更快的numpy直角坐标到球坐标的转换? [英] Faster numpy cartesian to spherical coordinate conversion?

问题描述

我有一个来自3轴加速度计(XYZ)的300万个数据点的数组，我想向该数组添加3列，其中包含等效的球坐标(r，theta，phi).以下代码有效，但似乎太慢了.我该怎么做?

I have an array of 3 million data points from a 3-axiz accellerometer (XYZ), and I want to add 3 columns to the array containing the equivalent spherical coordinates (r, theta, phi). The following code works, but seems way too slow. How can I do better?

import numpy as np
import math as m

def cart2sph(x,y,z):
    XsqPlusYsq = x**2 + y**2
    r = m.sqrt(XsqPlusYsq + z**2)               # r
    elev = m.atan2(z,m.sqrt(XsqPlusYsq))     # theta
    az = m.atan2(y,x)                           # phi
    return r, elev, az

def cart2sphA(pts):
    return np.array([cart2sph(x,y,z) for x,y,z in pts])

def appendSpherical(xyz):
    np.hstack((xyz, cart2sphA(xyz)))

推荐答案

这类似于

This is similar to Justin Peel's answer, but using just numpy and taking advantage of its built-in vectorization:

import numpy as np

def appendSpherical_np(xyz):
    ptsnew = np.hstack((xyz, np.zeros(xyz.shape)))
    xy = xyz[:,0]**2 + xyz[:,1]**2
    ptsnew[:,3] = np.sqrt(xy + xyz[:,2]**2)
    ptsnew[:,4] = np.arctan2(np.sqrt(xy), xyz[:,2]) # for elevation angle defined from Z-axis down
    #ptsnew[:,4] = np.arctan2(xyz[:,2], np.sqrt(xy)) # for elevation angle defined from XY-plane up
    ptsnew[:,5] = np.arctan2(xyz[:,1], xyz[:,0])
    return ptsnew

请注意，正如评论中所建议，我已经从您的原始功能中更改了仰角的定义.在我的计算机上，使用pts = np.random.rand(3000000, 3)进行测试，时间从76秒缩短到3.3秒.我没有Cython，因此无法将计时与该解决方案进行比较.

Note that, as suggested in the comments, I've changed the definition of elevation angle from your original function. On my machine, testing with pts = np.random.rand(3000000, 3), the time went from 76 seconds to 3.3 seconds. I don't have Cython so I wasn't able to compare the timing with that solution.

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