如何计算两个矩形框的IoU_mmcv box_iou_rotated

由于最近跟别人说我搞过目标检测，然后被问到一个问题，如何计算两个矩形框的IoU？  

一开始我回答说可以采用OpenCV的&运算和|运算来计算IoU，但他要我不采用OpenCV 的接口，自己写。

当时我回答的比较乱，分4种情况进行讨论2个矩形框之间的位置关系，最后他告诉我说，根本不需要讨论，采用max(),min()就可以实现。囧。

大致做法是先求出2个矩形框的交集，交集也是矩形框或空，计算相交部分面积再除以并集的面积即可。

关键在于怎么不分情况讨论就计算出交集部分的面积。

摘取SSD中bbox_util.cpp一部分代码如下：

void IntersectBBox(const NormalizedBBox& bbox1, const NormalizedBBox& bbox2,
                   NormalizedBBox* intersect_bbox) {
  if (bbox2.xmin() > bbox1.xmax() || bbox2.xmax() < bbox1.xmin() ||
      bbox2.ymin() > bbox1.ymax() || bbox2.ymax() < bbox1.ymin()) {
    // Return [0, 0, 0, 0] if there is no intersection.
    intersect_bbox->set_xmin(0);
    intersect_bbox->set_ymin(0);
    intersect_bbox->set_xmax(0);
    intersect_bbox->set_ymax(0);
  } else {
    intersect_bbox->set_xmin(std::max(bbox1.xmin(), bbox2.xmin()));
    intersect_bbox->set_ymin(std::max(bbox1.ymin(), bbox2.ymin()));
    intersect_bbox->set_xmax(std::min(bbox1.xmax(), bbox2.xmax()));
    intersect_bbox->set_ymax(std::min(bbox1.ymax(), bbox2.ymax()));
  }
}

float JaccardOverlap(const NormalizedBBox& bbox1, const NormalizedBBox& bbox2,
                     const bool normalized) {
  NormalizedBBox intersect_bbox;
  IntersectBBox(bbox1, bbox2, &intersect_bbox);
  float intersect_width, intersect_height;
  if (normalized) {
    intersect_width = intersect_bbox.xmax() - intersect_bbox.xmin();
    intersect_height = intersect_bbox.ymax() - intersect_bbox.ymin();
  } else {
    intersect_width = intersect_bbox.xmax() - intersect_bbox.xmin() + 1;
    intersect_height = intersect_bbox.ymax() - intersect_bbox.ymin() + 1;
  }
  if (intersect_width > 0 && intersect_height > 0) {
    float intersect_size = intersect_width * intersect_height;
    float bbox1_size = BBoxSize(bbox1);
    float bbox2_size = BBoxSize(bbox2);
    return intersect_size / (bbox1_size + bbox2_size - intersect_size);
  } else {
    return 0.;
  }
}

这个交集的计算采用max，min就计算出来了，真是巧妙啊。