如何转置矩阵

本文关键字：转置 何转置 | 更新日期: 2025-04-15 19:25:47

我已经用c#制作了8x8矩阵，现在我需要对矩阵进行转置。你能帮我把矩阵转置一下吗？这是我的矩阵

public double[,] MatriksT(int blok)
{
    double[,] matrixT = new double[blok, blok];
    for (int j = 0; j < blok ; j++)
    {
        matrixT[0, j] = Math.Sqrt(1 / (double)blok);
    }
    for (int i = 1; i < blok; i++)
    {
        for (int j = 0; j < blok; j++)
        {
            matrixT[i, j] = Math.Sqrt(2 / (double)blok) * Math.Cos(((2 * j + 1) * i * Math.PI) / 2 * blok);
        }
    }
    return matrixT;
}

如何转置矩阵

public double[,] Transpose(double[,] matrix)
{
    int w = matrix.GetLength(0);
    int h = matrix.GetLength(1);
    double[,] result = new double[h, w];
    for (int i = 0; i < w; i++)
    {
        for (int j = 0; j < h; j++)
        {
            result[j, i] = matrix[i, j];
        }
    }
    return result;
}

public class Matrix<T>
{
    public static T[,] TransposeMatrix(T[,] matrix)
    {
        var rows    = matrix.GetLength(0);
        var columns = matrix.GetLength(1);
        var result = new T[columns, rows];
        for (var c = 0; c < columns; c++)
        {
            for (var r = 0; r < rows; r++)
            {
                result[c, r] = matrix[r, c];
            }
        }
        return result;
    }
}

这就是它的名称：

int[,] matris = new int[5, 8]
        {
            {1  , 2 , 3 , 4 , 5 , 6 , 7 , 8 },
            {9  , 10, 11, 12, 13, 14, 15, 16},
            {17 , 18, 19, 20, 21, 22, 23, 24},
            {25 , 26, 27, 28, 29, 30, 31, 32},
            {33 , 34, 35, 36, 37, 38, 39, 40},
        };
var tMatrix = Matrix<int>.TransposeMatrix(matris);

您的矩阵不是一个"完美正方形"；一您需要一个结果（transposedMatris）矩阵。如果你的矩阵是一个完美的正方形，那么在同一个矩阵上交换（a[i，j]，a[j，i]）就足够了。

泛型类型类具有静态方法，该方法应按如下方式调用。

int[,] transposedMatris = Matrix.TransposeMatrix<int>(matris);
dumpMatrix(transposedMatris);

在控制台上转储转置矩阵。

// Using Generics (dump string, integer, decimal, string)
public static void dumpMatrix<T>(T[,] a)
{
    int m = a.GetLength(0);
    int n = a.GetLength(1);
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            Console.Write(a[i, j] + " ");
        }
        Console.WriteLine();
    }
    Console.WriteLine();
}

转置矩阵转储：

1 9 17 25 33 
2 10 18 26 34 
3 11 19 27 35 
4 12 20 28 36 
5 13 21 29 37 
6 14 22 30 38 
7 15 23 31 39 
8 16 24 32 40

我不是苏雷，但你需要按照以下步骤将图像顺时针旋转90度。

  • 步骤1：转座矩阵

  • 步骤2：水平翻转

     int[,] a = new int[5, 5] {
     {1, 2, 3, 4, 5},
     {6, 7, 8, 9, 10},
     {11, 12, 13, 14, 15},
     {16, 17, 18, 19, 20},
     {21, 22, 23, 24, 25}
 };
 int m = a.GetLength(0);
 int n = a.GetLength(1);
 //dumpMatrix(a);
 // Step 1: Transpose Matrix
 int temp = 0;
 for (int i = 0; i < m; i++)
 {
     for (int j = i; j < n; j++)
     {
         temp = a[i, j];
         a[i, j] = a[j, i];
         a[j, i] = temp;
     }
 }
 //Console.WriteLine("Step 1: Transpose Matrix");
 //dumpMatrix(a);
 // Step 2: Flip horizontally
 for (int i = 0; i < m; i++)
 {
     for (int j = 0; j < (n/2); j++)
     {
         temp = a[i, j];
         a[i, j] = a[i, n - 1 - j];
         a[i, n - 1 - j] = temp;
     }
 }
 //Console.WriteLine("Step 2: Flip horizontally");
 //dumpMatrix(a);
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