# -*- 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])