33

我正在尝试使用多维数组（[2][2]）制作一个简单的矩阵乘法方法。我对此有点陌生，我只是找不到我做错了什么。我真的很感激任何帮助告诉我它是什么。我宁愿不使用库或类似的东西，我主要是为了了解它是如何工作的。非常感谢你。

我在主要方法中声明我的数组如下：

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;
}
4

9 回答 9

34

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();
        }
    }
}
于 2013-07-12T21:22:50.327 回答
12

爪哇。矩阵乘法。

用不同大小的矩阵进行测试。

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
于 2014-05-22T22:01:35.050 回答
4 
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");
    }
}
于 2013-07-12T22:12:08.580 回答
2

尝试这个，

public static Double[][] multiplicar(Double A[][],Double B[][]){
    Double[][] C= new Double[2][2];
    int i,j,k;
     for (i = 0; i < 2; i++) {
         for (j = 0; j < 2; j++) {
             C[i][j] = 0.00000;
         }
     }
    for(i=0;i<2;i++){
        for(j=0;j<2;j++){
            for (k=0;k<2;k++){
                  C[i][j]+=(A[i][k]*B[k][j]);

            }

        }
    }
    return C;
}
于 2013-07-12T21:38:21.307 回答
1

方法mults是过程（Pascal）或子例程（Fortran）

该方法multMatrix是一个函数（Pascal，Fortran）

import java.util.*;

public class MatmultE
{
private static Scanner sc = new Scanner(System.in);
  public static void main(String [] args)
  {
    double[][] A={{4.00,3.00},{2.00,1.00}}; 
    double[][] B={{-0.500,1.500},{1.000,-2.0000}};
    double[][] C=multMatrix(A,B);
    printMatrix(A);
    printMatrix(B);    
    printMatrix(C);

    double a[][] = {{1, 2, -2, 0}, {-3, 4, 7, 2}, {6, 0, 3, 1}};
    double b[][] = {{-1, 3}, {0, 9}, {1, -11}, {4, -5}};
    double[][] c=multMatrix(a,b);
    printMatrix(a);
    printMatrix(b);    
    printMatrix(c);

    double[][] a1 = readMatrix();
    double[][] b1 = readMatrix();
    double[][] c1 = new double[a1.length][b1[0].length];
    mults(a1,b1,c1,a1.length,a1[0].length,b1.length,b1[0].length);
    printMatrix(c1);
    printMatrixE(c1);
  }

   public static double[][] readMatrix() {
       int rows = sc.nextInt();
       int cols = sc.nextInt();
       double[][] result = new double[rows][cols];
       for (int i = 0; i < rows; i++) {
           for (int j = 0; j < cols; j++) {
              result[i][j] = sc.nextDouble();
           }
       }
       return result;
   }


  public static void printMatrix(double[][] mat) {
  System.out.println("Matrix["+mat.length+"]["+mat[0].length+"]");
       int rows = mat.length;
       int columns = mat[0].length;
       for (int i = 0; i < rows; i++) {
           for (int j = 0; j < columns; j++) {
               System.out.printf("%8.3f " , mat[i][j]);
           }
           System.out.println();
       }
   System.out.println();
  }

  public static void printMatrixE(double[][] mat) {
  System.out.println("Matrix["+mat.length+"]["+mat[0].length+"]");
       int rows = mat.length;
       int columns = mat[0].length;
       for (int i = 0; i < rows; i++) {
           for (int j = 0; j < columns; j++) {
               System.out.printf("%9.2e " , mat[i][j]);
           }
           System.out.println();
       }
   System.out.println();
  }


   public static double[][] multMatrix(double a[][], double b[][]){//a[m][n], b[n][p]
   if(a.length == 0) return new double[0][0];
   if(a[0].length != b.length) return null; //invalid dims

   int n = a[0].length;
   int m = a.length;
   int p = b[0].length;

   double ans[][] = new double[m][p];

   for(int i = 0;i < m;i++){
      for(int j = 0;j < p;j++){
         ans[i][j]=0;
         for(int k = 0;k < n;k++){
            ans[i][j] += a[i][k] * b[k][j];
         }
      }
   }
   return ans;
   }

   public static void mults(double a[][], double b[][], double c[][], int r1, 
                        int c1, int r2, int c2){
      for(int i = 0;i < r1;i++){
         for(int j = 0;j < c2;j++){
            c[i][j]=0;
            for(int k = 0;k < c1;k++){
               c[i][j] += a[i][k] * b[k][j];
            }
         }
      }
   }
}

作为输入矩阵，您可以输入

在E.txt

4 4
1 1 1 1
2 4 8 16
3 9 27 81
4 16 64 256
4 3
4.0 -3.0 4.0
-13.0 19.0 -7.0
3.0 -2.0 7.0
-1.0 1.0 -1.0

在 unix 之类的 cmmd 行中执行命令：

$ java MatmultE < inE.txt > outE.txt

你得到输出

outC.txt

Matrix[2][2]
   4.000    3.000 
   2.000    1.000 

Matrix[2][2]
  -0.500    1.500 
   1.000   -2.000 

Matrix[2][2]
   1.000    0.000 
   0.000    1.000 

Matrix[3][4]
   1.000    2.000   -2.000    0.000 
  -3.000    4.000    7.000    2.000 
   6.000    0.000    3.000    1.000 

Matrix[4][2]
  -1.000    3.000 
   0.000    9.000 
   1.000  -11.000 
   4.000   -5.000 

Matrix[3][2]
  -3.000   43.000 
  18.000  -60.000 
   1.000  -20.000 

Matrix[4][3]
  -7.000   15.000    3.000 
 -36.000   70.000   20.000 
-105.000  189.000   57.000 
-256.000  420.000   96.000 

Matrix[4][3]
-7.00e+00  1.50e+01  3.00e+00 
-3.60e+01  7.00e+01  2.00e+01 
-1.05e+02  1.89e+02  5.70e+01 
-2.56e+02  4.20e+02  9.60e+01
于 2016-03-14T12:06:54.287 回答
1

试试这个，它可以帮助你

import java.util.Scanner;

public class MulTwoArray {

public static void main(String[] args) {

    int i, j, k;
    int[][] a = new int[3][3];
    int[][] b = new int[3][3];
    int[][] c = new int[3][3];



    Scanner sc = new Scanner(System.in);

    System.out.println("Enter size of array a");
    int rowa = sc.nextInt();
    int cola = sc.nextInt();

    System.out.println("Enter size of array b");
    int rowb = sc.nextInt();
    int colb = sc.nextInt();


    //read and b

    System.out.println("Enter elements of array a");
    for (i = 0; i < rowa; ++i) {
        for (j = 0; j < cola; ++j) {

            a[i][j] = sc.nextInt();

        }
        System.out.println();
    }
    System.out.println("Enter elements of array b");
    for (i = 0; i < rowb; ++i) {
        for (j = 0; j < colb; ++j) {

            b[i][j] = sc.nextInt();

        }
        System.out.println("\n");
    }

    //print a and b

    System.out.println("the elements of array a");
    for (i = 0; i < rowa; ++i) {
        for (j = 0; j < cola; ++j) {

            System.out.print(a[i][j]);
            System.out.print("\t");

        }
        System.out.println("\n");
    }
    System.out.println("the elements of array b");
    for (i = 0; i < rowb; ++i) {
        for (j = 0; j < colb; ++j) {

            System.out.print(b[i][j]);
            System.out.print("\t");

        }
        System.out.println("\n");

    }

    //multiply a and b

    for (i = 0; i < rowa; ++i) {

        for (j = 0; j < colb; ++j) {
            c[i][j] = 0;
            for (k = 0; k < cola; ++k) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    }


    //print multi result

    System.out.println("result of multiplication of array a and b is ");
    for (i = 0; i < rowa; ++i) {
        for (j = 0; j < colb; ++j) {

            System.out.print(c[i][j]);
            System.out.print("\t");
        }
        System.out.println("\n");
    }
}
}
于 2015-08-28T14:03:28.080 回答
0

导入 java.util.*；公共类 Mult {

public static int[][] C;

public static void main(String[] args) {

    Scanner s = new Scanner(System.in);

    System.out.println("Enter Row of Matrix A");
    int Rowa = s.nextInt();

    System.out.println("Enter Column of Matrix A");
    int Cola = s.nextInt();

    System.out.println("Enter Row of Matrix B");
    int Rowb = s.nextInt();

    System.out.println("Enter Column of Matrix B");
    int Colb = s.nextInt();

    int[][] A = new int[Rowa][Cola];
    int[][] B = new int[Rowb][Colb];

    C= new int[Rowa][Colb];
    //int[][] C = new int;
    System.out.println("Enter Values of Matrix A");

    for(int i =0 ; i< A.length ; i++) {
        for(int j = 0 ; j<A.length;j++) {
            A[i][j] = s.nextInt();
        }

    }

    System.out.println("Enter Values of Matrix B");

    for(int i =0 ; i< B.length ; i++) {
        for(int j = 0 ; j<B.length;j++) {
            B[i][j] = s.nextInt();
        }
    }




    if(Cola==Rowb) {
        for(int i = 0;i < A.length;i++){
              for(int j = 0;j < A.length;j++){
                 C[i][j]=0;
                 for(int k = 0;k < B.length;k++){
                    C[i][j] += A[i][k] * B[k][j];
                 }
              }
           }
    }
    else {
        System.out.println("Cannot multiply");
    }









    // Printing matrix A
    /*
    for(int i =0 ; i< A.length ; i++) {
        for(int j = 0 ; j<A.length;j++) {
        System.out.print(A[i][j]+ "\t");
        }
        System.out.println();
    }
    */

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

}

}

于 2018-11-17T17:15:56.510 回答
0

我的代码非常简单，适用于任何矩阵顺序

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    System.out.println(" Enter No. of rows in matrix 1 : ");
    int arows = sc.nextInt();
    System.out.println(" Enter No. of columns in matrix 1 : ");
    int acols = sc.nextInt();
    System.out.println(" Enter No. of rows in matrix 2 : ");
    int brows = sc.nextInt();
    System.out.println(" Enter No. of columns in matrix 2 : ");
    int bcols = sc.nextInt();
    if (acols == brows) {
        System.out.println(" Enter elements of matrix 1 ");
        int a[][] = new int[arows][acols];
        int b[][] = new int[brows][bcols];
        for (int i = 0; i < arows; i++) {
            for (int j = 0; j < acols; j++) {
                a[i][j] = sc.nextInt();
            }
        }
        System.out.println(" Enter elements of matrix 2 ");
        for (int i = 0; i < brows; i++) {
            for (int j = 0; j < bcols; j++) {
                b[i][j] = sc.nextInt();
            }
        }
        System.out.println(" The Multiplied matrix is : ");
        int sum = 0;
        int c[][] = new int[arows][bcols];
        for (int i = 0; i < arows; i++) {
            for (int j = 0; j < bcols; j++) {
                for (int k = 0; k < brows; k++) {
                    sum = sum + a[i][k] * b[k][j];
                    c[i][j] = sum;
                }

                System.out.print(c[i][j] + " ");
                sum = 0;
            }
            System.out.println();
        }
    } else {
        System.out.println("Order of matrix in invalid");
    }
}
于 2020-10-16T16:08:04.730 回答
0

乘以 4x4 矩阵

float[] mul(float[] l, float[] r) {
    float[] res = new float[16];
    for (int i = 0; i < 16; i++) {
        int y = i / 4;
        int x = i % 4;
        res[i] = l[x]      * r[y] +
                 l[x + 4]  * r[y + 4] +
                 l[x + 8]  * r[y + 8] +
                 l[x + 12] * r[y + 12];
    }
    return res;
}
于 2021-07-28T20:20:22.643 回答