如何在“快速近似 SIFT”中旋转方向？
论文“快速近似 SIFT”（M Grabner、H Grabner、ACCV 2006） http://www.icg.tu-graz.ac.at/publications /pubobjects/mgrabner06FastApproxSIFT 展示了一种使用积分直方图从图像中提取 SIFT 描述符的改进方法。
它说“对于描述符，我们相对于方向旋转每个子补丁的中点，并计算重叠子补丁的直方图，而不对齐平方区域，但相对于主方向移动子补丁直方图。”
在本文中，可以使用积分直方图轻松计算关键点周围4*4子补丁的直方图。然而，结果直方图不会随着关键点的方向旋转。传统的SIFT需要子块中的每个像素都随着方向旋转，然后计算直方图。但论文中的这种新方法似乎可以通过“相对于主方向移动子块直方图”，在获得非旋转直方图之后进行旋转。 我不明白如何“相对于主方向移动子补丁直方图”？
我在这里引用：“对于描述符，我们旋转每个子的中点相对于方向的补丁，并计算重叠子补丁的直方图，无需对齐平方区域，而是相对于主方向移动子补丁直方图。”
例如，如果非旋转子块直方图有 8 个从 0 到 2pi 的 bin，间隔为 pi/4，则每个 bin 的值为 2,4,5,3,6,8,7,1，并且关键点的方向是pi/6，如何知道旋转直方图中8个bin的新值？
The paper "Fast Approximated SIFT" (M Grabner, H Grabner, ACCV 2006)
http://www.icg.tu-graz.ac.at/publications/pubobjects/mgrabner06FastApproxSIFT
shows an improved method to extract SIFT descriptors from image using integral histograms.
It says "for the descriptor we rotate the midpoints of each sub-patch relative to the orientation and compute the histograms of overlapping sub-patches without aligning the squared region but shifting the sub-patch histogram relative to the main orientation."
In this paper, the histogram of the 4*4 sub-patches around the keypoint can be computed easily using integral histogram. However, the result histograms are not rotated with orientation of the keypoint. The conventional SIFT needs every pixel in the sub-patches to be rotated with an orientation, then compute the histogram. But it seems this new method in the paper can make the rotation after getting the non-rotated histogram by "shifting the sub-patch histogram relative to the main orientation". I do not understand how to "shifting the sub-patch histogram relative to the main orientation"?
I quote here:"for the descriptor we rotate the midpoints of each sub-patch relative to the orientation and compute the histograms of overlapping sub-patches without aligning the squared region but shifting the sub-patch histogram relative to the main orientation."
For example if a non-rotated sub-patch histogram has 8 bins from 0 to 2pi, with an interval pi/4, each bin's value 2,4,5,3,6,8,7,1, and the orientation of keypoint is pi/6, how to know the new value of 8 bins in the rotated histogram?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
据我了解：它们将方向舍入到下一个 Pi/4 区间。这样你就可以旋转整个数组，
2 4 5 3 6 8 7 1变成_ 4 5 3 6 8 7 1 2，它表示旋转补丁的直方图。As far as I understand it: They round the orientation to the next Pi/4 interval. That way you can just rotate the entire array and
2 4 5 3 6 8 7 1becomes_ 4 5 3 6 8 7 1 2which represents the histogram of the rotated patch.