C#，数值计算——插值和外推，三次样条插值（Spline_interp）的计算方法与源程序

本文主要是介绍C#，数值计算——插值和外推，三次样条插值（Spline_interp）的计算方法与源程序，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

1 文本格式

using System;

namespace Legalsoft.Truffer
{
    /// <summary>
    /// 三次样条插值
    /// Cubic Spline Interpolation
    /// Cubic spline interpolation object. Construct with x and y vectors, and
    /// (optionally) values of the first derivative at the endpoints, then call
    /// interp for interpolated values
    /// </summary>
    public class Spline_interp : Base_interp
    {
        private double[] y2 { get; set; }

        public Spline_interp(double[] xv, double[] yv, double yp1 = 1.0e99, double ypn = 1.0e99) : base(xv, yv[0], 2)
        {
            this.y2 = new double[xv.Length];
            sety2(xv, yv, yp1, ypn);
        }

        public Spline_interp(double[] xv, double yv, double yp1 = 1.0e99, double ypn = 1.0e99) : base(xv, yv, 2)
        {
            this.y2 = new double[xv.Length];
            sety2(xv, y2, yp1, ypn);
        }

        /// <summary>
        /// This routine stores an array y2[0..n - 1] with second derivatives of the
        /// interpolating function at the tabulated points pointed to by xv, using
        /// function values pointed to by yv.If yp1 and/or ypn are equal to 1.0E99 or
        /// larger, the routine is signaled to set the corresponding boundary condition
        /// for a natural spline, with zero second derivative on that boundary;
        /// otherwise, they are the values of the first derivatives at the endpoints.
        /// </summary>
        /// <param name="xv"></param>
        /// <param name="yv"></param>
        /// <param name="yp1"></param>
        /// <param name="ypn"></param>
        public void sety2(double[] xv, double[] yv, double yp1, double ypn)
        {
            double[] u = new double[n - 1];
            if (yp1 > 0.99e99)
            {
                y2[0] = u[0] = 0.0;
            }
            else
            {
                y2[0] = -0.5;
                u[0] = (3.0 / (xv[1] - xv[0])) * ((yv[1] - yv[0]) / (xv[1] - xv[0]) - yp1);
            }
            for (int i = 1; i < n - 1; i++)
            {
                double sig = (xv[i] - xv[i - 1]) / (xv[i + 1] - xv[i - 1]);
                double p = sig * y2[i - 1] + 2.0;
                y2[i] = (sig - 1.0) / p;
                u[i] = (yv[i + 1] - yv[i]) / (xv[i + 1] - xv[i]) - (yv[i] - yv[i - 1]) / (xv[i] - xv[i - 1]);
                u[i] = (6.0 * u[i] / (xv[i + 1] - xv[i - 1]) - sig * u[i - 1]) / p;
            }
            double qn;
            double un;
            if (ypn > 0.99e99)
            {
                qn = un = 0.0;
            }
            else
            {
                qn = 0.5;
                un = (3.0 / (xv[n - 1] - xv[n - 2])) * (ypn - (yv[n - 1] - yv[n - 2]) / (xv[n - 1] - xv[n - 2]));
            }
            y2[n - 1] = (un - qn * u[n - 2]) / (qn * y2[n - 2] + 1.0);
            for (int k = n - 2; k >= 0; k--)
            {
                y2[k] = y2[k] * y2[k + 1] + u[k];
            }
        }

        /// <summary>
        /// Given a value x, and using pointers to data xx and yy, this routine returns
        /// an interpolated value y, and stores an error estimate dy. The returned
        /// value is obtained by mm-point polynomial interpolation on the subrange
        /// xx[jl..jl + mm - 1].
        /// </summary>
        /// <param name="jl"></param>
        /// <param name="x"></param>
        /// <returns></returns>
        /// <exception cref="Exception"></exception>
        public override double rawinterp(int jl, double x)
        {
            int klo = jl;
            int khi = jl + 1;
            double h = xx[khi] - xx[klo];
            //if (h == 0.0)
            if (Math.Abs(h) <= float.Epsilon)
            {
                throw new Exception("Bad input to routine splint");
            }
            double a = (xx[khi] - x) / h;
            double b = (x - xx[klo]) / h;
            double y = a * yy[klo] + b * yy[khi] + ((a * a * a - a) * y2[klo] + (b * b * b - b) * y2[khi]) * (h * h) / 6.0;
            return y;
        }
    }
}
 

2 代码格式

using System;namespace Legalsoft.Truffer
{/// <summary>/// 三次样条插值/// Cubic Spline Interpolation/// Cubic spline interpolation object. Construct with x and y vectors, and/// (optionally) values of the first derivative at the endpoints, then call/// interp for interpolated values/// </summary>public class Spline_interp : Base_interp{private double[] y2 { get; set; }public Spline_interp(double[] xv, double[] yv, double yp1 = 1.0e99, double ypn = 1.0e99) : base(xv, yv[0], 2){this.y2 = new double[xv.Length];sety2(xv, yv, yp1, ypn);}public Spline_interp(double[] xv, double yv, double yp1 = 1.0e99, double ypn = 1.0e99) : base(xv, yv, 2){this.y2 = new double[xv.Length];sety2(xv, y2, yp1, ypn);}/// <summary>/// This routine stores an array y2[0..n - 1] with second derivatives of the/// interpolating function at the tabulated points pointed to by xv, using/// function values pointed to by yv.If yp1 and/or ypn are equal to 1.0E99 or/// larger, the routine is signaled to set the corresponding boundary condition/// for a natural spline, with zero second derivative on that boundary;/// otherwise, they are the values of the first derivatives at the endpoints./// </summary>/// <param name="xv"></param>/// <param name="yv"></param>/// <param name="yp1"></param>/// <param name="ypn"></param>public void sety2(double[] xv, double[] yv, double yp1, double ypn){double[] u = new double[n - 1];if (yp1 > 0.99e99){y2[0] = u[0] = 0.0;}else{y2[0] = -0.5;u[0] = (3.0 / (xv[1] - xv[0])) * ((yv[1] - yv[0]) / (xv[1] - xv[0]) - yp1);}for (int i = 1; i < n - 1; i++){double sig = (xv[i] - xv[i - 1]) / (xv[i + 1] - xv[i - 1]);double p = sig * y2[i - 1] + 2.0;y2[i] = (sig - 1.0) / p;u[i] = (yv[i + 1] - yv[i]) / (xv[i + 1] - xv[i]) - (yv[i] - yv[i - 1]) / (xv[i] - xv[i - 1]);u[i] = (6.0 * u[i] / (xv[i + 1] - xv[i - 1]) - sig * u[i - 1]) / p;}double qn;double un;if (ypn > 0.99e99){qn = un = 0.0;}else{qn = 0.5;un = (3.0 / (xv[n - 1] - xv[n - 2])) * (ypn - (yv[n - 1] - yv[n - 2]) / (xv[n - 1] - xv[n - 2]));}y2[n - 1] = (un - qn * u[n - 2]) / (qn * y2[n - 2] + 1.0);for (int k = n - 2; k >= 0; k--){y2[k] = y2[k] * y2[k + 1] + u[k];}}/// <summary>/// Given a value x, and using pointers to data xx and yy, this routine returns/// an interpolated value y, and stores an error estimate dy. The returned/// value is obtained by mm-point polynomial interpolation on the subrange/// xx[jl..jl + mm - 1]./// </summary>/// <param name="jl"></param>/// <param name="x"></param>/// <returns></returns>/// <exception cref="Exception"></exception>public override double rawinterp(int jl, double x){int klo = jl;int khi = jl + 1;double h = xx[khi] - xx[klo];//if (h == 0.0)if (Math.Abs(h) <= float.Epsilon){throw new Exception("Bad input to routine splint");}double a = (xx[khi] - x) / h;double b = (x - xx[klo]) / h;double y = a * yy[klo] + b * yy[khi] + ((a * a * a - a) * y2[klo] + (b * b * b - b) * y2[khi]) * (h * h) / 6.0;return y;}}
}

这篇关于C#，数值计算——插值和外推，三次样条插值（Spline_interp）的计算方法与源程序的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---