我正在测试Fast NMS algorithm by Malisiewicz et al。我在运行示例时注意到,在一个例子中,如果我输入两个没有重叠的特定框,并且IoU阈值低于大约0.75,那么仍然会抑制一个框。在
我误解NMS了吗?我认为如果它们之间没有重叠,就不应该丢弃任何框,不管IoU阈值设置在哪里。在
示例:
import numpy as np
def non_max_suppression_fast(boxes, overlapThresh):
# if there are no boxes, return an empty list
if len(boxes) == 0:
return []
# 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
return boxes[pick]
# Two test boxes
#xmin,ymin,xmax,ymax
boxes = np.vstack([[0.3, 0.2, 0.4, 0.5],
[0.1, 0.1, 0.2, 0.2]])
# no box suppression
print(non_max_suppression_fast(boxes, overlapThresh=.75))
# one box is suppressed
print(non_max_suppression_fast(boxes, overlapThresh=.74))
您的输入测试用例是不合法的,参数
boxes
期望框坐标为绝对格式,例如像素坐标。在您可以注意到,当计算所有长方体的面积时
+1
是一个添加的像素,确保area
是方框实际占用的像素数。在试试这个:
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