How to rearrange a two dimensional array - c

Question

I want to assign the values in one array to another, but in a different
order. For example the original array would be:
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
And the desired array is:
8 11 14 3 6 16 19
9 12 1 4 7 17 20
10 13 2 5 15 18 21
How would I do this? In the code below I moved a selected row to the center, then assigned each array individually. I can't figure out a better way.
int main()
{
int decide, i, j, k;
int save[3][7];
int shuffle[3][7];
int arr[3][7] = {
{1, 2, 3, 4, 5, 6, 7},
{8, 9, 10, 11, 12, 13, 14},
{15, 16, 17, 18, 19, 20, 21}
};
printf("Select a number between 1 and 21\n\n0-GroupA (1 - 7))\n1-GroupB ( 8 - 14)\n2-GroupC (15 - 21)\n\n");
for(k = 0 ; k < 1 ; ++k)
{
printf("Select which group the number falls into:");
scanf("%d", &decide);
for(j = 0 ; j < 7 ; ++j)
{/*moves selected row to middle row of array*/
save[0][j] = arr[1][j];
arr[1][j] = arr[decide][j];
arr[decide][j] = save[0][j];
}
/*this is a horrible method*/
shuffle[0][0] = arr[0][0];
shuffle[1][0] = arr[0][1];
shuffle[2][0] = arr[0][2];
shuffle[0][1] = arr[0][3];
shuffle[1][1] = arr[0][4];
shuffle[2][1] = arr[0][5];
shuffle[0][2] = arr[0][6];
shuffle[1][2] = arr[1][0];
shuffle[2][2] = arr[1][1];
shuffle[0][3] = arr[1][2];
shuffle[1][3] = arr[1][3];
shuffle[2][3] = arr[1][4];
shuffle[0][4] = arr[1][5];
shuffle[1][4] = arr[1][6];
shuffle[2][4] = arr[2][0];
shuffle[0][5] = arr[2][1];
shuffle[1][5] = arr[2][2];
shuffle[2][5] = arr[2][3];
shuffle[0][6] = arr[2][4];
shuffle[1][6] = arr[2][5];
shuffle[2][6] = arr[2][6];
}

Answer 1

like this ?
#include <stdio.h>
#define ROWS 3
#define COLS 7
int main(void){
int arr[ROWS][COLS] = {
{1, 2, 3, 4, 5, 6, 7},
{8, 9, 10, 11, 12, 13, 14},
{15, 16, 17, 18, 19, 20, 21}
};
int result[ROWS][COLS];
int order[ROWS] = {1, 0, 2};
int n = sizeof(arr)/sizeof(**arr);//number of elements
int r = 0, c = 0, i = 0, j = 0;
while(n--){
result[r++][c] = arr[order[i]][j++];
if(r == ROWS){
r = 0;
c += 1;
}
if(j == COLS){
j = 0;
i += 1;
}
}
//check print
for(r = 0; r < ROWS; ++r){
for(c = 0; c < COLS; ++c)
printf("%-2d ", result[r][c]);
puts("");
}
}

Answer 2

The transposition part can be done like this:
#include <stdio.h>
typedef struct Sequence2D
{
int idx_slow;
int idx_fast;
int max_slow;
int max_fast;
} Sequence2D;
static inline void next2D(Sequence2D *seq)
{
if (++seq->idx_fast == seq->max_fast)
{
seq->idx_fast = 0;
if (++seq->idx_slow == seq->max_slow)
seq->idx_slow = 0;
}
}
static inline void dump2D(const char *tag, int rows, int cols, int data[rows][cols])
{
printf("%s:\n", tag);
for (int r = 0; r < rows; r++)
{
for (int c = 0; c < cols; c++)
printf("%3d", data[r][c]);
putchar('\n');
}
}
enum { ROWS = 3, COLS = 7 };
int main(void)
{
int src[ROWS][COLS] =
{
{ 1, 2, 3, 4, 5, 6, 7 },
{ 8, 9, 10, 11, 12, 13, 14 },
{ 15, 16, 17, 18, 19, 20, 21 },
};
int dst[ROWS][COLS];
Sequence2D dst2D = { 0, 0, COLS, ROWS };
Sequence2D src2D = { 0, 0, ROWS, COLS };
dump2D("Source", ROWS, COLS, src);
for (int i = 0; i < ROWS * COLS; i++)
{
dst[dst2D.idx_fast][dst2D.idx_slow] = src[src2D.idx_slow][src2D.idx_fast];
printf("dst[%d][%d] = src[%d][%d] = %d\n",
dst2D.idx_fast, dst2D.idx_slow,
src2D.idx_slow, src2D.idx_fast,
dst[dst2D.idx_fast][dst2D.idx_slow]);
next2D(&dst2D);
next2D(&src2D);
}
dump2D("Target", ROWS, COLS, dst);
return 0;
}
The Sequence2D structure plus the next2D() function cycle through array indexes, changing the fast index on each call of next2D() and the slow index when necessary. This saves you writing out the sequence of 21 assignments, and immediately scales to other shapes without requiring much extra work on your part. Indeed, it would not take much to handle all the sizes as run-time inputs instead of compile-time constants.
The output of this code is:
Source:
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
dst[0][0] = src[0][0] = 1
dst[1][0] = src[0][1] = 2
dst[2][0] = src[0][2] = 3
dst[0][1] = src[0][3] = 4
dst[1][1] = src[0][4] = 5
dst[2][1] = src[0][5] = 6
dst[0][2] = src[0][6] = 7
dst[1][2] = src[1][0] = 8
dst[2][2] = src[1][1] = 9
dst[0][3] = src[1][2] = 10
dst[1][3] = src[1][3] = 11
dst[2][3] = src[1][4] = 12
dst[0][4] = src[1][5] = 13
dst[1][4] = src[1][6] = 14
dst[2][4] = src[2][0] = 15
dst[0][5] = src[2][1] = 16
dst[1][5] = src[2][2] = 17
dst[2][5] = src[2][3] = 18
dst[0][6] = src[2][4] = 19
dst[1][6] = src[2][5] = 20
dst[2][6] = src[2][6] = 21
Target:
1 4 7 10 13 16 19
2 5 8 11 14 17 20
3 6 9 12 15 18 21
If you rearrange the initial matrix (src) according to whatever arcane rules you're using before you copy the 'transpose' of src into the result (dst), then you should be able to get the result you desire.
As noted, the code can be generalized rather easily using C99 and variable-length arrays:
#include <assert.h>
#include <stdio.h>
typedef struct Sequence2D
{
int idx_slow;
int idx_fast;
int max_slow;
int max_fast;
} Sequence2D;
static inline void next2D(Sequence2D *seq)
{
if (++seq->idx_fast == seq->max_fast)
{
seq->idx_fast = 0;
if (++seq->idx_slow == seq->max_slow)
seq->idx_slow = 0;
}
}
static inline void dump2D(const char *tag, int rows, int cols, int data[rows][cols])
{
printf("%s:\n", tag);
for (int r = 0; r < rows; r++)
{
for (int c = 0; c < cols; c++)
printf("%3d", data[r][c]);
putchar('\n');
}
}
static void transpose(int rows, int cols)
{
assert(rows * cols < 1000000);
int src[rows][cols];
int dst[rows][cols];
Sequence2D dst2D = { 0, 0, cols, rows };
Sequence2D src2D = { 0, 0, rows, cols };
int cells = rows * cols;
for (int i = 0; i < cells; i++)
{
src[src2D.idx_slow][src2D.idx_fast] = i;
next2D(&src2D);
}
/* src2D is back to its initial state! */
dump2D("Source", rows, cols, src);
for (int i = 0; i < cells; i++)
{
dst[dst2D.idx_fast][dst2D.idx_slow] = src[src2D.idx_slow][src2D.idx_fast];
printf("dst[%d][%d] = src[%d][%d] = %d\n",
dst2D.idx_fast, dst2D.idx_slow,
src2D.idx_slow, src2D.idx_fast,
dst[dst2D.idx_fast][dst2D.idx_slow]);
next2D(&dst2D);
next2D(&src2D);
}
dump2D("Target", rows, cols, dst);
}
int main(void)
{
transpose(3, 7);
transpose(9, 4);
return 0;
}
The first section of the output is the same as before; the 9x4 part looks like:
Source:
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
16 17 18 19
20 21 22 23
24 25 26 27
28 29 30 31
32 33 34 35
dst[0][0] = src[0][0] = 0
dst[1][0] = src[0][1] = 1
dst[2][0] = src[0][2] = 2
dst[3][0] = src[0][3] = 3
dst[4][0] = src[1][0] = 4
dst[5][0] = src[1][1] = 5
dst[6][0] = src[1][2] = 6
dst[7][0] = src[1][3] = 7
dst[8][0] = src[2][0] = 8
dst[0][1] = src[2][1] = 9
dst[1][1] = src[2][2] = 10
dst[2][1] = src[2][3] = 11
dst[3][1] = src[3][0] = 12
dst[4][1] = src[3][1] = 13
dst[5][1] = src[3][2] = 14
dst[6][1] = src[3][3] = 15
dst[7][1] = src[4][0] = 16
dst[8][1] = src[4][1] = 17
dst[0][2] = src[4][2] = 18
dst[1][2] = src[4][3] = 19
dst[2][2] = src[5][0] = 20
dst[3][2] = src[5][1] = 21
dst[4][2] = src[5][2] = 22
dst[5][2] = src[5][3] = 23
dst[6][2] = src[6][0] = 24
dst[7][2] = src[6][1] = 25
dst[8][2] = src[6][2] = 26
dst[0][3] = src[6][3] = 27
dst[1][3] = src[7][0] = 28
dst[2][3] = src[7][1] = 29
dst[3][3] = src[7][2] = 30
dst[4][3] = src[7][3] = 31
dst[5][3] = src[8][0] = 32
dst[6][3] = src[8][1] = 33
dst[7][3] = src[8][2] = 34
dst[8][3] = src[8][3] = 35
Target:
0 9 18 27
1 10 19 28
2 11 20 29
3 12 21 30
4 13 22 31
5 14 23 32
6 15 24 33
7 16 25 34
8 17 26 35

Answer 3

You can't transpose a matrix in place. That is to say, it is not possible to write a function which will take a matrix and modify it so that it is its transpose, without creating a temporary matrix of equal size. At least without some very fancy programming.
Now your task isn't exactly the transpose, but it seems to be very similar. So I wouldn't expect there to be a nice, in place solution. You must simply create a temporary, apply the mapping from input to temporary, and write the values back.
What might help is this.
/* swap two rows of a matrix */
void swaprows(int *mtx, int width, int height, int rowa, int rowb)
{
int i;
for(i=0;i<width;i++)
{
int temp = mtx[rowa*width+i];
mtx[rowa*width+i] = mtx[rowb*width+i];
mtx[rowa*width+i] = temp;
}
}
/* transpose a matrix */
int transpose(int *mtx, int width, int height)
{
int *tmtx = malloc(width*height*sizeof(int));
if(!tmtx)
return -1;
for(y=0;y<width;y++)
for(x=0;x<height;x++)
tmtx[y*height+x] = mtx[x*width+y];
mempy(mtx, tmtx, width * height * sizeof(int));
free(tmtx);
}
/* your logic */
int arr[3][7] = {
{1, 2, 3, 4, 5, 6, 7},
{8, 9, 10, 11, 12, 13, 14},
{15, 16, 17, 18, 19, 20, 21}
};
swaprows((int *) arr, 7, 3, 0, 1);
transpose(int *) arr, 7, 3);
This seems to be what you are trying to do.