C#,数值计算——插值和外推,双线性插值(Bilin_interp)的计算方法与源程序

1 文本格式

using System;

namespace Legalsoft.Truffer
{
    ///


    /// 双线性插值
    /// interpolation routines for two dimensions
    /// Object for bilinear interpolation on a matrix.
    /// Construct with a vector of x1.
    /// values, a vector of x2 values, 
    /// and a matrix of tabulated function values yij
    /// Then call interp for interpolated values.
    ///

    public class Bilin_interp
    {
        private int m { get; set; }
        private int n { get; set; }
        private double[,] y { get; set; }
        private Linear_interp x1terp { get; set; } = null;
        private Linear_interp x2terp { get; set; } = null;

        public Bilin_interp(double[] x1v, double[] x2v, double[,] ym)
        {
            this.m = x1v.Length;
            this.n = x2v.Length;
            this.y = ym;
            this.x1terp = new Linear_interp(x1v, x1v);
            this.x2terp = new Linear_interp(x2v, x2v);
        }

        public double interp(double x1p, double x2p)
        {
            int i = x1terp.cor > 0 ? x1terp.hunt(x1p) : x1terp.locate(x1p);
            int j = x2terp.cor > 0 ? x2terp.hunt(x2p) : x2terp.locate(x2p);
            double t = (x1p - x1terp.xx[i]) / (x1terp.xx[i + 1] - x1terp.xx[i]);
            double u = (x2p - x2terp.xx[j]) / (x2terp.xx[j + 1] - x2terp.xx[j]);
            double yy = (1.0 - t) * (1.0 - u) * y[i, j] + t * (1.0 - u) * y[i + 1, j] + (1.0 - t) * u * y[i, j + 1] + t * u * y[i + 1, j + 1];
            return yy;
        }
    }
}
 

2 代码格式

using System;

namespace Legalsoft.Truffer
{
    /// 
    /// 双线性插值
    /// interpolation routines for two dimensions
    /// Object for bilinear interpolation on a matrix.
    /// Construct with a vector of x1.
    /// values, a vector of x2 values, 
    /// and a matrix of tabulated function values yij
    /// Then call interp for interpolated values.
    /// 
    public class Bilin_interp
    {
        private int m { get; set; }
        private int n { get; set; }
        private double[,] y { get; set; }
        private Linear_interp x1terp { get; set; } = null;
        private Linear_interp x2terp { get; set; } = null;

        public Bilin_interp(double[] x1v, double[] x2v, double[,] ym)
        {
            this.m = x1v.Length;
            this.n = x2v.Length;
            this.y = ym;
            this.x1terp = new Linear_interp(x1v, x1v);
            this.x2terp = new Linear_interp(x2v, x2v);
        }

        public double interp(double x1p, double x2p)
        {
            int i = x1terp.cor > 0 ? x1terp.hunt(x1p) : x1terp.locate(x1p);
            int j = x2terp.cor > 0 ? x2terp.hunt(x2p) : x2terp.locate(x2p);
            double t = (x1p - x1terp.xx[i]) / (x1terp.xx[i + 1] - x1terp.xx[i]);
            double u = (x2p - x2terp.xx[j]) / (x2terp.xx[j + 1] - x2terp.xx[j]);
            double yy = (1.0 - t) * (1.0 - u) * y[i, j] + t * (1.0 - u) * y[i + 1, j] + (1.0 - t) * u * y[i, j + 1] + t * u * y[i + 1, j + 1];
            return yy;
        }
    }
}

你可能感兴趣的:(C#数值计算,Numerical,Recipes,c#,数值计算,算法,开发语言)