Question

我想创建一个值表，其中对于每个“t”值，函数“TabRot”计算结果，并找到“TabRot”最大的“t”值。由于我的步进间隔在以下代码中为0.001（也可以是0.00001）;什么是最快的计算方法？

捕捉是可以有两个/更多相同的最大值，我只需要选择第一个。

到目前为止

代码：

// Devise optimize function
for (double t = -0.025; t <= 0.025; t = t + 0.001)//theta[{t, -.05, .05, .001}]
                    {
                        //Table[t,TabRot]
                        DataTable TabVal = new DataTable();
                        TabVal.Rows.Add(RotTab.TabRot(t, i1));

                    }

Answer 1

不清楚您的问题，但您需要实例化数据表外部循环，否则您将在每次循环迭代中创建数据表

DataTable TabVal = new DataTable();
for (double t = -0.025; t <= 0.025; t = t + 0.001)//theta[{t, -.05, .05, .001}]
                    {
                        TabVal.Rows.Add(RotTab.TabRot(t, i1));
                    }

根据您的评论，看起来您正在尝试存储单个标量值。在这种情况下，使用list<double>之类的

非常有效

    List<double> TabVal = new List<double>();
    for (double t = -0.025; t <= 0.025; t = t + 0.001)//theta[{t, -.05, .05, .001}]
   {
      TabVal.Add(RotTab.TabRot(t, i1));
   }

填好列表后，您可以通过调用

等MAX()方法找到最大值

TabVal.Max();

Answer 2

我假设RotTab.TabRot（double，object）;返回一个double。您的评论向我表明您希望tabVal保持（t，TabRot（t，i1）），这样就可以填充它。

//Defined elsewhere: i1, RotTab[.TabRot]
//populates tabVal with (t,trot) pairs
//maxt contains the t value that produced the maximum 
DataTable tabVal = new DataTable();//move this here, otherwise tabVal will only ever have the last item in it
//initialize what your columns are
tabVal.Columns.Add("t", double);
tabVal.Columns.Add("trot", double);

//These should probably be somewhere else, but for this code I'm putting them here. Magic numbers are bad, mmkay?
double start = -0.025;
double end = 0.025;
double step = 0.001;

//init to the first value, since that's definitely a valid possible max
double max = RotTab.TabRot(start, i1);
double maxt = start;

//used for holding what we're looking at.
double cur;
for (double t = start; t <= end; t += step) {
    cur = RotTab.TabRot(t, i1);
    if (cur > max) {
        max = cur;
        maxt = t;
    }
    tabVal.Rows.Add(new double[] {t, cur});
}

如果您要做的就是采取最大值并偶尔查找t-＆gt;小跑（不想再次计算小跑），您可以使用词典代替。可能会更有效率。

编辑：如果您需要的只是最大值，那么......

//These should probably be somewhere else, but for this code I'm putting them here. Magic numbers are bad, mmkay?
double start = -0.025;
double end = 0.025;
double step = 0.001;

double max = RotTab.TabRot(start, i1);
double maxt = start;
double cur;
for (int t = start; t <= end; t += step) {
    cur = RotTab.TabRot(t, i1);
    if (cur > max) {
        max = cur;
        maxt = t;
    }
}

您不需要存储任何东西，只需找到最大值。

制表数据并找到最大值c＃

2 个答案: