Implement 2d array coordinates in 1d array in C++

Question

The code inside the for loop is for the x and y (j and i) "coordinates" from a 2d array. How could I implement this neighbor/index finding in a 1d array? I think I could implement it for the first four equations. But i'm confused as how to implement up-left etc.

for(int i=0; i<cols*rows; i++){
        //Counts current index's 8 neigbour int values
      int count=0;
      int x = i%cols;
      int y = i/rows;
      //rows y i
      //cols x j

      count+= [grid][i][(j-1+cols)%cols] //left
         +[grid][i][(j+1+cols)%cols]    //right
         +[grid][(i-1+rows)%rows][j]    //up
         +[grid][(i+1+rows)%rows][j]    //down
         +[grid][(i-1+rows)%rows][ (j-1+cols)%cols]  //up-left
         +[grid][(i+1+rows)%rows][ (j+1+cols)%cols]  //down-right
         +[grid][(i+1+rows)%rows][ (j-1+cols)%cols]  //down-left
         +[grid][(i-1+rows)%rows][ (j+1+cols)%cols] ;//up-right
}

Do you mean you have a 1D array, example int grid[rows * cols]? You have calculated x/y where data should be grid[i] — Barmak Shemirani
– Barmak Shemirani, Commented Mar 6, 2018 at 6:43
Yes, the size of my 1d array is rows*cols. I have a 1d array, but the neighbor finding is for a 2d array I want to know how to do this equation(s) for a 1d array. — Philip Miller
– Philip Miller, Commented Mar 6, 2018 at 6:55

Barmak Shemirani · Accepted Answer · 2018-03-06 07:43:22Z

2

Starting with a 1-D vector:

int rows = 10;
int cols = 10;
vector<int> grid(rows * cols);

You can manage this in different ways, example

for(int y = 0; y < rows; y++)
{
    for(int x = 0; x < cols; x++)
    {
        int point = grid[y * rows + x];
    }
}

Where you can access any point at any given x and y in a 2-dimensional plane.

Top-left is:

x = 0;
y = 0;

bottom-right is

x = cols - 1;
y = rows - 1;

And so on.

answered Mar 6, 2018 at 7:43

Barmak Shemirani

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3 Comments

Philip Miller Over a year ago

Could you get the coordinates (x,y) from the index of the loop without making more for loops?

Barmak Shemirani Over a year ago

If you know the index i, then x = i%cols and y = i/rows, you already did that yourself! Conversely, if you know x and y, then index is i = y * rows + x as shown above.

Jeffrey Faust Over a year ago

That's not quite right to get from index to x, y. It should be x = i%cols and y = i/cols.

LSchueler · Accepted Answer · 2018-03-06 06:56:33Z

1

Use a function like this

inline int idx(const int i, const int j, const int rows) const
{
    return i * rows + j;
}

to convert the 2d indices to 1d indices. This way you don't have to change your algorithm.

Usage would be grid[idx(i, (j-1+cols)%cols, rows)].

answered Mar 6, 2018 at 6:56

LSchueler

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Comments

moooeeeep · Accepted Answer · 2018-03-06 11:24:57Z

The basic formula for computing the 1d coordinate from the 2d index pattern is usually one of the following:

row_index * row_length + column_index
column_index * column_length + row_index

Which one applies to your case depends on whether you would like to have a row-based or column-based memory layout for your 2d array. It makes sense to factor out the computation of this index into a separate function, as suggested in the other answer.

Then you just need to fill in the values somehow.

You could do it like this, for example:

// iterate big picture
// TODO: make sure to handle the edge cases appropriately
for (int i_row = 1; i_row < n_rows - 1; i_row++) {
    for (int i_col = 1; i_col < n_cols -1; i_col++) {
        // compute values
        dst[i_row*n_cols+i_col] = 0;
        for (int r = i_row-1; r < i_row+2; r++) {
            for (int c = i_col-1; c < i_col+2; c++) {
                dst[i_row*n_cols+i_col] += src[r*n_cols + c];
            }
        }
    }
}

Assuming src and dst are distinct 1d vectors of size n_rows*n_cols...

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