n球座標系到笛卡爾座標系

Question

是否有任何改變笛卡爾座標系和n-spherical one的有效方法？轉型如下：

以下是我的代碼，但我想擺脫循環：

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

您的原始代碼可以加快與記憶中間sin產品，即

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 

但速度還是輸給numpy的基礎的方法

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

我的建議是：組裝竇一個向量，就可以使用cumprod，然後用餘弦分別乘以每一個。