n-sphere coordinate system to Cartesian coordinate system

arrays, numpy, python

Solution

Your original code can be sped up by memoizing intermediate `sin` product, i.e.

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

Problem

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]))) ```

Original source