IOU计算-纯python

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)