2

我正在使用 JMathPlot 库来生成一个简单的图形，在这种情况下，它是一个 3D，它会针对 for 循环中的每次迭代进行更新。但是，我得到的循环速度是：

Exception in thread "AWT-EventQueue-0" java.lang.NullPointerException

和

Exception in thread "AWT-EventQueue-0" java.lang.IndexOutOfBoundsException

据我了解，这与 AWT 线程和主线程不相处有关。我知道我需要以特殊方式更新图形线程，只是不确定如何。这是我的代码，如果有人可以建议我如何更新情节（我猜是重新绘制）而不会出现错误，那就太好了。

import javax.swing.*;
import org.math.plot.*;
import static java.lang.Math.*;
import static org.math.array.DoubleArray.*;

public class GridPlotsExample {

public static void main(String[] args) throws InterruptedException {

    JFrame frame = new JFrame("a plot panel");
    frame.setSize(600, 600);
    frame.setVisible(true);

    // create your PlotPanel (you can use it as a JPanel) with a legend
    // at SOUTH
    Plot3DPanel plot = new Plot3DPanel("SOUTH");

    frame.setContentPane(plot);

    for (int i = 1; i < 10; i++) {

        // define your data
        double[] x = increment(0.0, 0.1, i); // x = 0.0:0.1:1.0
        double[] y = increment(0.0, 0.05, i);// y = 0.0:0.05:1.0
        double[][] z1 = f1(x, y);
        double[][] z2 = f2(x, y);

        // add grid plot to the PlotPanel
        plot.addGridPlot("z=cos(PI*x)*sin(PI*y)", x, y, z1);
        plot.addGridPlot("z=sin(PI*x)*cos(PI*y)", x, y, z2);

    }

}

// function definition: z=cos(PI*x)*sin(PI*y)
public static double f1(double x, double y) {
    double z = cos(x * PI) * sin(y * PI);
    return z;
}

// grid version of the function
public static double[][] f1(double[] x, double[] y) {
    double[][] z = new double[y.length][x.length];
    for (int i = 0; i < x.length; i++)
        for (int j = 0; j < y.length; j++)
            z[j][i] = f1(x[i], y[j]);
    return z;
}

// another function definition: z=sin(PI*x)*cos(PI*y)
public static double f2(double x, double y) {
    double z = sin(x * PI) * cos(y * PI);
    return z;
}

// grid version of the function
public static double[][] f2(double[] x, double[] y) {
    double[][] z = new double[y.length][x.length];
    for (int i = 0; i < x.length; i++)
        for (int j = 0; j < y.length; j++)
            z[j][i] = f2(x[i], y[j]);
    return z;
}
}
4

2 回答 2

0

您可以将 2 个代码语句移到 for 循环中并尝试：

for (int i = 1; i < 10; i++) {
    Plot3DPanel plot = new Plot3DPanel("SOUTH");
    frame.setContentPane(plot);

    // define your data
    double[] x = increment(0.0, 0.1, i); // x = 0.0:0.1:1.0
    double[] y = increment(0.0, 0.05, i);// y = 0.0:0.05:1.0
    double[][] z1 = f1(x, y);
    double[][] z2 = f2(x, y);

    // add grid plot to the PlotPanel
    plot.addGridPlot("z=cos(PI*x)*sin(PI*y)", x, y, z1);
    plot.addGridPlot("z=sin(PI*x)*cos(PI*y)", x, y, z2);

}

编辑：使用同一框架时，您必须使用刷新其内容的标准方式（无效和验证）。

于 2012-02-09T16:59:55.270 回答
0

如果您需要在每次交互后查看结果，请在结束时暂停线程以绘制其他结果：

for (int i = 1; i < 10; i++) {

    // define your data
    double[] x = increment(0.0, 0.1, i); // x = 0.0:0.1:1.0
    double[] y = increment(0.0, 0.05, i);// y = 0.0:0.05:1.0
    double[][] z1 = f1(x, y);
    double[][] z2 = f2(x, y);

    // add grid plot to the PlotPanel
    plot.addGridPlot("z=cos(PI*x)*sin(PI*y)", x, y, z1);
    plot.addGridPlot("z=sin(PI*x)*cos(PI*y)", x, y, z2);

    try {
      // Pause 1 ms
       Thread.sleep(1000);        
    } catch (InterruptedException e) {
      // TODO: handle exception
    }

}
于 2018-12-14T18:37:32.763 回答