1.1.33

上次更新:2019-04-17
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解答

这里矩阵使用交错数组实现(方便取行向量),不是普通的二维数组。

矩阵和矩阵、矩阵和向量、向量和矩阵都使用行向量点乘列向量的方式计算。

代码

public class Matrix
{
    /// <summary>
    /// 计算两个向量的点积
    /// </summary>
    /// <param name="x">需要点乘的向量</param>
    /// <param name="y">需要点乘的另一个向量</param>
    /// <returns>返回点乘的结果</returns>
    /// <exception cref="FormatException"></exception>
    public static double Dot(double[] x, double[] y)
    {
        //确保两向量等长
        if (x.Length != y.Length)
        {
            throw new FormatException("the length of two vectors must be equal");
        }

        //点乘
        double result = 0;
        for (int i = 0; i < x.Length; ++i)
        {
            result += x[i] * y[i];
        }

        return result;
    }

    /// <summary>
    /// 计算两个矩阵相乘的结果,返回一个矩阵
    /// </summary>
    /// <param name="a">用交错数组表示的 m * p 矩阵</param>
    /// <param name="b">用交错数组表示的 p * n 矩阵</param>
    /// <returns>返回 m * n 的矩阵</returns>
    /// <exception cref="FormatException"></exception>
    /// <example>
    ///     a = {(1,2,3),(4,5,6)}
    ///     b = {(1,4),(2,5),(3,6)}
    ///     Mult(a, b) = {(14,32),(32,77)}
    /// </example>
    public static double[][] Mult(double[][] a, double[][] b)
    {
        if (a[0].Length != b.GetLength(0))
        {
            throw new FormatException("a's column number must be equal to b's row number");
        }

        int m = a.GetLength(0);
        int n = b[0].Length;
        int p = a[0].Length;

        double[][] result = new double[m][];

        for (int i = 0; i < m; ++i)
        {
            double[] resultrow = new double[n];
            for (int j = 0; j < n; ++j)
            {
                //result[i][j] = 行向量 a[i] 与列向量 b[j] 的点积
                double[] row = a[i];
                double[] col = new double[p];
                //取得列向量
                for (int k = 0; k < p; ++k)
                {
                    col[k] = b[k][j];
                }
                //点积
                resultrow[j] = Dot(row, col);
            }
            result[i] = resultrow;
        }
        return result;
    }

    /// <summary>
    /// 将一个矩阵转置
    /// </summary>
    /// <param name="a">待转置的矩阵</param>
    /// <returns>返回转置后的数组</returns>
    public static double[][] Transpose(double[][] a)
    {
        double[][] trans = new double[a[0].Length][];
        for (int i = 0; i < a[0].Length; ++i)
        {
            double[] row = new double[a.GetLength(0)];
            for (int j = 0; j < a.GetLength(0); ++j)
            {
                row[j] = a[j][i];
            }
            trans[i] = row;
        }
        return trans;
    }

    /// <summary>
    /// 计算矩阵与向量的乘积
    /// </summary>
    /// <param name="a">左乘的矩阵</param>
    /// <param name="x">列向量</param>
    /// <returns>返回一个向量</returns>
    /// <exception cref="FormatException"></exception>
    public static double[] Mult(double[][] a, double[] x)
    {
        if (a[0].Length != x.Length)
        {
            throw new FormatException("a's column number must be equal to x's length");
        }

        double[] result = new double[a.GetLength(0)];

        for (int i = 0; i < a.GetLength(0); ++i)
        {
            result[i] = Dot(a[i], x);
        }

        return result;
    }

    /// <summary>
    /// 计算向量与矩阵的乘积
    /// </summary>
    /// <param name="x">行向量</param>
    /// <param name="a">矩阵</param>
    /// <returns>返回一个向量</returns>
    /// <exception cref="FormatException"></exception>
    public static double[] Mult(double[] x, double[][] a)
    {
        if (a.GetLength(0) != x.Length)
        {
            throw new FormatException("a's column number must be equal to x's length");
        }

        double[] result = new double[a[0].Length];

        for (int i = 0; i < a[0].Length; ++i)
        {
            double[] colVector = new double[a.GetLength(0)];
            for (int j = 0; j < colVector.Length; ++j)
            {
                colVector[j] = a[j][i];
            }
            result[i] = Dot(x, colVector);
        }

        return result;
    }

    /// <summary>
    /// 在控制台上输出矩阵
    /// </summary>
    /// <param name="a">需要输出的矩阵</param>
    public static void PrintMatrix(double[][] a)
    {
        for (int i = 0; i < a.GetLength(0); ++i)
        {
            for (int j = 0; j < a[i].Length; ++j)
            {
                Console.Write($"\t{a[i][j]}");
            }
            Console.Write("\n");
        }
    }

    /// <summary>
    /// 在控制台上输出一行向量
    /// </summary>
    /// <param name="a">需要输出的向量</param>
    public static void PrintVector(double[] a)
    {
        for (int i = 0; i < a.Length; ++i)
        {
            Console.Write($"\t{a[i]}");
        }
        Console.Write("\n");
    }
}
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