C# 多项式拟合函数

以下是使用C#实现多项式拟合函数的示例代码：

using System;
using System.Collections.Generic;
using System.Linq;

namespace PolynomialFitting
{
    class Program
    {
        static void Main(string[] args)
        {
            // 定义数据点
            List<double> x = new List<double> { 1, 2, 3, 4, 5 };
            List<double> y = new List<double> { 1, 4, 9, 16, 25 };

            // 多项式拟合
            int degree = 2; // 多项式次数
            var coefficients = FitPolynomial(x, y, degree);

            // 输出拟合结果
            Console.WriteLine("拟合结果：");
            for (int i = 0; i < coefficients.Length; i++)
            {
                Console.WriteLine($"A[{i}] = {coefficients[i]}");
            }

            // 预测新数据
            double newX = 6;
            double newY = Predict(newX, coefficients);
            Console.WriteLine($"x = {newX}, y = {newY}");
        }

        // 多项式拟合函数
        static double[] FitPolynomial(List<double> x, List<double> y, int degree)
        {
            int n = x.Count;
            int m = degree + 1;
            double[,] X = new double[n, m];
            double[] Y = new double[n];

            // 构造矩阵X和向量Y
            for (int i = 0; i < n; i++)
            {
                double xi = x[i];
                double yi = y[i];
                for (int j = 0; j < m; j++)
                {
                    X[i, j] = Math.Pow(xi, j);
                }
                Y[i] = yi;
            }

            // 求解线性方程组
            var XTranspose = Transpose(X);
            var XTX = Multiply(XTranspose, X);
            var XTY = Multiply(XTranspose, Y);
            var coefficients = Solve(XTX, XTY);

            return coefficients;
        }

        // 矩阵转置
        static double[,] Transpose(double[,] matrix)
        {
            int rows = matrix.GetLength(0);
            int cols = matrix.GetLength(1);
            double[,] result = new double[cols, rows];
            for (int i = 0; i < rows; i++)
            {
                for (int j = 0; j < cols; j++)
                {
                    result[j, i] = matrix[i, j];
                }
            }
            return result;
        }

        // 矩阵乘法
        static double[,] Multiply(double[,] matrix1, double[,] matrix2)
        {
            int rows1 = matrix1.GetLength(0);
            int cols1 = matrix1.GetLength(1);
            int rows2 = matrix2.GetLength(0);
            int cols2 = matrix2.GetLength(1);
            if (cols1 != rows2)
            {
                throw new ArgumentException("矩阵维度不匹配");
            }
            double[,] result = new double[rows1, cols2];
            for (int i = 0; i < rows1; i++)
            {
                for (int j = 0; j < cols2; j++)
                {
                    double sum = 0;
                    for (int k = 0; k < cols1; k++)
                    {
                        sum += matrix1[i, k] * matrix2[k, j];
                    }
                    result[i, j] = sum;
                }
            }
            return result;
        }

        // 解线性方程组
        static double[] Solve(double[,] A, double[] b)
        {
            int n = b.Length;
            double[,] Ab = new double[n, n + 1];
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    Ab[i, j] = A[i, j];
                }
                Ab[i, n] = b[i];
            }
            for (int i = 0; i < n; i++)
            {
                // 消元
                double pivot = Ab[i, i];
                for (int j = i + 1; j <= n; j++)
                {
                    Ab[i, j] /= pivot;
                }
                for (int j = i + 1; j < n; j++)
                {
                    double factor = Ab[j, i];
                    for (int k = i + 1; k <= n; k++)
                    {
                        Ab[j, k] -= factor * Ab[i, k];
                    }
                }
            }
            // 回代
            double[] x = new double[n];
            for (int i = n - 1; i >= 0; i--)
            {
                double sum = 0;
                for (int j = i + 1; j < n; j++)
                {
                    sum += Ab[i, j] * x[j];
                }
                x[i] = Ab[i, n] - sum;
            }
            return x;
        }

        // 预测新数据
        static double Predict(double x, double[] coefficients)
        {
            double y = 0;
            for (int i = 0; i < coefficients.Length; i++)
            {
                y += coefficients[i] * Math.Pow(x, i);
            }
            return y;
        }
    }
}