使用数组的矩阵乘法

Question

我正在尝试使用多维数组([2][2])进行简单的矩阵乘法.我对此有点新意见,而我却无法找到它我做错了什么.我非常感谢能告诉我它是什么的任何帮助.我宁愿不使用库或类似的东西,我主要是这样做以了解它是如何工作的.非常感谢你提前.

我在主方法中声明我的arays如下:

Double[][] A={{4.00,3.00},{2.00,1.00}}; 
Double[][] B={{-0.500,1.500},{1.000,-2.0000}};

A*B应该返回单位矩阵.它没有.

public static Double[][] multiplicar(Double[][] A, Double[][] B){
//the method runs and returns a matrix of the correct dimensions
//(I actually changed the .length function to a specific value to eliminate 
//it as a possible issue), but not the correct values

    Double[][] C= new Double[2][2];
    int i,j;

    ////I fill the matrix with zeroes, if I don't do this it gives me an error
    for(i=0;i<2;i++) {
        for(j=0;j<2;j++){
            C[i][j]=0.00000;
        }
    } 
    ///this is where I'm supposed to perform the adding of every element in
    //a row of A multiplied by the corresponding element in the
    //corresponding column of B, for all columns in B and all rows in A
    for(i=0;i<2;i++){
        for(j=0;j<2;j++)
            C[i][j]+=(A[i][j]*B[j][i]);
    }
    return C;
}

Answer 1

你可以试试这段代码:

public class MyMatrix {
    Double[][] A = { { 4.00, 3.00 }, { 2.00, 1.00 } };
    Double[][] B = { { -0.500, 1.500 }, { 1.000, -2.0000 } };

    public static Double[][] multiplicar(Double[][] A, Double[][] B) {

        int aRows = A.length;
        int aColumns = A[0].length;
        int bRows = B.length;
        int bColumns = B[0].length;

        if (aColumns != bRows) {
            throw new IllegalArgumentException("A:Rows: " + aColumns + " did not match B:Columns " + bRows + ".");
        }

        Double[][] C = new Double[aRows][bColumns];
        for (int i = 0; i < aRows; i++) {
            for (int j = 0; j < bColumns; j++) {
                C[i][j] = 0.00000;
            }
        }

        for (int i = 0; i < aRows; i++) { // aRow
            for (int j = 0; j < bColumns; j++) { // bColumn
                for (int k = 0; k < aColumns; k++) { // aColumn
                    C[i][j] += A[i][k] * B[k][j];
                }
            }
        }

        return C;
    }

    public static void main(String[] args) {

        MyMatrix matrix = new MyMatrix();
        Double[][] result = multiplicar(matrix.A, matrix.B);

        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++)
                System.out.print(result[i][j] + " ");
            System.out.println();
        }
    }
}

Answer 2

Java的.矩阵乘法.

测试不同大小的矩阵.

public class Matrix {

/**
 * Matrix multiplication method.
 * @param m1 Multiplicand
 * @param m2 Multiplier
 * @return Product
 */
    public static double[][] multiplyByMatrix(double[][] m1, double[][] m2) {
        int m1ColLength = m1[0].length; // m1 columns length
        int m2RowLength = m2.length;    // m2 rows length
        if(m1ColLength != m2RowLength) return null; // matrix multiplication is not possible
        int mRRowLength = m1.length;    // m result rows length
        int mRColLength = m2[0].length; // m result columns length
        double[][] mResult = new double[mRRowLength][mRColLength];
        for(int i = 0; i < mRRowLength; i++) {         // rows from m1
            for(int j = 0; j < mRColLength; j++) {     // columns from m2
                for(int k = 0; k < m1ColLength; k++) { // columns from m1
                    mResult[i][j] += m1[i][k] * m2[k][j];
                }
            }
        }
        return mResult;
    }

    public static String toString(double[][] m) {
        String result = "";
        for(int i = 0; i < m.length; i++) {
            for(int j = 0; j < m[i].length; j++) {
                result += String.format("%11.2f", m[i][j]);
            }
            result += "\n";
        }
        return result;
    }

    public static void main(String[] args) {
        // #1
        double[][] multiplicand = new double[][] {
                {3, -1, 2},
                {2,  0, 1},
                {1,  2, 1}
        };
        double[][] multiplier = new double[][] {
                {2, -1, 1},
                {0, -2, 3},
                {3,  0, 1}
        };
        System.out.println("#1\n" + toString(multiplyByMatrix(multiplicand, multiplier)));
        // #2
        multiplicand = new double[][] {
                {1, 2, 0},
                {-1, 3, 1},
                {2, -2, 1}
        };
        multiplier = new double[][] {
                {2},
                {-1},
                {1}
        };
        System.out.println("#2\n" + toString(multiplyByMatrix(multiplicand, multiplier)));
        // #3
        multiplicand = new double[][] {
                {1, 2, -1},
                {0,  1, 0}
        };
        multiplier = new double[][] {
                {1, 1, 0, 0},
                {0, 2, 1, 1},
                {1, 1, 2, 2}
        };
        System.out.println("#3\n" + toString(multiplyByMatrix(multiplicand, multiplier)));
    }
}

输出:

#1
      12.00      -1.00       2.00
       7.00      -2.00       3.00
       5.00      -5.00       8.00

#2
       0.00
      -4.00
       7.00

#3
       0.00       4.00       0.00       0.00
       0.00       2.00       1.00       1.00

Answer 3

static int b[][]={{21,21},{22,22}};

static int a[][] ={{1,1},{2,2}};

public static void mul(){
    int c[][] = new int[2][2];

    for(int i=0;i<b.length;i++){
        for(int j=0;j<b.length;j++){
            c[i][j] =0;
        }   
    }

    for(int i=0;i<a.length;i++){
        for(int j=0;j<b.length;j++){
            for(int k=0;k<b.length;k++){
            c[i][j]= c[i][j] +(a[i][k] * b[k][j]);
            }
        }
    }

    for(int i=0;i<c.length;i++){
        for(int j=0;j<c.length;j++){
            System.out.print(c[i][j]);
        }   
        System.out.println("\n");
    }
}