2012-04-24 81 views
0

我有两个特征和权重元素矩阵。我正在实现一个学习算法。我想更新arraylist的元素(用于表示一个特征样本的向量)。以下是代码。但我的矩阵元素(矢量元素不)更新。我也提供了样本解决方案。预计不会更新前后的相同值。请让我知道代码中的缺陷在哪里?无法更新arrayList中的元素

for(int i =0 ; i< N ; i++){ //N is a large real number 
    ArrayList<Double> featureVector = new ArrayList<Double>(); 
    featureVector = FeatureMatrix.get(i); 
    System.out.println("Before::"+ featureVector); 
    if(testList.contains(i)){ 
    for(int j=0 ; j< testList.size(); j++){ 
     if(i == testList.get(j)){  
     int indexInTestList= j; 
     List<Double> subListNextCandidate ; 
     subListNextCandidate = weightVectorNextCandidate.subList((10*indexIntTestList),((10)*(indexInTestList+1))); //clips a portion of member from long list of members 
     List<Double> approxWeight = new ArrayList<Double>(); 
     approxWeight = getApproxWeight(FeatureVector, indexInTestList, FeatureMatrix,WeightMatrix, bias); //approxWeight is a vector of same dimension as of featureVector 

     for(int l=0 ; l< 10;l++){     
      double Update = featureVector.get(l)+ Rate*((subListCandidate.get(l)-approxWeight.get(l))-(lambda*featureVector.get(l)*(1/M)));//M is large real number 
      featureVector.set(l,Update); 

     }     
     } 
    } 
    } 

    else{ 
    for(int l=0 ; l< 10;l++){ 
     double Update = featureVector.get(l) -Rate*(lambda*featureVector.get(l)*(1/M)); 
     featureVector.set(l, Update); 
    }     
    } 
    System.out.println("After:::"+ FeatureMatrix.get(i)); 
} 

示例输出::

Before::[0.04539928251182193, -0.16233604402485394, 0.905018369795912, -1.2817141994528614, 0.7065420460225843, -0.8946090188977665, -1.74892020689701, -2.1539901172158187, 1.8229765478806985, -1.8109945435256574] 
After:::[0.04539928251182193, -0.16233604402485394, 0.905018369795912, -1.2817141994528614, 0.7065420460225843, -0.8946090188977665, -1.74892020689701, -2.1539901172158187, 1.8229765478806985, -1.8109945435256574] 
+3

您应该使用调试器并逐步执行代码中的行。 – 2012-04-24 20:16:37

+1

1.格式化您的代码2.“testList”来自何处? (在您发布的代码中没有初始化。) – aviad 2012-04-24 20:28:52

+0

测试列表作为方法的参数。 – thetna 2012-04-24 20:35:33

回答

2

我能想到的只有一对夫妇的合理原因,这会发生的:

  1. 率== 0
  2. testList.contains (i)始终为假

我强烈建议使用断点来调试。至少,放置一个System.out.println,其中调用featureVector.set()以确保它被调用。我猜测它永远不会被调用,因为条件永远不会成真。

确实使用断点,这将是一个生命的救星......

0

什么是testList.get(j)返回类型?你正在比较一个整数,我猜想是一个整数。这不太可能顺利......