python：计算两个方框的IOU的实现

朴灿烈づ我的快乐病毒、 2023-07-25 11:21 26阅读 0赞

> 在目标检测中，常常需要计算不同方框的并交比来选取合适的方框，做跟踪用到匹配也遇到了并交比。

结果图：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1ODk2MTM2_size_16_color_FFFFFF_t_70]  
代码如下：

import numpy as np
    import cv2
    
    
    def compute_iou(rec1, rec2):
        """ computing IoU rec1: (x0, y0, x1, y1) rec2: (x0, y0, x1, y1) :return: scala value of IoU """
        # computing area of each rectangle
        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])
        print(top_line, left_line, right_line, bottom_line)
    
        # judge if there is an intersect area
        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__':
        sector = np.zeros((500, 500, 3), np.uint8)
        rect1 = (120, 120, 224, 224)
        # (top, left, bottom, right)
        rect2 = (150, 150, 385, 385)
        cv2.rectangle(sector, rect1[:2], rect1[2:], (0, 255, 0), 4)
        cv2.rectangle(sector, rect2[:2], rect2[2:], (0, 0, 255), 4)
        iou = compute_iou(rect1, rect2)
        cv2.putText(sector, 'iou: %.6f' % iou, (10,30), cv2.FONT_HERSHEY_SIMPLEX, 1, (0,0,255), 2)
        print(iou)
        cv2.imshow("img", sector)
        cv2.waitKey(0)
        cv2.destroyAllWindows()

[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1ODk2MTM2_size_16_color_FFFFFF_t_70]: /images/20230528/dc9306b9ed2e4fa192a4fc54f04ac73e.png