一系列双打

Question

如何修复此代码以使用1.0到15.0的双精度填充数组x？

int temp = 15;

string[] titles = new string[] { "Alpha", "Beta", "Gamma", "Delta" };
List<double[]> x = new List<double[]>();
for (int i = 0; i < titles.Length; i++)
{
    for (int j = 0; j < temp; j++)
    {
        x.Add[i][j]((double) j);
    }       
}

Answer 1

int temp = 15;

string[] titles = new string[] { "Alpha", "Beta", "Gamma", "Delta" };
List<double[]> x = new List<double[]>();

for (int i = 0; i < titles.Length; i++)
{
    double[] y = new double[temp];

    for (int j = 0; j < temp; j++)
    {
        y[j] = j + 1;
    }

    x.Add(y);
}

作为备注，x可以是Array。

或许你真的不需要List Array double个double[]。您可以简单地使用int temp = 15; string[] titles = new string[] { "Alpha", "Beta", "Gamma", "Delta" }; double[][] x = new double[titles.Length][]; for (int i = 0; i < titles.Length; i++) { double[] y = new double[temp]; for (int j = 0; j < temp; j++) { y[j] = j + 1; } x[i] = y; } 的锯齿状数组。

{{1}}

Answer 2

如果上面我的问题的回答是“是”，那么这里是一个答案：

        int temp = 15;

        string[] titles = new string[] { "Alpha", "Beta", "Gamma", "Delta" };
        List<double[]> x = new List<double[]>();
        for (int i = 0; i < titles.Length; i++)
        {
            double[] test = new double[15];
            for (int j = 1; j <= temp; j++)
            {
                test[j-1] = j;
            }
            x.Add(test);
        }