python实现交并比IOU

逃离我推掉我的手 2022-03-26 02:20 297阅读 0赞

交并比（Intersection-over-Union，IoU），目标检测中使用的一个概念，是产生的候选框（candidate bound）与原标记框（ground truth bound）的交叠率，即它们的交集与并集的比值。最理想情况是完全重叠，即比值为1。

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2d1eXVlYWxpYW4_size_16_color_FFFFFF_t_70][]

计算公式：

![20190115101357262.png][]

Python实现代码：

def cal_iou(box1, box2):
        """
        :param box1: = [xmin1, ymin1, xmax1, ymax1]
        :param box2: = [xmin2, ymin2, xmax2, ymax2]
        :return: 
        """
        xmin1, ymin1, xmax1, ymax1 = box1
        xmin2, ymin2, xmax2, ymax2 = box2
        # 计算每个矩形的面积
        s1 = (xmax1 - xmin1) * (ymax1 - ymin1)  # C的面积
        s2 = (xmax2 - xmin2) * (ymax2 - ymin2)  # G的面积
    
        # 计算相交矩形
        xmin = max(xmin1, xmin2)
        ymin = max(ymin1, ymin2)
        xmax = min(xmax1, xmax2)
        ymax = min(ymax1, ymax2)
    
        w = max(0, xmax - xmin)
        h = max(0, ymax - ymin)
        area = w * h  # C∩G的面积
        iou = area / (s1 + s2 - area)
        return iou

参考:[https://blog.csdn.net/qq\_30622831/article/details/85144849][https_blog.csdn.net_qq_30622831_article_details_85144849]

# -*-coding: utf-8 -*-
    """
        @Project: IOU
        @File   : IOU.py
        @Author : panjq
        @E-mail : pan_jinquan@163.com
        @Date   : 2018-10-14 10:44:06
    """
    def calIOU_V1(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
        """
        # 计算每个矩形的面积
        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 calIOU_V2(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
        """
        # cx1 = rec1[0]
        # cy1 = rec1[1]
        # cx2 = rec1[2]
        # cy2 = rec1[3]
        # gx1 = rec2[0]
        # gy1 = rec2[1]
        # gx2 = rec2[2]
        # gy2 = rec2[3]
        cx1,cy1,cx2,cy2=rec1
        gx1,gy1,gx2,gy2=rec2
        # 计算每个矩形的面积
        S_rec1 = (cx2 - cx1) * (cy2 - cy1)  # C的面积
        S_rec2 = (gx2 - gx1) * (gy2 - gy1)  # G的面积
    
        # 计算相交矩形
        x1 = max(cx1, gx1)
        y1 = max(cy1, gy1)
        x2 = min(cx2, gx2)
        y2 = min(cy2, gy2)
    
        w = max(0, x2 - x1)
        h = max(0, y2 - y1)
        area = w * h  # C∩G的面积
    
        iou = area / (S_rec1 + S_rec2 - area)
        return iou
    
    if __name__=='__main__':
        rect1 = (661, 27, 679, 47)
        # (top, left, bottom, right)
        rect2 = (662, 27, 682, 47)
        iou1 = calIOU_V1(rect1, rect2)
        iou2 = calIOU_V2(rect1, rect2)
        print(iou1)
        print(iou2)

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2d1eXVlYWxpYW4_size_16_color_FFFFFF_t_70]: /images/20220326/9d5fdf50da4e45f79911c25ee08b76ee.png
[20190115101357262.png]: /images/20220326/6003e76cdae243ba9d325f9a7cea3644.png
[https_blog.csdn.net_qq_30622831_article_details_85144849]: https://blog.csdn.net/qq_30622831/article/details/85144849