IOU的计算

def IouDo(box,boxs,mode='UNIUM'):
    #获得这一批建议框的坐标
    _x1 = boxs[:,0]
    _y1 = boxs[:,1]
    _x2 = boxs[:,2]
    _y2 = boxs[:,3]
    #获取交集的坐标
    x1 = np.maximum(box[0],_x1)
    y1 = np.maximum(box[1],_y1)
    x2 = np.minimum(box[2],_x2)
    y2 = np.minimum(box[3],_y2)
    #计算交集的宽和高
    w = np.maximum(0,x2-x1)
    h = np.maximum(0,y2-y1)
    #计算交集的面积
    area = w*h
    #计算并集的面积
    Area = (box[2]-box[0])*(box[3]-box[1])+(_x2-_x1)*(_y2-_y1)-area
    #计算交并比
    if mode == 'UNIUM':
        unium_area = np.true_divide(area,Area)
        return unium_area
    #计算最小值除以并集
    else:
        min_area = np.true_divide(np.minimum((box[2] - box[0]) * (box[3] - box[2]), (_x2 - _x1) * (_y2 - _y1)),Area)
        return min_area