IOU 计算方法
#!/usr/bin/env python# encoding: utf-8"""@version: 1.0@author: levio@contact: levio123@163.com@file: IoU.py"""def compute_iou(rec1, rec2):"""computing IoU:param rec1: (y0, ...
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#!/usr/bin/env python
# encoding: utf-8
"""
@version: 1.0
@author: levio
@contact: levio123@163.com
@file: IoU.py
"""
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)
def test_iou():
rect1 = (661, 27, 679, 47)
# (top, left, bottom, right)
rect2 = (662, 27, 682, 47)
iou = compute_iou(rect1, rect2)
print(iou)
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