在笛卡尔坐标系和n-spherical one之间有任何有效的转换方法吗?转换如下:
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])))
最佳答案
通过记住中间sin
产品(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
但是仍然会失去基于numpy的方法的速度
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
关于python - n球坐标系到笛卡尔坐标系,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/20133318/