如何转置非方阵？

Question

I have this code for transposing a matrix:

for(j=0; j<col; j++) { 
    for(k=0; k<row; k++) {
        mat2[j][k] = mat[k][j];
    }

It seems to work on a square matrix but not on a nonsquare matrix. Help me!

Answer 1

It will work for a non-square matrix, but you have to ensure that the number of rows in mat2 matches the number of columns in mat , and vice versa. Ie, if mat is an NxM matrix, then mat2 must be an MxN matrix.

Answer 2

Is that the actual code you have used in your application? Because it's wrong.
The syntax for the for statement is:

for (Initialization; Condition to continue with the loop; Step Operation) {}

In your case you should use something like this:

#define COLS 10
#define ROWS 5

int mat[COLS][ROWS];
int mat2[ROWS][COLS];

int i, j;

for (i = 0; i < COLS; i ++) {
    for (j = 0; j < ROWS; j++) {
        mat2[j][i] = mat[i][j];
    }
}

This way this could transpose your matrix. Naturally this way you need to know beforehand the dimensions for the matrix. Another way could be to dinamically initialize your matrix by using some User provided data, like this:

int ** mat;
int ** mat2;

int cols, rows;
int i, j;

/* Get matrix dimension from the user */

mat = (int **) malloc (sizeof(int *) * cols);

for (i = 0; i < cols; i++) {
    mat[i] = (int *) malloc (sizeof(int) * rows);
}

This way you'll dinamically initialize a matrix and then you can transpose it the same way as before.

Answer 3

Code for the transpose of a non square matrix with predefined dimensions in c++ would look something as follows :

#include <iostream>

using namespace std;

int main()
{
int a[7][6],b[6][7],x,i,j,k;
/*Input Matrix from user*/
cout<<"Enter elements for matrix A\n";
for(i=0;i<7;i++)
    {for(j=0;j<6;j++)
       {
         cin>>a[i][j];
       }
    }
 /*Print Input Matrix A*/
 cout<<"MATRIX A:-"<<endl;
 for(i=0;i<7;i++)
    {for(j=0;j<6;j++)
       {
         cout<<a[i][j];
       }
       cout<<endl;
    }
/*calculate TRANSPOSE*/
for(i=0;i<7;i++)
    {for(j=0,k=5;j<6;j++,k--)
        {
         x=j+(k-j);
         b[x][i]=a[i][j];
        }
    }
/*Print Output Matrix B*/
cout<<"matrix B:-\n";
for(i=0;i<6;i++)
  {for(j=0;j<7;j++)
   {
       cout<<b[i][j];
   }
   cout<<endl;
  }
}

You can replace with the appropriate header and print/scan statements to write the code in C and alternatively take input from the user and implement the same.

Answer 4

If col is the number of rows in mat2 (and columns in mat) and row is the number of columns in mat2 (and rows in mat), then this should work.

You need an extra close curly, but I assume you have that in your code.

Answer 5

  #include<conio.h>
  #include<ctype.h>
 #include<iostream.h>
 void trans(int [][10], int , int );
 int main()
 {
     int a[10][10],n,m,i,j;
  clrscr();
  cout<<"Enter the no. of rows and Columns: ";
  cin>>n>>m;
  cout<<"\n Enter the Matrix now:";
  for(i=0;i<n;i++)
   for(j=0;j<m;j++)
     cin>>a[i][j];  
  trans(a,n,m);
  getch();
  return 0;
 }
 void trans( int a[][10] , int n , int m )
 {  
   int i,b[10][10],j;
    for(i=0;i<n;i++)
     for(j=0;j<m;j++)
      { 
       b[j][i]=a[i][j];
      }
     cout<<"\n\nSize before Transpose "<<n<<"x"<<m<<"\n\n";
     for(i=0;i<m;i++)
       {
         for(j=0;j<n;j++)
          cout<<b[i][j]<<"\t";
     cout<<"\n";
      }
  cout<<"\n\nSize after transpose "<<m<<"x"<<n;  
   }

Answer 6

#include<conio.h>
#include<stdio.h>
main()
    int a[5][5],i,j,t;
    clrscr();
    for(i=1;i<=4;i++)
    { 
        for(j=1;j<=5;j++)
            scanf("%d",&a[i][j]);
    }

    for(i=1;i<=4;i++)
    { 
        for(j=i;j<=5;j++)
        {  
            if(i<=4&&j<=4)
            {
                t=a[j][i];
                a[j][i]=a[i][j];
                a[i][j]=t;
            }
            else
            a[j][i]=a[i][j];
        }
    }

    for(i=1;i<=5;i++)
    {
        for(j=1;j<=4;j++)
        {
            printf("%d ",a[i][j]);
        }
         printf("\n");
    }

    getch();
}