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问题描述

有直角的改变坐标系和正球有一个?转化如下:

以下是我的code,但我想摆脱循环的:

 导入numpy的是NP
进口scipy.sparse    高清coord_transform_n(R,阿尔法):
        阿尔法:[0,\\ PI)和间最后一个之间的n-2个值[0,2 \\ PI)的
        
        X = []
        因为我在范围内(alpha.shape [0]):
            x.append(R * np.prod(np.sin(阿尔法[0:I]))* np.cos(阿尔法[I]))
        返回np.asarray(X)
    打印coord_transform_n(1,np.asarray(np.asarray([1,2])))


您原来的code能与记忆中间产品,即可以加快

 高清ct_dynamic(R,阿尔法):
    阿尔法:[0,\\ PI)和间最后一个之间的n-2个值[0,2 \\ PI)的
    
    X = np.zeros(LEN(阿尔法)+ 1)
    S = 1
    对于E,A在枚举(阿尔法):
        X [E] = S * np.cos(一)
        S * = np.sin(一)
    X [LEN(阿尔法)= S
    返回X * R

但在速度还是输给numpy的基础的方法

 高清CT(R,编曲):
    A = np.concatenate((np.array([2 * np.pi]),ARR))
    SI = np.sin(一)
    SI [0] = 1
    SI = np.cumprod(SI)
    共= np.cos(一)
    合= np.roll(CO,-1)
    返回SI *合* R>>> N = 10
>>> C = np.random.random_sample(N)* np.pi
>>>所有(克拉(1,C)== ct_dynamic(1,C))
真正>>> timeit.timeit('从__main__进口coord_transform_n为F,C,F(2.4,C),数= 10000)
2.213547945022583>>> timeit.timeit('从__main__进口ct_dynamic为F,C,F(2.4,C),数= 10000)
0.9227950572967529>>> timeit.timeit('从__main__进口CT为F,C,F(2.4,C),数= 10000)
0.5197498798370361

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])))
解决方案

Your original code can be speeded up with memorizing 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

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05-31 02:47