Question

我正在努力比较边界框和组合重叠太多的框。我从另一篇文章中得到了这段代码：

def non_max_suppression_fast(boxes, overlapThresh):
   # if there are no boxes, return an empty list
   if len(boxes) == 0:
      return []

   # if the bounding boxes integers, convert them to floats --
   # this is important since we'll be doing a bunch of divisions
   if boxes.dtype.kind == "i":
      boxes = boxes.astype("float")
#  
   # initialize the list of picked indexes   
   pick = []

   # grab the coordinates of the bounding boxes
   x1 = boxes[:,0]
   y1 = boxes[:,1]
   x2 = boxes[:,2]
   y2 = boxes[:,3]

   # compute the area of the bounding boxes and sort the bounding
   # boxes by the bottom-right y-coordinate of the bounding box
   area = (x2 - x1 + 1) * (y2 - y1 + 1)
   idxs = np.argsort(y2)

   # keep looping while some indexes still remain in the indexes
   # list
   while len(idxs) > 0:
      # grab the last index in the indexes list and add the
      # index value to the list of picked indexes
      last = len(idxs) - 1
      i = idxs[last]
      pick.append(i)

      # find the largest (x, y) coordinates for the start of
      # the bounding box and the smallest (x, y) coordinates
      # for the end of the bounding box
      xx1 = np.maximum(x1[i], x1[idxs[:last]])
      yy1 = np.maximum(y1[i], y1[idxs[:last]])
      xx2 = np.minimum(x2[i], x2[idxs[:last]])
      yy2 = np.minimum(y2[i], y2[idxs[:last]])

      # compute the width and height of the bounding box
      w = np.maximum(0, xx2 - xx1 + 1)
      h = np.maximum(0, yy2 - yy1 + 1)

      # compute the ratio of overlap
      overlap = (w * h) / area[idxs[:last]]

      # delete all indexes from the index list that have
      idxs = np.delete(idxs, np.concatenate(([last],
         np.where(overlap > overlapThresh)[0])))

   # return only the bounding boxes that were picked using the
   # integer data type
   return boxes[pick].astype("int")

如何修改此代码以用两个边界框周围的矩形替换两个重叠的框？此代码仅删除其中一个框并保留另一个框。

提前谢谢！

Answer 1

不需要区域。边界矩形是坐标点的最小值/最大值。如果我们假设x1 < x2和y1 < y2，那么我们可以将x1中的min用作边界框x1，将y1中的min作为边界框{{1}对于边界框y1，x2，所有y2，x2中的最大值同样最高。

对于更复杂的边界形状，请查看凸包算法。

OpenCV：组合/合并重叠的边界框，由周围的边界框替换

1 个答案: