3D IoU (Intersection over Union)

3DIoU.py

# 3D IoU python Implementation
# [IoU](https://www.pyimagesearch.com/2016/11/07/intersection-over-union-iou-for-object-detection/)

def IoU(box0, box1):
  # box0: [x, y, z, d]
  r0 = box0[3] / 2
  s0 = box0[:3] - r0
  e0 = box0[:3] + r0
  r1 = box1[3] / 2
  s1 = box1[:3] - r1
  e1 = box1[:3] + r1
  
  overlap = [max(0, min(e0[i], e1[i]) - max(s0[i], s1[i])) for i in range(3)]
  intersection = reduce(lambda x,y:x*y, overlap)
  union = pow(box0[3], 3) + pow(box1[3], 3) - intersection
  return intersection / union

Can you elaborate how is the box encoded?
Thank you!

Not able to understand since box0[3]=d what does this mean?

Not able to understand since box0[3]=d what does this mean?

This box is only a cube which it has a same width, length, and height.
The d in each box represents a length while (x,y,z) represents a centroid of the box.
Therefore, r = box[3] = d / 2 gives you the half of the dimension.

For example, if a box has a centroid at (0,0,0) with d = 4, r = d/2 = 2 and you simply add r into the centroid to get the upper bound, and subtract to get the lower bound of a given box.