Sorting rectangular contours opencv

Question

i have a picture of words , each word is bounded by a rectangular contour. i need the rectangular contour vector to be sorted so i can do some operation on it.

Here is the picture of words(with contours). image with contours

i have tried to sort it and my compare function is

bool my_compare(Rect2d rect1, Rect2d rect2)
{  if(((rect1.tl().y) <= rect2.tl().y) &&(rect1.tl().y < rect2.br().y)) 
     return (rect1.tl().x < rect2.tl().x);
  return (rect1.tl().y < rect2.tl().y);
}

also i tried this

 if((rect1.tl().y == rect2.tl().y))
   return (rect1.tl().x < rect2.tl().x); 
   return (rect1.tl().y < rect2.tl().y);

the result is not sorted properly (it doesnt start from very top left or it starts but it skips a contour in the same line ).

Answer 1

The reason is probably because rectangles in the same row has different height. Let's say a rectangle is followed by a taller rectangle in the same row, the taller rectangle is going to be sorted earlier than the left one.

+--------------> x
|          ****           
| ****     *  *
| *  *     *  *
| ****     ****
| rect1    rect2
v
y

Using your first compare function, rect1.tl().y <= rect2.tl().y is false and rect1.tl().y < rect2.br().y) is true, thus return (rect1.tl().y < rect2.tl().y) which will put the right rectangle smaller than the left one

One suggestion to overcome this is if the rectangles top left corner difference in y is within a threshold, consider them to be on the same row and compare by x instead.

bool my_compare(Rect2d rect1, Rect2d rect2)
{   
    if (fabs(rect1.tl().y - rect2.tl().y) < threshold) 
        return (rect1.tl().x < rect2.tl().x);
    return (rect1.tl().y < rect2.tl().y);
}

fabs is the floating point absolute function under <cmath> . One way to calculate threshold is by using average height of the rectangle.

threshold = ((rect1.br().y - rect1.tl().y) + (rect2.br().y - rect2.tl().y)) / 2 * K;

K is a constant which you can experiment to see which perform best, it should be between 0 and 1.