目标检测中常用到NMS,在faster R-CNN中,每一个bounding box都有一个打分,NMS实现逻辑是:
1,按打分最高到最低将BBox排序 ,例如:A B C D E F
2,A的分数最高,保留。从B-E与A分别求重叠率IoU,假设B、D与A的IoU大于阈值,那么B和D可以认为是重复标记去除
3,余下C E F,重复前面两步。
#coding:utf-8
import numpy as np
def py_cpu_nms(dets, thresh):
"""Pure Python NMS baseline."""
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
scores = dets[:, 4] #bbox打分
areas = (x2 - x1 + 1) * (y2 - y1 + 1)
#打分从大到小排列,取index
order = scores.argsort()[::-1]
#keep为最后保留的边框
keep = []
while order.size > 0:
#order[0]是当前分数最大的窗口,肯定保留
i = order[0]
keep.append(i)
#计算窗口i与其他所有窗口的交叠部分的面积
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
#交/并得到iou值
ovr = inter / (areas[i] + areas[order[1:]] - inter)
#inds为所有与窗口i的iou值小于threshold值的窗口的index,其他窗口此次都被窗口i吸收
inds = np.where(ovr <= thresh)[0]
#order里面只保留与窗口i交叠面积小于threshold的那些窗口,由于ovr长度比order长度少1(不包含i),所以inds+1对应到保留的窗口
order = order[inds + 1]
return keep