Question

我有一个数组[160,160]，有27个测量数据点，想要插入整个数组。我找到了以下代码，但是没有得到它必须移交哪些参数，以便此方法有效：

 public static double BilinearInterpolation(double[] x, double[] y, double[,] z, double xval, double yval)
            {
                //calculates single point bilinear interpolation
                double zval = 0.0;
                for (int i = 0; i < x.Length - 1; i++)
                {
                    for (int j = 0; j < y.Length - 1; j++)
                    {   
                        if(xval>=x[i] && xval<x[i+1] && yval>=y[j] && yval<y[j+1])
                        {
                        zval = z[i,j]*(x[i+1]-xval)*(y[j+1]-yval)/(x[i+1]-x[i])/(y[j+1]-y[j])+
                               z[i+1,j]*(xval-x[i])*(y[j+1]-yval)/(x[i+1]-x[i])/(y[j+1]-y[j])+
                               z[i,j+1]*(x[i+1]-xval)*(yval-y[j])/(x[i+1]-x[i])/(y[j+1]-y[j])+
                               z[i+1,j+1]*(xval-x[i])*(yval-y[j])/(x[i+1]-x[i])/(y[j+1]-y[j]);
                        }
                    }
                }
                return zval;
            }

           public static double[] BilinearInterpolation(double [] x, double[] y, double[,]z,double[] xvals, double[]yvals)
           {
               //calculates multiple point bilinear interpolation
               double[] zvals = new double[xvals.Length];
               for (int i = 0; i < xvals.Length; i++)
                   zvals[i] = BilinearInterpolation(x, y, z, xvals[i], yvals[i]);
               return zvals;
           }

double [] x - ＆gt;我的点在数组中的x坐标？

double [] y - ＆gt;我的点在数组中的y坐标？

double [，] z - ＆gt;一个数组，其中存储插值？

double xval - ＆gt; ??

double yval - ＆gt; ??

这是对的吗？或者你可以更容易地插入数组双线性。

Answer 1

我认为xval和yval是您要插入的值。这些是您要评估的点的坐标。

数组的双线性插值c＃

1 个答案: