 14:33
  • I have Update the complete code , I am actually getting a list of ID's from Those list of ID's i am select random 25 – Yahoo May 25 '12 at 14:52
0

Take a big number wich not divide numFriends or just a big prime number (like one : 702038, 727699, 992700, 1201046, 1232255, 2312734, 3136255, 4235414, 6090515) then goes

var result=[] ;
var K=Math.floor((Math.random()*bigUnNumFreindsDivider) ;

for (var i=0; i<numFriends; i++)
{
    result[i]=(i*bigUnNumFreindsDivider+K)%numFreinds ;
}

This should work fine.

Aurel
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