使用markov链时获取ArrayIndexOutOfBoundsException

我有一个数组来存储绘制一条线的一组坐标。所以下面是一些示例坐标使用markov链时获取ArrayIndexOutOfBoundsException

double[][] plotMatrix = {{10,20},{55,80}, 
         {120,40},{225,30}, 
         {327.5,100}, 
         {427.5,30}, 
         {529,60}};

下一步是创建一个二维马尔科夫矩阵。

enter image description here

首先，我算其中从左列中的一个点，然后在上面列点时间。由于我想要一条线，每个点都跟着另一个点。这意味着如果我们有{10,20}作为输入，则{55,80}作为下一个点的可行性是100％。

我不太确定所有这些，请纠正我！

所以这是我的矩阵

double[][] markovMatrix = { {0.0,1.0,0.0,0.0,0.0,0.0,0.0}, 
            {0.0,0.0,1.0,0.0,0.0,0.0,0.0}, 
            {0.0,0.0,0.0,1.0,0.0,0.0,0.0}, 
            {0.0,0.0,0.0,0.0,1.0,0.0,0.0}, 
            {0.0,0.0,0.0,0.0,0.0,1.0,0.0}, 
            {0.0,0.0,0.0,0.0,0.0,0.0,1.0}, 
            {0.0,0.0,0.0,0.0,0.0,0.0,0.0}};

我的算法：

int seed = 0; 
    int output = 0; 

    for(int i = 0; i < 40;i++){ 
     double choice = r.nextDouble(); 

     double currentSum = 0.0; 

     for(;output < markovMatrix.length;output++){ 

      currentSum += markovMatrix[seed][output]; 

      if(choice <= currentSum){ 
       break; 
      } 
     } 

     System.out.println(output); 
     polygon.lineTo(plotMatrix[output][0], plotMatrix[output][1]); 

     seed = output; 

     output = 0; 
    }

我的问题是，我得到一个ArrayOutOfBoundsException:7当我试图同时访问plotMatrix和markovMatrix。但是在每个循环结束时输出设置为0。任何想法如何解决这个问题？

来源

2011-05-30 Lost in OWL

你能告诉你'markovMatrix'和'plotMatrix'的声明吗？当你逐步调试调试器中的代码时，矩阵中有多大，你期望它有多大？ – 2011-05-30 15:09:13

我不太清楚，如果它的正确的答案，

但（;输出< markovMatrix.length;输出++）将逐步从0到7，而你只有在markovMatrix 0-6项。

采用与（;输出< markovMatrix.length-1;输出++）修复了从1步进到6

不过，我怀疑你真的想一步从0到6而ArrayIndexOutOfBoundsException异常其实存在着你的问题。

来源

2011-05-30 15:07:57

何时完成循环与内部循环输出= 7这是数组的长度。你应该跳过最后一次迭代，因为你的数组索引是从0到6.

来源

2011-05-30 15:08:53 VirtualTroll

使用markov链时获取ArrayIndexOutOfBoundsException

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