问题
The binning process, which is part of the point feature histogram estimation, results in b^3 bins if only the three angular features (alpha, phi, theta) are used, where b is the number of bins.
Why is it b^3 and not b * 3?
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.
Where am I wrong?
回答1:
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.
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
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.)
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.
来源：https://stackoverflow.com/questions/31608611/pcl-point-feature-histograms-binning