python实现IOU计算案例

时间：2021-09-15 10:43:41|栏目：Python代码|点击：次

计算两个矩形的交并比，通常在检测任务里面可以作为一个检测指标。你的预测bbox和groundtruth之间的差异，就可以通过IOU来体现。很简单的算法实现，我也随便写了一个，嗯，很简单。

1. 使用时，请注意bbox四个数字的顺序(y0,x0,y1,x1)，顺序不太一样。

#!/usr/bin/env python
# encoding: utf-8
 
def compute_iou(rec1, rec2):
  """
  computing IoU
  :param rec1: (y0, x0, y1, x1), which reflects
      (top, left, bottom, right)
  :param rec2: (y0, x0, y1, x1)
  :return: scala value of IoU
  """
  # computing area of each rectangles
  S_rec1 = (rec1[2] - rec1[0]) * (rec1[3] - rec1[1])
  S_rec2 = (rec2[2] - rec2[0]) * (rec2[3] - rec2[1])
 
  # computing the sum_area
  sum_area = S_rec1 + S_rec2
 
  # find the each edge of intersect rectangle
  left_line = max(rec1[1], rec2[1])
  right_line = min(rec1[3], rec2[3])
  top_line = max(rec1[0], rec2[0])
  bottom_line = min(rec1[2], rec2[2])
 
  # judge if there is an intersect
  if left_line >= right_line or top_line >= bottom_line:
    return 0
  else:
    intersect = (right_line - left_line) * (bottom_line - top_line)
    return (intersect / (sum_area - intersect))*1.0
 
 
if __name__=='__main__':
  rect1 = (661, 27, 679, 47)
  # (top, left, bottom, right)
  rect2 = (662, 27, 682, 47)
  iou = compute_iou(rect1, rect2)
  print(iou)

补充知识：基于Python实现的IOU算法---最简单易懂的代码实现

概念介绍：