Question

有一个解决Lorentz系统的代码，我需要在相应的数组解决方案的每次迭代中保存所有解x，y，z，即5000个解x，y，z放在一个数组中，如何这是最好的吗？那么如何将数组转换为字符串呢？

public class Butterfly {

public static double dx(double x, double y, double z) {
    return -10*(x - y);
}

public static double dy(double x, double y, double z) {
    return -x*z + 28*x - y;
}

public static double dz(double x, double y, double z) {
    return x*y - 8*z/3;
}


public static void main(String[] args) {
    double x = 0.0, y = 20.0, z = 25.0; //
    double dt = 0.001;



    // uses Euler method
    for (int i = 0; i < 5000; i++) {


        // 
        double xnew = x + dx(x, y, z) * dt;
        double ynew = y + dy(x, y, z) * dt;
        double znew = z + dz(x, y, z) * dt;
        x = xnew; 
        y = ynew;
        z = znew;
        double[][] xyzArray = new double[5000][3];
        for (i = 0; i < xyzArray.length; i++) {
            for (int j = 0; i < xyzArray.length; j++) {
                xyzArray[i][j] = x;

            }
        }
        for (i = 0; i < xyzArray.length; i++) {
            for (int j = 0; i < xyzArray.length; j++) {
                System.out.println(xyzArray[i][j]);
            }
            System.out.println();
        }
       }
   }
}

- 编辑代码后，程序显示5000倍的值x，y，z，而不是所有值。我认为以这种方式失去了其他解决方案。毕竟，每个变量我有5000个决定，我需要保存每个......

 // uses Euler method
       double[][] xyzArray = new double[5000][3];
       for (int i = 0; i < xyzArray.length; i++) {
            for (i = 0; i < 5000; i++) {

            double xnew = x + dx(x, y, z) * dt;
            double ynew = y + dy(x, y, z) * dt;
            double znew = z + dz(x, y, z) * dt;

            xyzArray[i][0] = xnew;
            xyzArray[i][1] = ynew;
            xyzArray[i][2] = znew;
        }
    }
    for (int i = 0; i < xyzArray.length; i++) {
        System.out.println(xyzArray[i][0] + ", " + xyzArray[i][1] + ", " + xyzArray[i][2]);

    }

}

编辑后生成以下代码。仍会输出第一次迭代的5,000个解决方案：它在IDEAS或jshell中不起作用。我知道代码是正确的，但我无法理解为什么我得到错误的结果

public class Butterfly {

public static double dx(double x, double y, double z) {
    return -10 * (x - y);
}

public static double dy(double x, double y, double z) {
    return -x * z + 28 * x - y;
}

public static double dz(double x, double y, double z) {
    return x * y - 8 * z / 3;
}


public static void main(String[] args) {
   double x = 0.0, y = 20.0, z = 25.0;
   double dt = 0.001;
   double[][] xyzArray = new double[5000][3];
   for (int i = 0; i < xyzArray.length; i++) {
        double xnew = x + dx(x, y, z) * dt;
        double ynew = y + dy(x, y, z) * dt;
        double znew = z + dz(x, y, z) * dt;
        xyzArray[i][0] = xnew;
        xyzArray[i][1] = ynew;
        xyzArray[i][2] = znew;
   }
   for (int i = 0; i < xyzArray.length; i++) {
        System.out.println(xyzArray[i][0] + ", " + xyzArray[i][1] + ", " + xyzArray[i][2]);


    }

 }
}

Answer 1

您已在循环中声明了您的数组。您不想创建5000个数据点阵列的5000个副本！您只需要1个5000个数据点的数组，因此您需要在循环外声明和创建数组。

double[][] xyzArray = new double[5000][3];

// uses Euler method
for (int i = 0; i < xyzArray.length; i++) {

    // ... compute xnew, ynew, znew ... etc ...

获得当前步骤的新值后，您需要将它们保存在数组的[i]条目中。您将x存储在[i][0]子条目中，y存储在[i][1]子条目中，z存储在[i][2]中子条目。

    xyzArray[i][0] = xnew;
    xyzArray[i][1] = ynew;
    xyzArray[i][2] = znew;

在计算完所有值后打印出值：

} // end of Euler method loop

for (int i = 0; i < xyzArray.length; i++) {
    System.out.println(xyzArray[i][0] + ", " + xyzArray[i][1] +", " + xyzArray[i][2]);
}

您的revision 3 code有效。

查看修订版3代码
复制程序文本
将其粘贴到jshell
使用Butterfly.main(new String[] {})

你会看到你的5000 x，y，z值 - 都是不同的。

C:\>"\Program Files\Java\jdk-10\bin\jshell.exe"
|  Welcome to JShell -- Version 10
|  For an introduction type: /help intro

jshell> public class Butterfly {
   ...>
   ...> public static double dx(double x, double y, double z) {
   ...>     return -10*(x - y);

         [... many lines omitted for brevity ...]

   ...>     }
   ...>    }
   ...> }
|  created class Butterfly

jshell> Butterfly.main(new String[] {})
0.2, 19.98, 24.933333333333334
0.39780000000000004, 19.960633333333334, 24.870840444444443
0.5934283333333333, 19.9419174796712, 24.81245854319926
0.786913224796712, 19.92386713960567, 24.758126085937494

         [... many lines omitted for brevity ...]

0.381817425662861, 0.5879585365342771, 12.654916756967012
0.38387883677157514, 0.59322959817818, 12.621394805096582
0.3859723443856412, 0.5985398896533907, 12.587965480571079
0.3880980198383187, 0.6038899688589537, 12.554628592467306

jshell>

将一系列数字保存到数组中

1 个答案: