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。
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;
}
答案 0 :(得分：28)
您可以尝试以下代码：
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();
}
}
}
答案 1 :(得分：11)
<强>爪哇。矩阵乘法。
使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)));
}
}
<强>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
答案 2 :(得分：5)
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");
}
}
答案 3 :(得分：2)
试试这个，
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;
}
答案 4 :(得分：1)
试试这个，它可能有所帮助
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");
}
}
}
答案 5 :(得分：1)
方法mults是一个过程（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];
}
}
}
}
}
作为输入矩阵，您可以输入
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，如cmmd行执行命令：
$ java MatmultE＆lt; inE.txt＆gt; 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
答案 6 :(得分：0)
导入java.util。*; 公共课程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();
}
}
}
答案 7 :(得分：0)
我的代码非常简单，可以处理任意阶数的矩阵
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");
}
}
答案 8 :(得分：0)
使用 For 循环
import java.util.Scanner;
public class MatrixMultiplicationExample{
public static void main(String args[]){
int row1, col1, row2, col2;
Scanner s = new Scanner(System.in);
System.out.print("Enter number of rows in first matrix:");
row1 = s.nextInt();
System.out.print("Enter number of columns in first matrix:");
col1 = s.nextInt();
System.out.print("Enter number of rows in second matrix:");
row2 = s.nextInt();
System.out.print("Enter number of columns in second matrix:");
col2 = s.nextInt();
if (col1 != row2) {
System.out.println("Matrix multiplication is not possible");
}
else {
int a[][] = new int[row1][col1];
int b[][] = new int[row2][col2];
int c[][] = new int[row1][col2];
System.out.println("Enter values for matrix A : \n");
for (int i = 0; i < row1; i++) {
for (int j = 0; j < col1; j++)
a[i][j] = s.nextInt();
}
System.out.println("Enter values for matrix B : \n");
for (int i = 0; i < row2; i++) {
for (int j = 0; j < col2; j++)
b[i][j] = s.nextInt();
}
System.out.println("Matrix multiplication is : \n");
for(int i = 0; i < row1; i++) {
for(int j = 0; j < col2; j++){
c[i][j]=0;
for(int k = 0; k < col1; k++){
c[i][j] += a[i][k] * b[k][j];
}
System.out.print(c[i][j] + " ");
}
System.out.println();
}
}
}
}
程序输出
Enter number of rows in first matrix:2
Enter number of columns in first matrix:2
Enter number of rows in second matrix:2
Enter number of columns in second matrix:2
Enter values for matrix A :
3 4
2 1
Enter values for matrix B :
1 5
3 7
Matrix multiplication is :
15 43
5 17
答案 9 :(得分：0)
将 4x4 矩阵相乘
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;
}