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

{

/// <summary>

/// 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.

/// </summary>

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.xxi) / (x1terp.xxi + 1 - x1terp.xxi);

double u = (x2p - x2terp.xxj) / (x2terp.xxj + 1 - x2terp.xxj);

double yy = (1.0 - t) * (1.0 - u) * yi, j + t * (1.0 - u) * yi + 1, j + (1.0 - t) * u * yi, j + 1 + t * u * yi + 1, j + 1;

return yy;

}

}

}

2 代码格式

cs 复制代码
using System;

namespace Legalsoft.Truffer
{
    /// <summary>
    /// 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.
    /// </summary>
    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;
        }
    }
}
相关推荐
huangdong_31 分钟前
1688商品图片采集技术解析:登录态处理与SKU图自动分类
开发语言
youngerwang40 分钟前
【从搬运工到协处理器:网卡芯片架构、算法、验证与边缘演进深度剖析】
网络·算法·架构·芯片
chase_my_dream43 分钟前
C++ + SLAM 高频面试问题整理
开发语言·c++·面试
KaMeidebaby1 小时前
卡梅德生物技术快报|纯化重组蛋白实操详解
人工智能·python·tcp/ip·算法·机器学习
Cloud_Shy6181 小时前
解读《Effective Python 3rd Edition》:从练气到老魔(第五章 Item 30 - 32)
开发语言·人工智能·笔记·python·学习方法
天佑木枫2 小时前
15天Python入门系列 · 序
开发语言·python
手写码匠2 小时前
从零实现 Prompt 工程引擎:结构化提示、自动优化与多轮自省体系
人工智能·深度学习·算法·aigc
无限码力2 小时前
阿里算法岗 0530笔试真题 - 多约束条件下的元素匹配统计
算法·阿里笔试真题·阿里机试真题·阿里算法岗笔试
lqqjuly2 小时前
MLA — 多头潜在注意力深度解析
深度学习·神经网络·算法
宋拾壹2 小时前
同时添加多个类目
android·开发语言·javascript