python实现IOU计算

https://blog.csdn.net/leviopku/article/details/81629492

计算两个矩形的交并比，通常在检测任务里面可以作为一个检测指标。你的预测bbox和groundtruth之间的差异，就可以通过IOU来体现。

#!/usr/bin/env python
# encoding: utf-8

import numpy as np

'''
函数说明：计算两个框的重叠面积
输入：
rec1  第一个框xmin ymin xmax ymax
rec2  第二个框xmin ymin xmax ymax
输出：
iouv 重叠比例 0 没有
'''
def compute_iou(rec1, rec2):
    
    # computing area of each rectangles
    S_rec1 = (rec1[2] - rec1[0]) * (rec1[3] - rec1[1]) # H1*W1
    S_rec2 = (rec2[2] - rec2[0]) * (rec2[3] - rec2[1]) # H2*W2
  
    # computing the sum_area
    sum_area = S_rec1 + S_rec2 #总面积
  
    # find the each edge of intersect rectangle
    left_line = max(rec1[0], rec2[0])
    right_line = min(rec1[2], rec2[2])
    top_line = max(rec1[1], rec2[1])
    bottom_line = min(rec1[3], rec2[3])
   
    # judge if there is an intersect
    if left_line >= right_line or top_line >= bottom_line:
        #print("没有重合区域")
        return 0
    else:
	#print("有重合区域")
        intersect = (right_line - left_line) * (bottom_line - top_line)
        iouv=(float(intersect) / float(sum_area - intersect))*1.0

        return iouv

'''
函数说明：获取两组匹配结果
输入：
rectA  车位
rectB  车辆
threod 重叠面积最小数值界限 默认0.6
输出：
CarUse 一维数组保存是否占用  1 占用 0 没有

'''
def TestCarUse(rectA,rectB,threod=0.6，debug=0):
    #threod=0.8#设定最小值
    ALength=len(rectA)
    BLength=len(rectB)

    #创建保存匹配结果的矩阵
    recIOU=np.zeros((ALength,BLength),dtype=float,order='C')
    #用于记录车位能够使否占用    
    CarUse=np.zeros((1,ALength),dtype=int,order='C')

    for i in range(0,ALength):
    	for j in range(0,BLength):
            iou = compute_iou(rectA[i], rectB[j])
            recIOU[i][j]=format(iou,'.3f')
            if iou>=threod:         
                CarUse[0,i]=1  #有一个超过匹配认为车位i被占用
    if debug==1:
	    print('----匹配矩阵----')
	    print(recIOU)
	    '''
	    print('----车位占用情况----')
	    for i in range(0,ALength):
		msg='车位'+str(i)+"-"+str(CarUse[0][i])
		print(msg)
	    '''
    return CarUse

 
  
if __name__=='__main__':
    #A代表车位
    rectA1 = (30, 10, 70, 20)
    rectA2 = (70, 10, 80, 20)
    rectA =[rectA1,rectA2]
    #B代表检测车辆
    rectB1 = (20, 10, 35, 20)
    rectB2 = (30, 15, 70, 25)
    rectB3 = (70, 10, 80, 20)
    rectB =[rectB1,rectB2,rectB3]
    
    #获取车位占用情况 rectA车位  rectB车辆  0.6占面积最小比
    CarUse=TestCarUse(rectA,rectB,0.6,1)

    print('----车位占用情况----')
    for i in range(0,len(CarUse)+1):
	msg='车位'+str(i)+"-"+str(CarUse[0][i])
        print(msg)