将数字序列保存到数组中
将数字序列保存到数组中
提问于 2018-06-01 02:09:08
有一个用于求解Lorentz系统的代码，我需要在每次迭代时将所有的解x，y，z保存在相应的数组解决方案中，也就是说，将5000个解x，y，z放入一个数组中，怎么做才是最好的？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();
}
}
}
}А在编辑代码后，程序显示一个值x，y，z的5000次，而不是所有的值。我认为这样就失去了其他的解决方案。毕竟，我有每个变量的5000个决策，我需要保存每个变量…
// 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，将生成以下代码。第一次迭代的5,000个解决方案仍然是输出的:它不能在IDEAS或jshell中工作。我知道代码是正确的，但我不明白为什么我得到了错误的结果
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]);
}
}
}回答 1
Stack Overflow用户
回答已采纳
发布于 2018-06-01 03:28:33
您已经在循环中声明了数组。您不希望为包含5000个数据点的数组创建5000个副本！您只需要一个包含5000个数据点的数组，因此需要在循环外部声明和创建该数组。
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，需要将它们保存在数组的[i]-th条目中。您将把x存储在[i][0]子项中，将y存储在[i][1]子项中，将z存储在[i][2]子项中。
xyzArray[i][0] = xnew;
xyzArray[i][1] = ynew;
xyzArray[i][2] = znew;要在计算完所有值后打印值，请执行以下操作：
} // 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]);
}你的revision 3 code起作用了。
使用Butterfly.main(new String[] {})
- 查看修订版3的代码
- 复制程序的文本并将其粘贴到
jshell - Execute中
你会看到5000x，y，z的值--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>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50630353
复制相关文章
相似问题