更快的numpy笛卡尔到球面坐标转换?

这与 Justin Peel numpy 并利用其内置的矢量化:

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,正如评论中所建议的,我已经 更改了仰角的定义 从你原来的功能。在我的机器上,用 pts = np.random.rand(3000000, 3) ,时间从76秒变为3.3秒。我没有Cython,所以无法将时间与解决方案进行比较。

下面是我为这个写的一个简单的Cython代码:

cdef extern from "math.h":
    long double sqrt(long double xx)
    long double atan2(long double a, double b)

import numpy as np
cimport numpy as np
cimport cython

ctypedef np.float64_t DTYPE_t

@cython.boundscheck(False)
@cython.wraparound(False)
def appendSpherical(np.ndarray[DTYPE_t,ndim=2] xyz):
    cdef np.ndarray[DTYPE_t,ndim=2] pts = np.empty((xyz.shape[0],6))
    cdef long double XsqPlusYsq
    for i in xrange(xyz.shape[0]):
        pts[i,0] = xyz[i,0]
        pts[i,1] = xyz[i,1]
        pts[i,2] = xyz[i,2]
        XsqPlusYsq = xyz[i,0]**2 + xyz[i,1]**2
        pts[i,3] = sqrt(XsqPlusYsq + xyz[i,2]**2)
        pts[i,4] = atan2(xyz[i,2],sqrt(XsqPlusYsq))
        pts[i,5] = atan2(xyz[i,1],xyz[i,0])
    return pts

我花了300万分从62.4秒降到1.22秒。那不算太寒酸。我相信还有其他的改进可以做。

TLDR编号: 我已经用VPython测试过了,用atan2表示theta(elev)是错误的,用 我推荐sympy1.0acos函数(它甚至不抱怨r=0的acos(z/r))。

http://mathworld.wolfram.com/SphericalCoordinates.html

如果我们把它转换成物理系统(r,θ,phi)=(r,elev,方位角),我们得到:

r = sqrt(x*x + y*y + z*z)
phi = atan2(y,x)
theta = acos(z,r)

未优化但 对的

from sympy import *
def asCartesian(rthetaphi):
    #takes list rthetaphi (single coord)
    r       = rthetaphi[0]
    theta   = rthetaphi[1]* pi/180 # to radian
    phi     = rthetaphi[2]* pi/180
    x = r * sin( theta ) * cos( phi )
    y = r * sin( theta ) * sin( phi )
    z = r * cos( theta )
    return [x,y,z]

def asSpherical(xyz):
    #takes list xyz (single coord)
    x       = xyz[0]
    y       = xyz[1]
    z       = xyz[2]
    r       =  sqrt(x*x + y*y + z*z)
    theta   =  acos(z/r)*180/ pi #to degrees
    phi     =  atan2(y,x)*180/ pi
    return [r,theta,phi]

您可以使用以下函数自己测试它:

test = asCartesian(asSpherical([-2.13091326,-0.0058279,0.83697319]))

一些象限的其他测试数据:

[[ 0.          0.          0.        ]
 [-2.13091326 -0.0058279   0.83697319]
 [ 1.82172775  1.15959835  1.09232283]
 [ 1.47554111 -0.14483833 -1.80804324]
 [-1.13940573 -1.45129967 -1.30132008]
 [ 0.33530045 -1.47780466  1.6384716 ]
 [-0.51094007  1.80408573 -2.12652707]]

test   = v.arrow(pos = (0,0,0), axis = vis_ori_ALA , shaftwidth=0.05, color=v.color.red)

为了完成前面的答案,这里有一个 Numexpr 实现(可能回退到Numpy),

import numpy as np
from numpy import arctan2, sqrt
import numexpr as ne

def cart2sph(x,y,z, ceval=ne.evaluate):
    """ x, y, z :  ndarray coordinates
        ceval: backend to use: 
              - eval :  pure Numpy
              - numexpr.evaluate:  Numexpr """
    azimuth = ceval('arctan2(y,x)')
    xy2 = ceval('x**2 + y**2')
    elevation = ceval('arctan2(z, sqrt(xy2))')
    r = eval('sqrt(xy2 + z**2)')
    return azimuth, elevation, r

对于大型数组,这允许比纯的Numpy实现提高2倍的速度,并且可以与C或Cython速度相媲美。当前的numpy解决方案(与 ceval=eval appendSpherical_np 在 @mtrw 对于大型阵列,

In [1]: xyz = np.random.rand(3000000,3)
   ...: x,y,z = xyz.T
In [2]: %timeit -n 1 appendSpherical_np(xyz)
1 loops, best of 3: 397 ms per loop
In [3]: %timeit -n 1 cart2sph(x,y,z, ceval=eval)
1 loops, best of 3: 280 ms per loop
In [4]: %timeit -n 1 cart2sph(x,y,z, ceval=ne.evaluate)
1 loops, best of 3: 145 ms per loop

尽管尺寸较小, 实际上更快,

In [5]: xyz = np.random.rand(3000,3)
...: x,y,z = xyz.T
In [6]: %timeit -n 1 appendSpherical_np(xyz)
1 loops, best of 3: 206 µs per loop
In [7]: %timeit -n 1 cart2sph(x,y,z, ceval=eval)
1 loops, best of 3: 261 µs per loop
In [8]: %timeit -n 1 cart2sph(x,y,z, ceval=ne.evaluate)
1 loops, best of 3: 271 µs per loop