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'))

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!

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?

Is this the line that is problematic?
interArea = max(0, xB - xA + 1) * max(0, yB - yA + 1)

Thanks meyerjo!

interArea = abs(max((xB - xA, 0)) * max((yB - yA), 0))

Should be
interArea = max((xB - xA), 0) * max((yB - yA), 0)

Because max(x, 0) the absolute value is unnecessary.

interArea = abs(max((xB - xA, 0)) * max((yB - yA), 0))

Should be interArea = max((xB - xA), 0) * max((yB - yA), 0)

Because max(x, 0) the absolute value is unnecessary.

yep the abs is useless, which should be remove.