使用numpy中的旋转矩阵有效旋转一组点 [英] Efficiently rotate a set of points with a rotation matrix in numpy

问题描述

我有一个3D点列表，它们存储在形状为(N,3)的numpy数组A和形状为(3,3)的旋转矩阵R中.我想就地为A中的每个点x计算R.x的点积.我可以天真地做到这一点:

I have a list of 3D points stored in numpy array A with shape (N,3) and a rotation matrix R with shape (3,3). I'd like to compute the dot product of R.x for each point x in A in-place. Naively I can do this:

for n in xrange(N):
    A[n,:] = dot(R, A[n,:]) 

是否可以通过本机numpy调用将其向量化?如果重要的话，N大约是几千.

Is there a way to vectorize this with a native numpy call? If it matters, N is on order of a couple thousand.

推荐答案

您可以将A与旋转矩阵的转置相乘:

You can multiply A with the transpose of the rotation matrix:

A = dot(A, R.T)

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