python calculate intersect of union area between two box

 refer to code:

..

def IoU(box1, box2):

# box = (x1, y1, x2, y2)

box1_area = (box1[2] - box1[0] + 1) * (box1[3] - box1[1] + 1)

box2_area = (box2[2] - box2[0] + 1) * (box2[3] - box2[1] + 1)

# obtain x1, y1, x2, y2 of the intersection

x1 = max(box1[0], box2[0])

y1 = max(box1[1], box2[1])

x2 = min(box1[2], box2[2])

y2 = min(box1[3], box2[3])

x11, y11, x12, y12 = box1

x21, y21, x22, y22 = box2

if x21<x11 and y21<y11 and x22>x12 and y22>y12:

return 1.0

if x21>x11 and y21>y11 and x22<x12 and y22<y12:

return 1.0

# compute the width and height of the intersection

w = max(0, x2 - x1 + 1)

h = max(0, y2 - y1 + 1)

inter = w * h

iou = inter / (box1_area + box2_area - inter)

return iou

..

Thank you

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rect intersector check using image (python)

1. make image as numpy
2. add 1 all element for each inside of box
3. check if there is bigger element value than 1

Thank you.

#draw new image

img = np.zeros((int(height), int(width), 1), dtype = "uint8")

#draw new image

img.fill(0)

for i, v in enumerate(box_list):

left = int(v['bbox'][0])

top = int(v['bbox'][1])

right = int(v['bbox'][2])

bottom = int(v['bbox'][3])

img[top:bottom,left:right] += 1

#check overlap

imgv = img.reshape(height*width)

if sum(imgv > 1) > 0:

print('intersector found')