Question

给定一个[未旋转的]矩形列表，获取该列表中任何其他矩形未包含的矩形列表的最快方法是什么？

例如：

[{x1:0, y1:0, x2:10, y2:10}, {x1:0, y1:0, x2:11, y2: 11}, {x1:5, y1:100, x2:5, y2:100}]

将返回：

[{x1:0, y1:0, x2:11, y2: 11}, {x1:5, y1:100, x2:5, y2:100}]

Answer 1

我使用的是Python，我使用的函数将矩形列表作为输入。像：

[[0,0, 10, 10], [0, 0, 11, 11], [5, 100, 5, 100]]

def non_max_suppression_fast(boxes, overlapThresh=1):
   # 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 == 1)[0])))

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

查找列表中的矩形的最快方法（列表

1 个答案: