Python 组合重叠矩形,将重叠与两个矩形进行比较,并用边界矩形替换两个矩形
我正在比较边界框和重叠过多的组合框。我在另一篇文章中使用了以下代码:Python 组合重叠矩形,将重叠与两个矩形进行比较,并用边界矩形替换两个矩形,python,numpy,opencv,image-processing,Python,Numpy,Opencv,Image Processing,我正在比较边界框和重叠过多的组合框。我在另一篇文章中使用了以下代码: 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 i
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")
我有两个问题:这段代码只比较从第一个框到边界框的重叠,这意味着如果我检查大框有多少重叠,百分比要比小框的重叠百分比小得多。我解决了这个问题,用这个替换了重叠功能:
这似乎奏效了
但现在我有另一个问题:如果发现两个重叠的框,代码不会用一个与两个框的外角一样大的新框替换这两个框。它只是删除其中一个框,并保留另一个框,即使它是较小的一个。
我怎样才能解决这个问题?我想用一个更大的边界框替换这些框
提前谢谢你 您是否将此代码与
findContours
函数一起使用?我使用Canny和findContours的组合:edge=cv.Canny(灰色、thrs1、thrs2、光圈大小=5)…..cnts、=cv.findContours(edge.copy()、cv.RETR\u外部、cv.CHAIN\u近似简单)
overlap = np.maximum((w * h) / area[idxs[:last]],(w * h) / area[i])