# -*- coding: utf-8 -*-
import numpy as np def IOU1(A,B): #左上右下坐标(x1,y1,x2,y2)
w=max(0,min(A[2],B[2])-max(A[0],B[0])) h=max(0,min(A[3],B[3])-max(A[1],B[1])) areaA=(A[2]-A[0]+1)*(A[3]-A[1]+1) areaB=(B[2]-B[0]+1)*(B[3]-B[1]+1) inter=w*h union=areaA+areaB-inter return inter/union def nms(dets, thresh): """Pure Python NMS baseline."""
#x1、y1、x2、y2、以及score赋值
x1 = dets[:, 0] y1 = dets[:, 1] x2 = dets[:, 2] y2 = dets[:, 3] scores = dets[:, 4] areas = (x2 - x1 + 1) * (y2 - y1 + 1) order = scores.argsort()[::-1] keep = [] while order.size > 0:#还有数据
i = order[0] keep.append(i) #计算当前概率最大矩形框与其他矩形框的相交框的坐标
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:]]) #计算相交框的面积
w = np.maximum(0.0, xx2 - xx1 + 1) h = np.maximum(0.0, yy2 - yy1 + 1) inter = w * h #计算重叠度IOU：重叠面积/（面积1+面积2-重叠面积）
IOU = inter / (areas[i] + areas[order[1:]] - inter) #找到重叠度不高于阈值的矩形框索引
left_index = np.where(IOU <= thresh)[0] #将order序列更新，由于前面得到的矩形框索引要比矩形框在原order序列中的索引小1，所以要把这个1加回来
order = order[left_index + 1] print(keep) if __name__ == '__main__': dets=[[0,0,100,101,0.9],[5,6,90,110,0.7],[17,19,80,120,0.8],[10,8,115,105,0.5]] dets=np.array(dets) nms(dets,0.5) print IOU1(dets[0],dets[2])
