N-spherical coordinate system with a Cartesian coordinate system

Question

Is there an effective way to change between a Cartesian coordinate system and an n-spherical 1 ? The conversion is as follows:

Below 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 source code can be accelerated by remembering the intermediate sin , 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 losing speed until 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

My suggestion: Collect the sine in one vector, use cumprod on it, then multiply each with its cosine.