使用数组的矩阵乘法

我正在尝试使用多维数组（ [2][2] ）制作一个简单的矩阵乘法。 我对此有点新意，我无法找到它我做错了什么。 在告诉我它是什么时，我真的很感激。 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main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 

Answer 4

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; } 

Answer 5

方法mults是一个程序（Pascal）或子程序（Fortran）

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]; } } } } } 

您可以输入输入矩阵的位置

inE.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中像cmmd一行执行命令：

$ java MatmultE outE.txt

你得到了输出

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 

Answer 6

试试这个，它可能对你有帮助

 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"); } } }