Shuffle array except for even index's javascript

Question

I am able to shuffle an array with this function

function getBadlyShuffledList(list) {
  var length = list.length;
  var i;
  var j;

  while (length) {
    i = Math.floor(Math.random() * length--);


      j = list[length];
      list[length] = list[i];
      list[i] = j;

  }
  return list;
}

console.log(getBadlyShuffledList([1,2,3,4,5,6]));

The problem is I need to leave the index's that are divisible by two in their places. I tried using an if statement and it get's the index if it is 2 or 4, but then I don't know what to do to make it stay put.

For example if I have an array [1, 2, 3, 4, 5, 6] When shuffled, 3 and 5 should remain put. ex; [2, 4, 3, 1, 5, 6] How do I go about doing this?

isn't that an endless while loop, since length never changes — depperm, Jul 31 '15 at 16:25
Your example and your requirements don't match. If it's indices (or indexes, but not index's :)) divisible by 2, it should be indices 0, 2, and 4, or numbers 1, 3, and 5. — Heretic Monkey, Jul 31 '15 at 16:29
@MikeMcCaughan Your correct it is indices 0, 2, 4 or numbers 1, 3, 5 — rdeg, Jul 31 '15 at 16:41

score 2 · Accepted Answer · edited May 23 '17 at 12:14

I borrowed a shuffle function from the following link... How to randomize (shuffle) a JavaScript array?

Build an array that contains the elements you want to shuffle. Shuffle them. And finally add each element one by one back into your original array skipping every other element. It works but there is room for optimization.

/**
 * Randomize array element order in-place.
 * Using Fisher-Yates shuffle algorithm.
 */
function shuffleArray(array) {
    for (var i = array.length - 1; i > 0; i--) {
        var j = Math.floor(Math.random() * (i + 1));
        var temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
    return array;
}


function getBadlyShuffledList(list) {

  listToShuffle = []

  for(var i=0; i < list.length; i++){
    if (i%2!==0){
      listToShuffle.push(list[i]);
    }
  }

  shuffleArray(listToShuffle);

  for(var i=0; i < list.length; i++){
    if (i%2!==0){
      list[i] = listToShuffle[0];
      listToShuffle.shift();
    }
  }

  return list;
}

console.log(getBadlyShuffledList([3, 1, 2, 3, 4, 5, 6, 6, 6, 9, 2]));

I'm not sure how OP intented the result, but I think this is wrong: `[3, 1, 2, 3, 4, 5, 6, 6, 6, 9, 2]` => `[3, 5, 2, 9, 4, 1, 6, 3, 6, 6, 2]` — Rudie, Jul 31 '15 at 16:54
The way I read the OP it should be every other element stays the same starting with the first one. In your array. 3 . 2 . 4. 6. 6. 2 are all in the same position in the output with the other elements shuffled around. — abaldwin99, Jul 31 '15 at 17:02

Shuffle array except for even index's javascript

1 Answers1