n球坐标系到笛卡尔坐标系

Question

Is there any efficient way of changing between Cartesian coordinate system and n-spherical one ? The transformation is as follows:

The following is my code but I want to get rid of the loop:

import numpy as np
import scipy.sparse

    def coord_transform_n(r,alpha):
        """alpha: the n-2 values between [0,\pi) and last one between [0,2\pi)
        """
        x=[]
        for i in range(alpha.shape[0]):
            x.append(r*np.prod(np.sin(alpha[0:i]))*np.cos(alpha[i]))
        return np.asarray(x)
    print coord_transform_n(1,np.asarray(np.asarray([1,2])))

Answer 1

Your original code can be sped up by memoizing intermediate sin product, ie

def ct_dynamic(r, alpha):
    """alpha: the n-2 values between [0,\pi) and last one between [0,2\pi)
    """
    x = np.zeros(len(alpha) + 1)
    s = 1
    for e, a in enumerate(alpha):
        x[e] = s*np.cos(a)
        s *= np.sin(a)
    x[len(alpha)] = s
    return x*r

But still loses in speed to numpy based approach

def ct(r, arr):
    a = np.concatenate((np.array([2*np.pi]), arr))
    si = np.sin(a)
    si[0] = 1
    si = np.cumprod(si)
    co = np.cos(a)
    co = np.roll(co, -1)
    return si*co*r

>>> n = 10
>>> c = np.random.random_sample(n)*np.pi
>>> all(ct(1,c) == ct_dynamic(1,c))
True

>>> timeit.timeit('from __main__ import coord_transform_n as f, c; f(2.4,c)', number=10000)
2.213547945022583

>>> timeit.timeit('from __main__ import ct_dynamic as f, c; f(2.4,c)', number=10000)
0.9227950572967529

>>> timeit.timeit('from __main__ import ct as f, c; f(2.4,c)', number=10000)
0.5197498798370361

Answer 2

我的建议：在一个载体中组装窦，在其上使用cumprod，然后将每个与cosinus相乘（预先组装在另一个载体中）。