计算两个矩阵之差的程序

17 Mar 2025 | 4 分钟阅读

说明

在这个程序中，我们需要减去两个矩阵并打印结果矩阵。

矩阵相减

只有当矩阵 B 和矩阵 A 具有相同的维度，即行数和列数都相同时，才能从矩阵 A 中减去矩阵 B，反之亦然。不能将一个 2x3 的矩阵减去一个 3x2 的矩阵。两个矩阵的相减可以通过相减它们对应的元素来完成，如下所示：

Program to calculate the subtraction of 2 matrices

可以通过遍历第一个和第二个矩阵来执行两个矩阵的相减。计算它们对应元素之间的差值，并将结果存储在第三个矩阵中。

算法

声明并初始化两个二维数组 a 和 b。
计算数组 a 中的行数和列数（因为两个数组的维度相同），并将它们分别存储在变量 rows 和 cols 中。
声明另一个维度为 rows 和 columns 的数组 diff。
遍历数组 a 和 b，计算对应元素之间的差值
例如：a11 - b11 = diff11
显示数组 diff 的元素。

解决方案

Python

#Initialize matrix a
a = [
        [4, 5, 6],
        [3, 4, 1],
        [1, 2, 3]
    ];
 
#Initialize matrix b
b = [
          [2, 0, 3],
         [2, 3, 1],
         [1, 1, 1]
     ];
 
#Calculates number of rows and columns present in given matrix
rows = len(a);
cols = len(a[0]);
 
#Array diff will hold the result and is initialized with zeroes.
diff = [[0]*rows for i in range(cols)];
 
#Performs subtraction of matrices a and b. Store the result in matrix diff
for i in range(0, rows):
    for j in range(0, cols):
        diff[i][j] = a[i][j] - b[i][j];
 
print("Subtraction of two matrices: ");
for i in range(0, rows):
    for j in range(0, cols):
       print(diff[i][j]),
 
    print(" ");

输出

Subtraction of two matrices: 
2 5 3 
1 1 0 
0 1 2

C

#include <stdio.h>
 
int main()
{
    int rows, cols;
        
    //Initialize matrix a
    int a[][3] = {
                    {4, 5, 6},
                    {3, 4, 1},
                    {1, 2, 3}
                 };
    
    //Initialize matrix b
    int b[][3] = {
                      {2, 0, 3},
                     {2, 3, 1},
                     {1, 1, 1}
                 };
    
    //Calculates number of rows and columns present in given matrix
    rows = (sizeof(a)/sizeof(a[0]));
    cols = (sizeof(a)/sizeof(a[0][0]))/rows;
    
    //Array diff will hold the result
    int diff[rows][cols];
    
    //Performs subtraction of matrices a and b. Store the result in matrix diff
    for(int i = 0; i < rows; i++){
        for(int j = 0; j < cols; j++){
            diff[i][j] = a[i][j] - b[i][j];
        }
    }
    
    printf("Subtraction of two matrices: \n");
    for(int i = 0; i < rows; i++){
        for(int j = 0; j < cols; j++){
           printf("%d ", diff[i][j]);
        }
        printf("\n");
    }
 
    return 0;
}

输出

Subtraction of two matrices: 
2 5 3 
1 1 0 
0 1 2

JAVA

public class SubMatrix
{
    public static void main(String[] args) {
        int rows, cols;
        
        //Initialize matrix a
          int a[][] = {
                          {4, 5, 6},
                          {3, 4, 1},
                          {1, 2, 3}
                       };
          
          //Initialize matrix b
          int b[][] = {
                          {2, 0, 3},
                         {2, 3, 1},
                         {1, 1, 1}
                     };
          
          //Calculates number of rows and columns present in given matrix
          rows = a.length;
        cols = a[0].length;
        
          //Array diff will hold the result
        int diff[][] = new int[rows][cols];
        
        //Performs subtraction of matrices a and b. Store the result in matrix diff
        for(int i = 0; i < rows; i++){
            for(int j = 0; j < cols; j++){
                diff[i][j] = a[i][j] - b[i][j];
            }
        }
        
        System.out.println("Subtraction of two matrices: ");
        for(int i = 0; i < rows; i++){
            for(int j = 0; j < cols; j++){
               System.out.print(diff[i][j] + " ");
            }
            System.out.println();
        }
    }
}

输出

Subtraction of two matrices: 
2 5 3 
1 1 0 
0 1 2

C#

using System;
                    
public class SubMatrix
{
    public static void Main()
    {
        int rows, cols;
        
        //Initialize matrix a
          int[,] a = {
                          {4, 5, 6},
                          {3, 4, 1},
                          {1, 2, 3}
                     };
          
          //Initialize matrix b
          int[,] b = {
                          {2, 0, 3},
                         {2, 3, 1},
                         {1, 1, 1}
                    };
          
          //Calculates number of rows and columns present in given matrix
          rows = a.GetLength(0);
        cols = a.GetLength(1);
        
          //Array diff will hold the result
        int[,] diff = new int[rows, cols];
        
        //Performs subtraction of matrices a and b. Store the result in matrix diff
        for(int i = 0; i < rows; i++){
            for(int j = 0; j < cols; j++){
                diff[i,j] = a[i,j] - b[i,j];
            }
        }
        
        Console.WriteLine("Subtraction of two matrices: ");
        for(int i = 0; i < rows; i++){
            for(int j = 0; j < cols; j++){
               Console.Write(diff[i,j] + " ");
            }
            Console.WriteLine();
        }
    }
}

输出

Subtraction of two matrices: 
2 5 3 
1 1 0 
0 1 2

PHP

<!DOCTYPE html>
<html>
<body>
<?php
//Initialize matrix a
$a = array(
            array(4, 5, 6),
            array(3, 4, 1),
            array(1, 2, 3)
          );
 
//Initialize matrix b
$b = array(
              array(2, 0, 3),
             array(2, 3, 1),
             array(1, 1, 1)
           );
 
//Calculates number of rows and columns present in given matrix
$rows = count($a);
$cols = count($a[0]);
 
//Array diff will hold the result and initialize it with 0
$diff = array_fill(0, $cols, array_fill(0, $rows, 0));
 
//Performs subtraction of matrices a and b. Store the result in matrix diff
for($i = 0; $i < $rows; $i++){
    for($j = 0; $j < $cols; $j++){
        $diff[$i][$j] = $a[$i][$j] - $b[$i][$j];
    }
}
 
print("Subtraction of two matrices: <br>");
for($i = 0; $i < $rows; $i++){
    for($j = 0; $j < $cols; $j++){
       print($diff[$i][$j] . " ");
    }
    print("<br>");
}
?>
</body>
</html>

输出

Subtraction of two matrices: 
2 5 3 
1 1 0 
0 1 2

下一主题#

计算两个矩阵之差的程序

说明

矩阵相减

算法

解决方案

Python

C

JAVA

C#

PHP

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程序

计算两个矩阵之差的程序

说明

矩阵相减

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JAVA

C#

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