nms
def cal_iou(bbox1,bbox2):
# x1,y1,x2,y2
# min_x - max_x
inter_x = min(bbox1[2],bbox2[2]) - max(bbox1[0],bbox2[0])
# min_y - max_y
inter_y = min(bbox1[3],bbox2[3]) - max(bbox1[1],bbox2[1])
if inter_x <=0 or inter_y <=0:
return 0
inter_area = inter_x * inter_y
area1 = (bbox1[2] - bbox1[0] +1)* (bbox1[3] - bbox1[1] +1)
area2 = (bbox2[2] - bbox2[0] +1)* (bbox2[3] - bbox2[1] +1)
#分母不会是0,面积最小是1
iou = inter_area / (area1+area2 - inter_area)
return iou
def nms(bboxes,scores,iou_thre=0.3):
infos = list(map(list,zip(scores,bboxes)))
print(infos)
infos.sort(key=lambda x:x[0],reverse=True)
#print(infos)
for i in range(len(infos)):
for j in range(i+1,len(infos)):
iou = cal_iou(infos[i][1],infos[j][1])
print(iou)
if iou > iou_thre:
#score置-1,后面清空
infos[j][0] = -1
#剔除score=-1的框
new_bboxes = []
for i in range(len(infos)):
if infos[i][0] != -1:
new_bboxes.append(infos[i][1])
return new_bboxes
test_bboxes = [[0,0,100,100],[10,10,100,100],[100,100,200,200]]
test_scores = [0.98,0.3,0.7]
res = nms(test_bboxes,test_scores)
print(res)
标签:bbox2,bbox1,代码,bboxes,min,深度,test,100,高频
From: https://www.cnblogs.com/forrestr/p/18233507