PCL点特征直方图-分箱 [英] PCL Point Feature Histograms - binning

问题描述

如果仅使用三个角度特征(alpha，phi，theta)，则作为点特征直方图估计一部分的分箱过程将导致b^3个分箱，其中b是分箱数.

The binning process, which is part of the point feature histogram estimation, results in b^3 bins if only the three angular features (alpha, p theta) are used, where b is the number of bins.

为什么是b^3而不是b * 3?

比方说，我们考虑alpha. 特征值范围细分为b个间隔.您遍历查询点的所有邻居，并计算一个间隔内的alpha值的数量.所以您有b个垃圾箱用于alpha.当对其他两个功能重复此操作时，会得到3 * b回收箱.

Let's say we consider alpha. The feature value range is subdivided into b intervals. You iterate over all neighbors of the query point and count the amount of alpha values which lie in one interval. So you have b bins for alpha. When you repeat this for the other two features, you get 3 * b bins.

我在哪里错了?

推荐答案

为简单起见，我将首先以2D进行解释，即具有两个角度特征.在这种情况下，您将拥有b ^ 2个容器，而不是b * 2.

For simplicity, I'll first explain it in 2D, i.e. with two angular features. In that case, you would have b^2 bins, not b*2.

要素空间分为规则网格.要素根据其在2D(或3D)空间中的位置进行分类，而不是沿每个维度独立进行分类.请参见以下具有两个要素尺寸和b = 4的示例，其中该要素被装箱到标有#的单元格中:

The feature space is divided into a regular grid. Features are binned according to their position in the 2D (or 3D) space, not independently along each dimension. See the following example with two feature dimensions and b=4, where the feature is binned into the cell marked with #:

^ phi
|
+-+-+-+-+
| | | | |
+-+-+-+-+
| | | | |
+-+-+-+-+
| | | |#|
+-+-+-+-+
| | | | |
+-+-+-+-+-> alpha

该特征被合并到单元格中，其中alpha在给定间隔中，而phi在另一个间隔中.您所理解的主要区别在于，尺寸未未独立处理.每个像元都在所有维度上指定一个间隔，而不是一个. (这在3D中的工作方式相同，只是要为theta和3D网格设置另一个尺寸，而不是2D尺寸.)

The feature is binned into the cell where alpha is in a given interval AND phi in another interval. The key difference to your understanding is that the dimensions are not treated independently. Each cell specifies an interval on all the dimensions, not a single one. (This would work the same way in 3D, only that you would have another dimension for theta and a 3D grid instead of a 2D one.)

这种分箱方式在2D情况下会导致b ^ 2个分箱，因为alpha维度中的每个间隔都与phi维度中的所有间隔组合在一起，导致数字的平方，而不是加倍.添加另一个尺寸，您将得到问题的提示，而不是三重.

This way of binning results in b^2 bins for the 2D case, since each interval in the alpha dimension is combined with ALL intervals in the phi dimension, resulting in a squaring of the number, not a doubling. Add another dimension, and you get the cubing instead of the tripling, as in your question.

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