iou计算代码，纯python
from：https://blog.csdn.net/leviopku/article/details/81629492
#!/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)
标签：right,bottom,python,IOU,计算,rec2,line,rec1,iou From： https://www.cnblogs.com/chentiao/p/16870277.html