Java 2d数组比较

您好，我想问一个关于2d数组中比较的问题。 我有以下2d数组，每行有两个参数。 我想比较前5行（即周期0）和第5行（即周期1） 我想比较一下 比较将是一行一行作为一维数组我需要两个一维数组来相互比较

榜样

时段0（第0行）与时段1（第0行、第1行、第2行、第3行、第4行）

时段0第1行与时段1（第0行、第1行、第2行、第3行、第4行）

时段0第2行与时段1（第0行、第1行、第2行、第3行、第4行）

时段0第3行与时段1（第0行、第1行、第2行、第3行、第4行）

时段0第4行与时段1（第0行、第1行、第2行、第3行、第4行）

---

然后

时段1第0行与时段2（第0行、第1行、第2行、第3行、第4行）

时段1第1行与时段2（第0行、第1行、第2行、第3行、第4行）

时段1第2行与时段2（第0行、第1行、第2行、第3行、第4行）

时段1第3行与时段2（第0行、第1行、第2行、第3行、第4行）

时段1第4行与时段2（第0行、第1行、第2行、第3行、第4行）

---

然后

时段2第0行与时段3（第0行、第1行、第2行、第3行、第4行）

时段2第1行与时段3（第0行、第1行、第2行、第3行、第4行）

时段2第2行与时段3（第0行、第1行、第2行、第3行、第4行）

时段2第3行与时段3（第0行、第1行、第2行、第3行、第4行）

时段2第4行与时段3（第0行、第1行、第2行、第3行、第4行）

---

我的目标是获取周期0的第一行，然后将其转换为1d数组，然后获取周期1的第一行，然后将其转换为1d数组，依此类推

二维阵列如下所示

double[]myDistributions=新的double[]{

        row0 {0.15250886479593964,0.2610516793197853,0.11441768814194446,0.1265241345428162,0.3454976331995246},
        row1 {0.14389124837314887,0.10513281153155449,0.1833462741873425,0.36788054102686596,0.1997491248810186},
Period0 row2 {0.1111207312911868,0.17499901413730568,0.2914581757577288,0.20433331657432438,0.2180887622394655},
        row3 {0.0948730457966342,0.19288720600625753,0.19471332499886804,0.18018001280247228,0.3373464103957629},
        row4 {0.18877271931210932,0.26521449714587747,0.13230019262559328,0.27631809895720494,0.13739449195931552},

        row0 {0.1274517393962726,0.10526928843184565,0.35751329613481436,0.12240396832200726,0.2873617077151316},
        row1 {0.3016290729137225,0.09902028076323677,0.17515717333485062,0.35620664945852193,0.06798682352968743},
Period1 row2 {0.0948730457966342,0.19288720600625753,0.19471332499886804,0.18018001280247228,0.3373464103957629},
        row3 {0.18877271931210932,0.26521449714587747,0.13230019262559328,0.27631809895720494,0.13739449195931552},
        row4 {0.17804585265772793,0.25651982583759037,0.3860867129515085,0.09453797159458521,0.08480963695844175},

        row0 {0.2311120677942418,0.07901250493611567,0.15795189397863776,0.24546018732208122,0.28646334596884354},
        row1 {0.08823944830766299,0.26313933789756516,0.10406419933285384,0.3365866979847223,0.20797031647719286},
Period2 row2 {0.1274517393962726,0.10526928843184565,0.35751329613481436,0.12240396832200726,0.2873617077151316},
        row3 {0.3016290729137225,0.09902028076323677,0.17515717333485062,0.35620664945852193,0.06798682352968743},
        row4 {0.11962680738039468,0.1590225091909952,0.24009305610431117,0.11189649648370673,0.36936113084052996},

        row0 {0.17804585265772793,0.25651982583759037,0.3860867129515085,0.09453797159458521,0.08480963695844175},
        row1 {0.08823944830766299,0.26313933789756516,0.10406419933285384,0.3365866979847223,0.20797031647719286},
Period3 row2 {0.1274517393962726,0.10526928843184565,0.35751329613481436,0.12240396832200726,0.2873617077151316},
        row3 {0.0948730457966342,0.19288720600625753,0.19471332499886804,0.18018001280247228,0.3373464103957629},
        row4 {0.3962773409768214,0.12002724792315751,0.15722439889764284,0.11854502459707339,0.20792598760527947}};

我找到了解决办法

int numOfTopicsInEachPeriod=5;

double[][] myFirstRow=new  double [numOfTopicsInEachPeriod][myDistributions[0].length];
double[][] mySecondRow=new double[numOfTopicsInEachPeriod][myDistributions[0].length];


int count =0;
int counter =0;
int temp=0;

for(int j =0;j<myDistributions.length-numOfTopicsInEachPeriod;j++){

   int mod=j%numOfTopicsInEachPeriod;
   int countOfTopics = j/numOfTopicsInEachPeriod;
   count=0;

   if(mod==0){
       temp=1;
       count=numOfTopicsInEachPeriod* countOfTopics; 

   }

   if(temp==1&&mod!=0){count=numOfTopicsInEachPeriod* countOfTopics; }

     for(int i=counter;i<numOfTopicsInEachPeriod;i++){

       myFirstRow[i]=myDistributions[count];
       mySecondRow[i]=myDistributions[count+5];

       System.out.print("myFunction(Row["+j+"],Row ["+(count+5)+"]),"+"\t" );

       count++;

   }

   System.out.println();
}

int numotopicsineachperiod=5；
double[][]myFirstRow=新的双精度[numOfTopicsInEachPeriod][myDistributions[0]。长度]；
double[]mySecondRow=new double[numotopicsineachperiod][myDistributions[0].length]；
整数计数=0；
int计数器=0；
内部温度=0；
因为（int j=0；j既然你在一篇评论中提到了你在做什么，我们可以理解你在这里想要实现什么。坦率地说，没有任何不尊重，你的解释很糟糕。：p

按照我收集的方式，您正在尝试从两个特定的二维（2D）中获取行双类型数组并从那些特定的双类型数据行中检索分歧值。当然，您还没有透露分歧的类型。如果您这样做，我将不得不假设它是一个。无论哪种方式，您都可以使用我在下面提供的代码，并对其进行修改，以调用您希望收集所需分歧的任何Java方法

为了完成这项任务，我们需要两种特定的方法，其中最重要的一种当然是为我们提供分歧的方法。我下面提供的方法就是这样做的，它是由马萨诸塞州阿默斯特大学计算机科学系的Charles Sutton不久前写的方法是“MALLET”（MALLET）的一部分。以下是开源的：

/**
* Returns the Kullback–Leibler (KL) Divergence, K(p1 || p2).
*
* The log is w.r.t. base 2. <p>
*
* *Note*: If any value in <tt>p2</tt> is <tt>0.0</tt> then the KL-divergence
* is <tt>infinite</tt>. Limin changes it to zero instead of infinite. 
* 
*/
public static double klDivergence(double[] p1, double[] p2) {
    double log2 = Math.log(2);
    double klDiv = 0.0;

    for (int i = 0; i < p1.length; ++i) {
        if (p1[i] == 0) { continue; }
        if (p2[i] == 0.0) { continue; } // Limin

        klDiv += p1[i] * Math.log( p1[i] / p2[i] );
    }

    return klDiv / log2; // moved this division out of the loop -DM
}

现在我们需要做的就是调用我们的kldFromDoubleArrays（）方法来处理所需的时段。在帖子中，您指定要处理的行为：


周期0的每一行与周期1的每一行相对
周期1的每一行与周期2的每一行相对
周期2的每一行与周期3的每一行相对

因此，知道了这一点，我们将调用我们的方法三次：

//Create a Underline for Console window display.
String ul = String.join("", Collections.nCopies(100, "=")) + "\n";

//Period0 To Period1 Comparison:
kldFromDoubleArrays(period0, "Period 0", period1, "Period 1");
System.out.println(ul);

//Period1 To Period2 Comparison:
kldFromDoubleArrays(period1, "Period 1", period2, "Period 2");
System.out.println(ul);

//Period2 To Period3 Comparison:
kldFromDoubleArrays(period2, "Period 2", period3, "Period 3");
System.out.println(ul);

以下是一些示例输出：

The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642

到此为止。我希望这对您有所帮助。祝您的项目好运。
我不太明白您希望如何比较每一行。您想要实现什么类型的比较（等于、大于、小于、在范围内、全部）？感谢Jacob G的回复，我假设第一个5行是我的周期0，第二个5行是我的周期1，第三个5行是我的周期2，剩下的5行是我的周期3。我有一个函数，它获取2 1d数组作为输入。因此，首先我的函数输入是（周期0行0）和（周期1行0）您不需要签名
newdouble[]
当您像那样初始化它时，感谢您的回复，我有一个比较函数，该函数获取2个1d数组作为输入。因此，我会将每个1d数组我的意思是行发送到该函数Period0 row0 vs Period1 row0 Period0 vs Period1 row1 Period0 row0 vs Period1 row2 Period0 row0 vs Period1 row3周期0第0行vs周期1第4行---谢谢：）ım使用mallet方法，但我有一个问题，我的周期是动态的，我可能有数百个周期，所以我不能创建这样的周期，这真的很奇怪……因为我没有创建它们……你在最初的帖子中创建了。这真的不重要，因为你可以将你喜欢的任何2D双类型数组传递给KLDFromDoubleArray（）方法，只要它们在两个数组中都包含相同数量的列。听起来您需要的是帮助从您正在接收的特定双类型数据创建1D或2D数组…我不确定，因为此时我仍然不清楚您的实际需求是什么。为了进一步帮助您，我需要了解一些真实的需求数据，以及你想用它做什么。你想处理的数据是以稳定的流的形式输入的，并且在接收时需要实时发散吗？或者，它是以数据文件的形式输入的吗？我不知道…现在只是猜测一下。我从数据库中填充这个矩阵，我只是在这里制作了一个示例。我在mallet中使用主题建模，我有5个topic在上面的矩阵中，我的意思是，每个周期的前5行指周期0，周期0的第一行指主题0，第二个主题1等等。然后周期1的主题相同，我想我理解你在这里想说什么。所以，你从数据库表（或多个表）中提取这些数据使用一些SQL查询，现在需要将该查询的结果集放入2D双类型数组中
//Create a Underline for Console window display.
String ul = String.join("", Collections.nCopies(100, "=")) + "\n";

//Period0 To Period1 Comparison:
kldFromDoubleArrays(period0, "Period 0", period1, "Period 1");
System.out.println(ul);

//Period1 To Period2 Comparison:
kldFromDoubleArrays(period1, "Period 1", period2, "Period 2");
System.out.println(ul);

//Period2 To Period3 Comparison:
kldFromDoubleArrays(period2, "Period 2", period3, "Period 3");
System.out.println(ul);

The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 1 of Period 1 is: -->  -0.16008580289377392
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 2 of Period 1 is: -->  0.025341952815786797
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642
The Divergence between Row 2 of Period 0 and Row 3 of Period 1 is: -->  -0.08495427575998642