參考博文:https://blog.csdn.net/shuzfan/article/details/52711706
https://blog.csdn.net/leviopku/article/details/80886386
Non-Maximum Suppression
翻譯:非極大值抑制
說人話:不是極大值的不要
作用:在目標檢測中去掉多餘的候選框(candidate boxes)
= 候選框三千,只取一個(分最高的)
過程:迭代-遍歷-消除
1、所有boxes按照foreground score排序,選中得分最高的框(圖中紅色,foreground score:0.98)
2、遍歷其餘的框,如果和當前最高分框的重疊面積(IOU)大於一定閾值,我們就將框刪除。
3、從未處理的框中繼續選一個得分最高的,重複上述過程。
代碼:
# --------------------------------------------------------
# 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閾值
:return:
'''
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
scores = dets[:, 4]
areas = (x2 - x1 + 1) * (y2 - y1 + 1) # 面積,這裏的+1是因爲計算像素點個數,如3-0=3,但0到3是4個數
order = scores.argsort()[::-1] # foreground score大小順序
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大於閾值的box,只留小於閾值的box
order = order[inds + 1] # 繼續下一輪
return keep