Fisher–Yates随机表行

Question

I intend to use the Fisher-Yates shuffle to shuffle existing rows of a given table such as:

<table>
    <tbody id="parent">
        <tr id="node1">
            <td>1</td>
            <td>A</td>
        </tr>
        <tr id="node2">
            <td>2</td>
            <td>B</td>
        </tr>
        <tr id="node3">
            <td>3</td>
            <td>C</td>
        </tr>
        <tr id="node4">
            <td>4</td>
            <td>D</td>
        </tr>
        <tr id="node5">
            <td>5</td>
            <td>E</td>
        </tr>
        <tr id="node6">
            <td>6</td>
            <td>F</td>
        </tr>
    </tbody>
</table>

With some help of JQuery, I have the following:

var parent = $("#parent");

function shuffleRows(parent) {
    var rows = parent.children();
    for (var i = rows.length - 1; i > 0; i--) {
        var j = Math.floor(Math.random() * (i + 1));
        var temp = rows[i];
        rows.eq(i - 1).after(rows[j]);
        rows.eq(j - 1).after(temp);
    }
}

shuffleRows(parent);

A working example is here: http://jsfiddle.net/98q9S/

However, as I test the above, I can't help but notice that row A gets the top spot far more often than other rows, and sometimes two consecutive shuffle produces the exact sequence of rows.

I wonder if there is anything wrong in my implementation, or it is just that my sample size is too small.

Answer 1

I noticed something similar in a toy app I was making using the algorithm. I ended up going with another solution I found on this site. I hope it helps:

makeArrRandom = function(array) {

  var counter = array.length, temp, index;
  while (counter > 0) {
    index = Math.floor(Math.random() * counter);
    counter--;
    temp = array[counter];
    array[counter] = array[index];
    array[index] = temp;
  }
  return array;
};