目标检测：IOU - 北京网站建设

IOU（Intersection over Union）交并比：

它计算的是“预测的边框”和“真实的边框”的交叠率，即它们的交集和并集的比值。这个比值用于衡量预测边框与真实边框的重叠程度，从而评估目标检测的准确性。

在目标检测任务中，通常使用bounding box与ground truth之间的IoU值大小来判断预测结果的好坏。一般情况下，认为IoU>0.5就是一个不错的预测结果。

两个box区域的交集除以并集

总的来说：IOU是一个在目标检测中非常重要的评估指标，用于量化预测边框与真实边框的匹配程度。

常规重叠框

代码示例

import cv2
import numpy as npimg = np.zeros((300,300,3),dtype=np.uint8)
img[:,:,:] = 255# 下方红色框
x1,y1,x2,y2 = 20,20,180,180
# 下方蓝色框
x3,y3,x4,y4 = 90,90,280,280cv2.rectangle(img, (x1, y1), (x2, y2), (0,0,255), 2)
cv2.rectangle(img, (x3, y3), (x4, y4), (255,0,0), 2)cv2.imshow('img',img)
cv2.waitKey(10000)
cv2.destroyAllWindows()

当前一个框由左上角 + 右下角共 2 个坐标点组成

如果 2 个框有交集，如下

交集也是一个矩形框，由左上角 + 右下角 2个坐标点组成

已知红框和篮框的左上角 + 右下角如何计算交集绿色框的左上角 + 右下角坐标

计算方法

绿色区域 (交集) 左上角与右下角

左上角：取红框和篮框左上角点的横坐标 (x) 与纵坐标 (y) 最大值

# 红框
a = [x1, y1, x2, y2]
# 蓝框
b = [x3, y3, x4, y4]c = np.array([a,b])
l_x1,l_y1 = np.max(c[:,0:2],axis=0)

右下角：取红框和篮框右下角点的横坐标 (x) 与纵坐标 (y) 最小值

# 红框
a = [x1, y1, x2, y2]
# 蓝框
b = [x3, y3, x4, y4]c = np.array([a,b])
r_x2,r_y2 = np.min(c[:,2:],axis=0)

代码实现

import cv2
import numpy as npimg = np.zeros((300,300,3),dtype=np.uint8)
img[:,:,:] = 255def inter(a,b):c = np.array([a,b])l_x1,l_y1 = np.max(c[:,0:2],axis=0)r_x2,r_y2 = np.min(c[:,2:],axis=0)return l_x1,l_y1,r_x2,r_y2x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 90,90,280,280up_x,up_y,bo_x,bo_y = inter([x1,y1,x2,y2],[x3,y3,x4,y4])
cv2.rectangle(img, (up_x, up_y), (bo_x, bo_y), (0,255,0), -1)
cv2.rectangle(img, (x1, y1), (x2, y2), (0,0,255), 2)
cv2.rectangle(img, (x3, y3), (x4, y4), (255,0,0), 2)cv2.imshow('img',img)
cv2.waitKey(10000)
cv2.destroyAllWindows()
cv2.imwrite('iou.jpg',img)

方法验证

实验不同的重叠形式

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 90,10,280,280

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,10,150,200

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,10,280,150

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,50,150,200

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,10,150,150

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,50,150,150

x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,50,200,150

计算IoU

先计算重叠区域面积，再计算红/蓝框的并集，最后计算占比

代码实现

import cv2
import numpy as npimg = np.zeros((300,300,3),dtype=np.uint8)
img[:,:,:] = 255def inter(a,b):c = np.array([a,b])l_x1,l_y1 = np.max(c[:,0:2],axis=0)r_x2,r_y2 = np.min(c[:,2:],axis=0)return l_x1,l_y1,r_x2,r_y2x1,y1,x2,y2 = 20,20,180,180
x3,y3,x4,y4 = 40,50,200,150up_x,up_y,bo_x,bo_y = inter([x1,y1,x2,y2],[x3,y3,x4,y4])# 红框 / 蓝框 / 重叠-绿色区域 面积
S_red = (x2-x1) * (y2-y1)
S_blue = (x4-x3) * (y4-y3)
S_green = (bo_x-up_x) * (bo_y-up_y)
print('面积-红框:',S_red,', 面积-蓝框:',S_blue,', 面积-绿色:',S_green)# 红框 与 蓝框 的 并集
S_union = S_red + S_blue - S_green
print('红框与蓝框的并集-面积:',S_union)iou = S_green / S_union
print('交集 / 并集: ',iou)cv2.rectangle(img, (up_x, up_y), (bo_x, bo_y), (0,255,0), -1)
cv2.rectangle(img, (x1, y1), (x2, y2), (0,0,255), 2)
cv2.rectangle(img, (x3, y3), (x4, y4), (255,0,0), 2)cv2.imshow('img',img)
cv2.waitKey(10000)
cv2.destroyAllWindows()
cv2.imwrite('iou.jpg',img)

结果

面积-红框: 25600 , 面积-蓝框: 16000 , 面积-绿色: 14000
红框与蓝框的并集-面积: 27600
交集 / 并集:  0.5072463768115942

无重叠情况

import cv2
import numpy as npimg = np.zeros((300,300,3),dtype=np.uint8)
img[:,:,:] = 255def inter(a,b):c = np.array([a,b])l_x1,l_y1 = np.max(c[:,0:2],axis=0)r_x2,r_y2 = np.min(c[:,2:],axis=0)return l_x1,l_y1,r_x2,r_y2x1,y1,x2,y2 = 20,20,120,120
x3,y3,x4,y4 = 140,150,260,240up_x,up_y,bo_x,bo_y = inter([x1,y1,x2,y2],[x3,y3,x4,y4])print(up_x,up_y,bo_x,bo_y)cv2.rectangle(img, (x1, y1), (x2, y2), (0,0,255), 2)
cv2.rectangle(img, (x3, y3), (x4, y4), (255,0,0), 2)cv2.imshow('img',img)
cv2.waitKey(10000)
cv2.destroyAllWindows()

结果

140 150 120 120

左上角：(140, 150)

右下角：(120, 120)

按照上面计算方式，在 2框无交集的情况的下，也能算出重叠区域的坐标

但无交集计算的坐标特点：左上角 > 右下角

代码实现

import cv2
import numpy as npimg = np.zeros((300,300,3),dtype=np.uint8)
img[:,:,:] = 255def inter(a,b):c = np.array([a,b])l_x1,l_y1 = np.max(c[:,0:2],axis=0)r_x2,r_y2 = np.min(c[:,2:],axis=0)return l_x1,l_y1,r_x2,r_y2x1,y1,x2,y2 = 20,20,120,120
x3,y3,x4,y4 = 140,150,260,240
# x1,y1,x2,y2 = 20,20,180,180
# x3,y3,x4,y4 = 40,50,200,150up_x,up_y,bo_x,bo_y = inter([x1,y1,x2,y2],[x3,y3,x4,y4])print(up_x,up_y,bo_x,bo_y)# 红框 / 蓝框 / 重叠-绿色区域 面积
S_red = (x2-x1) * (y2-y1)
S_blue = (x4-x3) * (y4-y3)if bo_x > up_x and bo_y > up_y: S_green = (bo_x-up_x) * (bo_y-up_y)
else: S_green = 0print('面积-红框:',S_red,', 面积-蓝框:',S_blue,', 面积-绿色:',S_green)# 红框 与 蓝框 的 并集
S_union = S_red + S_blue - S_green
print('红框与蓝框的并集-面积:',S_union)iou = S_green / S_union
print('交集 / 并集: ',iou)if bo_x > up_x and bo_y > up_y: cv2.rectangle(img, (up_x, up_y), (bo_x, bo_y), (0,255,0), -1)cv2.rectangle(img, (x1, y1), (x2, y2), (0,0,255), 2)
cv2.rectangle(img, (x3, y3), (x4, y4), (255,0,0), 2)cv2.imshow('img',img)
cv2.waitKey(10000)
cv2.destroyAllWindows()
cv2.imwrite('iou.jpg',img)

结果

140 150 120 120
面积-红框: 10000 , 面积-蓝框: 10800 , 面积-绿色: 0
红框与蓝框的并集-面积: 20800
交集 / 并集:  0.0