我有一个来自3轴accellerometer(XYZ)的300万个数据点的数组,我想在包含等效球面坐标(r,theta,phi)的数组中添加3列。以下代码有效,但似乎太慢了。我怎样才能做得更好?
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)))
答案 0 :(得分:31)
这类似于Justin Peel的答案,但只使用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那样,我从原始函数更改了仰角的定义。在我的机器上,使用pts = np.random.rand(3000000, 3)
进行测试,时间从76秒变为3.3秒。我没有Cython,所以我无法将时间与该解决方案进行比较。
答案 1 :(得分:11)
这是我为此编写的快速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3,000,000分将时间从62.4秒降低到1.22秒。那不是太破旧。我确信还有其他一些改进。
答案 2 :(得分:6)
!上面的所有代码中仍然存在错误..这是谷歌的最高结果.. TLDR: 我用VPython对此进行了测试,使用atan2进行theta(elev)是错误的,使用 ACOS!它对于phi(azim)是正确的。 我推荐使用sympy1.0 acos功能(它甚至不会抱怨acos(z / r),r = 0)。
http://mathworld.wolfram.com/SphericalCoordinates.html
如果我们将其转换为物理系统(r,theta,phi)=(r,elev,azimuth),我们有:
r = sqrt(x*x + y*y + z*z)
phi = atan2(y,x)
theta = acos(z,r)
右手物理系统的非优化但正确代码:
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]
您可以使用以下功能自行测试:
test = asCartesian(asSpherical([-2.13091326,-0.0058279,0.83697319]))
某些象限的其他一些测试数据:
[[ 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]]
我还使用VPython轻松地可视化矢量:
test = v.arrow(pos = (0,0,0), axis = vis_ori_ALA , shaftwidth=0.05, color=v.color.red)
答案 3 :(得分:5)
要完成之前的答案,这里有一个Numexpr实现(可能会回退到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
对于大型阵列,与纯Numpy实现相比,这可以使速度提高2倍,并且可以与C或Cython速度相媲美。当前numpy解决方案(与ceval=eval
参数一起使用时)也比@mtrw对大数组大小的appendSpherical_np
函数快25%,
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
虽然对于较小的尺寸,appendSpherical_np
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
答案 4 :(得分:0)
Octave具有一些用于坐标转换的内置功能,可使用oct2py软件包进行访问,以将笛卡尔坐标中的numpy数组转换为球面或极坐标(并返回):
from oct2py import octave
xyz = np.random.rand(3000000,3)
%timeit thetaphir = octave.cart2sph(xyz)
724 ms ± 206 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)