有没有有效的方法在笛卡尔坐标系和
n-spherical one之间进行切换？转型如下：
以下是我的代码,但我想摆脱循环：
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])))
解决方法
您可以通过记忆中间产品来加速您的原始代码,即
def ct_dynamic(r,alpha):
"""alpha: the n-2 values 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
但仍然在速度上失去基于numpy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,number=10000)
0.9227950572967529
>>> timeit.timeit('from __main__ import ct as f,number=10000)
0.5197498798370361