# -*- 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])