倾斜目标检测RRPN （Arbitrary-Oriented Scene Text Detection via Rotation Proposals）

RRPN可以采用一阶段的目标检测流程或二阶段的目标检测流程（Faster RCNN）

一阶段的目标检测只需要计算RRPN IOU, 不需要ROI pooling，

两阶段的需要计算RRPN IOU和RRPN pooling

1.RRPN中的IOU计算

(1)给定预测框和标注框的中心点坐标和宽高，角度。即region1和region2包含5个元素，（cx,cy,w,h,angle）

float computeRectIoU(T const * const region1, T const * const region2)

(2) 根据region得到4个端点的（x,y）坐标，共8个值，得到rect_pts1，rect_pts2。注意：倾斜框需要考虑角度的影响

convert_region_to_pts(region1, rect_pts1);
convert_region_to_pts(region2, rect_pts2);

template <typename T>
__DEVICE__ inline void convert_region_to_pts(T const * const roi, T * pts)
{
    T cx = roi[0];
    T cy = roi[1];
    T w = roi[2];
    T h = roi[3];
    T angle = deg2rad(roi[4]);

    T b = cos(angle)*0.5f;
    T a = sin(angle)*0.5f;

    pts[0] = cx - a*h - b*w;
    pts[1] = cy + b*h - a*w;
    pts[2] = cx + a*h - b*w;
    pts[3] = cy - b*h - a*w;
    pts[4] = 2*cx - pts[0];
    pts[5] = 2*cy - pts[1];
    pts[6] = 2*cx - pts[2];
    pts[7] = 2*cy - pts[3];
}

（3）根据两个区域region1,region2的4个端点rect_pts1,rect_pts2计算两个区域的相交的面积

float area_inter = computeRectInterArea(rect_pts1, rect_pts2);

因为预测框和标注的边界框有很多个，需要计算预测框和边界框两两之间的相交的面积：

N是预测框的个数，M是实际的标注框的个数，关键函数computeRectIoU
 for (int64_t i = 0; i < N; i++)
  {
    rect_1[0] = xc1[i];
    rect_1[1] = yc1[i];
    rect_1[2] = w1[i];
    rect_1[3] = h1[i];
    rect_1[4] = angle1[i];

    scalar_t rect_2[5];
    for (int64_t j = 0; j < M; j++)
    {
      rect_2[0] = xc2[j];
      rect_2[1] = yc2[j];
      rect_2[2] = w2[j];
      rect_2[3] = h2[j];
      rect_2[4] = angle2[j];

      float iou = computeRectIoU(rect_1, rect_2);
      iou_data[i*M+j] = iou;
    }

computeRectIoU:

分几种情况计算两个区域相交的面积：两个区域没有相交，一个区域完全包含在另一个区域中，两个区域间有一定的相交

float computeRectInterArea(const T* rect_pts1, const T* rect_pts2)
{
    T inter_pts_f[MAX_RECT_INTERSECTIONS*2];
    // int res = rotatedRectangleIntersection2(rect, pixel_rect, inter_pts);
    int num_intersects = 0;
    int res = rotatedRectangleIntersection(rect_pts1, rect_pts2, inter_pts_f, num_intersects);

    float interArea = 0.0f;
    if (res == RectIntersectTypes::INTERSECT_NONE)
    {
        interArea = 0.0f;
    } else if (res == RectIntersectTypes::INTERSECT_FULL_1 || res == RectIntersectTypes::INTERSECT_EQ)
    {
        interArea = contourArea(4, rect_pts1);
    } else if (res == RectIntersectTypes::INTERSECT_FULL_2)
    {
        interArea = contourArea(4, rect_pts2);
    } else {
        T order_pts_f2[MAX_RECT_INTERSECTIONS * 2];
        size_t npoints = convexHull(num_intersects, inter_pts_f, order_pts_f2);
        interArea = contourArea(npoints, order_pts_f2);
    }

countArea(两个区域相交的点的数目，两个区域相交的点的坐标)：根据相交区域的点的坐标来计算相交区域的面积

根据相交区域的点的坐标来计算相交区域的面积，注意计算(float)prev[0] * p[1] - (float)prev[1] * p[0]时是有正有负，所以这种计算方式看着不合理，其实是对的
float contourArea(const int npoints, const T* contour, bool oriented=false)
{
    if( npoints == 0 )
        return 0.;

    float a00 = 0.0;
    const T* prev = contour + (npoints-1)*2;
    for( int i = 0; i < npoints; i++ )
    {
        const T* p = contour + i*2;
        a00 += (float)prev[0] * p[1] - (float)prev[1] * p[0];
        prev = p;
    }

    a00 *= 0.5;
    if( !oriented )
        a00 = fabs(a00);

    return a00;
}

得到两个区域的相交的面积后，便可以计算两个区域的iou

mmdetection-rotated/mmdet/csrc_custom/rotate_rect_ops.h

float area1 = region1[2] * region1[3];
float area2 = region2[2] * region2[3];
float iou = area_inter / (area1 + area2 - area_inter + 1e-8);

RRPN中的ROI pooling

(1) 将proposals分为若干个subregion,通过中心点坐标，宽高，角度计算各个subregion的左上角的坐标

(2)根据subregion的左上角坐标，通过取整计算获得该subregion中每个点再特征图上的坐标，获得每个点的特征。遍历该subreigon中的每个点的特征，采用maxpooling对该区域的特征做池化。

结果：

Alibration study:

(1) 周围的上下文的信息：

保留proposals的中心点和角度，将长宽扩大1.x倍数，来查看结果

文字检测结果展示：

来源：CSDN

作者：大梦初醒123

链接：https://blog.csdn.net/qq_32425195/article/details/104148086

标签

const

rect

pts