参考博文:https://blog.csdn.net/shuzfan/article/details/52711706
https://blog.csdn.net/leviopku/article/details/80886386
Non-Maximum Suppression
翻译:非极大值抑制
说人话:不是极大值的不要
作用:在目标检测中去掉多余的候选框(candidate boxes)
= 候选框三千,只取一个(分最高的)
过程:迭代-遍历-消除
1、所有boxes按照foreground score排序,选中得分最高的框(图中红色,foreground score:0.98)
2、遍历其余的框,如果和当前最高分框的重叠面积(IOU)大于一定阈值,我们就将框删除。
3、从未处理的框中继续选一个得分最高的,重复上述过程。
代码:
# --------------------------------------------------------
# Fast R-CNN
# Copyright (c) 2015 Microsoft
# Licensed under The MIT License [see LICENSE for details]
# Written by Ross Girshick
# --------------------------------------------------------
import numpy as np
def py_cpu_nms(dets, thresh):
'''
:param dets: [x1 y1 x2 y2 score]
:param thresh: IOU阈值
:return:
'''
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
scores = dets[:, 4]
areas = (x2 - x1 + 1) * (y2 - y1 + 1) # 面积,这里的+1是因为计算像素点个数,如3-0=3,但0到3是4个数
order = scores.argsort()[::-1] # foreground score大小顺序
keep = []
while order.size > 0:
i = order[0]
keep.append(i) # 保留得分最大的box
xx1 = np.maximum(x1[i], x1[order[1:]])
yy1 = np.maximum(y1[i], y1[order[1:]])
xx2 = np.minimum(x2[i], x2[order[1:]])
yy2 = np.minimum(y2[i], y2[order[1:]])
w = np.maximum(0.0, xx2 - xx1 + 1)
h = np.maximum(0.0, yy2 - yy1 + 1)
inter = w * h
ovr = inter / (areas[i] + areas[order[1:]] - inter) # IoU
inds = np.where(ovr <= thresh)[0] # 消除掉IOU大于阈值的box,只留小于阈值的box
order = order[inds + 1] # 继续下一轮
return keep