Javascript 查找列表中未包含在列表中的其他矩形中的矩形的最快方法

给定一个[not rotated]矩形列表，从该列表中获取任何其他矩形都不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}]

---

我使用的是Python，我使用的函数将以矩形列表作为输入。 比如：

[[0,0,10,10]，[0,0,11,11]，[5100,5100]

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")

---

我使用的是Python，我使用的函数将以矩形列表作为输入。 比如：

[[0,0,10,10]，[0,0,11,11]，[5100,5100]

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")

---

澄清一下：它看起来像是“完全在……之上或之内”。。。例如，如果两个矩形具有相同的坐标，它们将相互“包含”对方，并且两个矩形都不在输出列表中。正确吗？另外，您只是在寻找一个算法的名称（现实世界中最快的还是最佳大O最快的）？如果您想要代码（JavaScript或Python？），到目前为止您尝试了什么？我在这里没有看到任何研究成果，这听起来像是一个家庭作业或面试问题。需要澄清的是，可能的重复：它看起来像是“通过”方式“完全包含在”。。。例如，如果两个矩形具有相同的坐标，它们将相互“包含”对方，并且两个矩形都不在输出列表中。正确吗？另外，您只是在寻找一个算法的名称（现实世界中最快的还是最佳大O最快的）？如果您想要代码（JavaScript或Python？），到目前为止您尝试了什么？我在这里没有看到任何研究成果，这听起来像是一个家庭作业或面试问题。这里可能重复一些讨论：这里的一些讨论：