Python code to compute the intersection of two boundingboxes

iou.py

def bb_intersection_over_union(boxA, boxB):
    # determine the (x, y)-coordinates of the intersection rectangle
    xA = max(boxA[0], boxB[0])
    yA = max(boxA[1], boxB[1])
    xB = min(boxA[2], boxB[2])
    yB = min(boxA[3], boxB[3])

    # compute the area of intersection rectangle
    interArea = abs(max((xB - xA, 0)) * max((yB - yA), 0))
    if interArea == 0:
        return 0
    # compute the area of both the prediction and ground-truth
    # rectangles
    boxAArea = abs((boxA[2] - boxA[0]) * (boxA[3] - boxA[1]))
    boxBArea = abs((boxB[2] - boxB[0]) * (boxB[3] - boxB[1]))

    # compute the intersection over union by taking the intersection
    # area and dividing it by the sum of prediction + ground-truth
    # areas - the interesection area
    iou = interArea / float(boxAArea + boxBArea - interArea)

    # return the intersection over union value
    return iou


if __name__ == '__main__':
    # Pointing out a wrong IoU implementation in https://www.pyimagesearch.com/2016/11/07/intersection-over-union-iou-for-object-detection/
    boxA = [0., 0., 10., 10.]
    boxB = [1., 1., 11., 11.]

    correct = bb_intersection_over_union(boxA, boxB)
    print('Correct solution - also analytical: {0}\n'
          'Solution by published function: {1}\n'
          'Solution by correction (ptyshevs): {2}'.format(correct, '0.704225352113', '0.680672268908'))

    print('Normalizing coordinates in a 100x100 coordinate system')
    boxA = [a / 100. for a in boxA]
    boxB = [b / 100. for b in boxB]

    correct = bb_intersection_over_union(boxA, boxB)

    print('Correct solution - also analytical: {0}\n'
          'Solution by published function: {1}\n'
          'Solution by correction: {2}'.format(correct, '0.964445166004', '0.680672268908'))
    
    print('Two boxes with no overlap')
    boxA = [0., 0., 10., 10.]
    boxB = [12., 12., 22., 22.]
    correct = bb_intersection_over_union(boxA, boxB)

    print('Correct solution - also analytical: {0}\n'
          'Solution by published function: {1}\n'
          'Solution by correction (ptyshevs): {2}'.format(correct, '0.0', '0.0204081632653'))


    print('Example in the comments from ptyshevs')
    boxA = [0., 0., 2., 2.]
    boxB = [1., 1., 3., 3.]
    correct = bb_intersection_over_union(boxA, boxB)

    print('Correct solution - also analytical: {0}\n'
          'Solution by published function: {1}\n'
          'Solution by correction (ptyshevs): {2}'.format(correct, '0.285714285714', '0.142857142857'))

Thank you Meyerjo

@meyejo Thanks for the implementation . will this work if rectangle inside the other rectangle?

@meyejo what if there are multiple bounding boxes?

@meyejo what if there are multiple bounding boxes?

Just iterate over all possible combinations

@meyejo Thanks for the implementation . will this work if rectangle inside the other rectangle?

This shouldn't be an issue. The union space of both boxes is 1. It depends on the respective size of the boxes, how large the IoU becomes.

Pointing out a wrong IoU implementation in https://www.pyimagesearch.com/2016/11/07/intersection-over-union-iou-for-object-detection/
is that still actual? if so, what the problem with his code?

are the bounding box coordinates in this format [x1 y1 x2 y2] ? Using this code sometimes I am getting iou>1, What could be the reason?

Why is the abs in line 9 necessary? Isn't the two maxima always at least zero? Thanks!