说我有一个数组：

  myList上：阵列=新的Array（）;
myList上= [1,2,3,4,5,6,7,8,9]

myRandomList：阵列=新的Array（）;

对于（VAR我：UINT = 0; I＆LT; myList中，我++）{
            VAR项目：数= Math.floor（的Math.random（）* myList.length-1）+ 1;
            myRandomList.push（项目）;
      }
 

唯一的一点是，我想myRandomList到没有任何重复的数字......是有办法从第一个列表中选择一个随机数，然后减去它，所以我不选择这个数字两倍的

更新：

我刚刚看到的洗牌从shadetyler.blogspot.com/2008/12/array-shuffle-as3.html

  Array.prototype.shuffle =功能（）{
对于（VAR I = 0; I＆LT; this.length;我++）{
VAR一个=这[一]
变种B = Math.floor（的Math.random（）* this.length）;
这个[我] =此[B]。
该[B] = A;
}
 

然而，有没有办法重写这是一个功能呢？     }

的标题说，洗牌数组，所以如果你正在寻找一个理想的洗牌，您可能希望的费雪耶茨算法是公正的。

所以，如果你想用/保留原来的，你会初始化 myRandomList

  VAR myRandomList：阵列=新阵列（myList.length）;
 

然后创建一个随机数的范围内说 然后交换 myRandomList [A] 与 myRandomList [I] 其中i是当前元素。

  //随机数
VAR一个= Math.floor（的Math.random（）* myList.length）;
// A掉
myRandomList [I] = myRandomList并[a];
//放无论是在指数中的第i个位置
myRandomList并[a] = myList中[I];
//恢复无论是在第i个位置索引
 

Say I have an array:

myList:Array = new Array();
myList = [1,2,3,4,5,6,7,8,9];

myRandomList:Array = new Array();

for (var i:uint = 0; i < myList; i++) {
            var item:Number = Math.floor(Math.random() * myList.length-1) + 1;
            myRandomList.push(item);
      }

The only thing is, I'd like myRandomList to not have any duplicate numbers...is there a way to select a random number from the first list and then SUBTRACT it so I don't select that number twice?

UPDATE:

I just saw this method of shuffling an array from shadetyler.blogspot.com/2008/12/array-shuffle-as3.html

Array.prototype.shuffle = function(){
for(var i = 0; i < this.length; i++){
var a = this[i];
var b = Math.floor(Math.random() * this.length);
this[i] = this[b];
this[b] = a;
}

However, is there a way to rewrite this as a function? }

The title says shuffle an array so if you are looking for an ideal shuffle you may want the Fisher–Yates algorithm that is unbiased.

So if you wanted to use/keep your original, you would initialize myRandomList

var myRandomList: Array = new Array( myList.length );

Then create a random number with the range say a and then swap myRandomList[a] with myRandomList[i] where i is the current element.

// Random number
var a = Math.floor(Math.random() * myList.length);
// A swap
myRandomList[i] = myRandomList[a];
// put whatever is in index a in the ith position
myRandomList[a] = myList[i];
// restore whatever was in the ith position to index a