概述

在[14]中：将scipy.special导入sp
在[16]中：sp.jv(1,范围(3))#n = 1,[x = 0,1,2]
Out [16]：array([0.,0.44005059,0.57672481])

但是没有相应的矢量化形式的球形贝塞尔函数,sp.sph_jn：

In [19]: sp.sph_jn(1,range(3)) 

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-19-ea59d2f45497> in <module>()
----> 1 sp.sph_jn(1,range(3)) #n=1,3 value array

/home/glue/anaconda/envs/fibersim/lib/python2.7/site-packages/scipy/special/basic.pyc in sph_jn(n,z)
    262     """
    263     if not (isscalar(n) and isscalar(z)):
--> 264         raise ValueError("arguments must be scalars.")
    265     if (n != floor(n)) or (n < 0):
    266         raise ValueError("n must be a non-negative integer.")

ValueError: arguments must be scalars.

此外,球形贝塞尔函数在一次通过中计算N的所有阶数.因此,如果我想要参数x = 10的n = 5 Bessel函数,则返回n = 1,2,3,4,5.它实际上在一次传递中返回jn及其衍生物：

In [21]: sp.sph_jn(5,10)
Out[21]: 
(array([-0.05440211,0.07846694,0.07794219,-0.03949584,-0.10558929,-0.05553451]),array([-0.07846694,-0.0700955,0.05508428,0.09374053,0.0132988,-0.07226858]))

为什么API中存在这种不对称性,并且有没有人知道一个库将返回矢量化的球形贝塞尔函数,或者至少更快(即在cython中)？

from scipy import special as sp
sphj = sp.specfun.sphj

import ctypes
addr = ctypes.pythonapi.PyCObject_AsVoidPtr(ctypes.py_object(sphj._cpointer))

然后你可以直接在Cython中调用fortran函数,注意我使用prange()来使用multicore来加速计算：

%%cython -c-Ofast -c-fopenmp --link-args=-fopenmp
from cpython.mem cimport PyMem_Malloc,PyMem_Free
from cython.parallel import prange
import numpy as np
import cython
from cpython cimport PyCObject_AsVoidPtr
from scipy import special

ctypedef void (*sphj_ptr) (const int *n,const double *x,const int *nm,const double *sj,const double *dj) nogil

cdef sphj_ptr _sphj=<sphj_ptr>PyCObject_AsVoidPtr(special.specfun.sphj._cpointer)


@cython.wraparound(False)
@cython.boundscheck(False)
def cython_sphj2(int n,double[::1] x):
    cdef int count = x.shape[0]
    cdef double * sj = <double *>PyMem_Malloc(count * sizeof(double) * (n + 1))
    cdef double * dj = <double *>PyMem_Malloc(count * sizeof(double) * (n + 1))
    cdef int * mn    = <int *>PyMem_Malloc(count * sizeof(int))
    cdef double[::1] res = np.empty(count)
    cdef int i
    if count < 100:
        for i in range(x.shape[0]):
            _sphj(&n,&x[i],mn + i,sj + i*(n+1),dj + i*(n+1))
            res[i] = sj[i*(n+1) + n]    #choose the element you want here        
    else:
        for i in prange(count,nogil=True):
            _sphj(&n,dj + i*(n+1))
            res[i] = sj[i*(n+1) + n]    #choose the element you want here
    PyMem_Free(sj)
    PyMem_Free(dj)
    PyMem_Free(mn)
    return res.base

比较一下,这是在forloop中调用sphj()的Python函数：

import numpy as np
def python_sphj(n,x):
    sphj = special.specfun.sphj
    res = np.array([sphj(n,v)[1][n] for v in x])
    return res

以下是10个元素的％timit结果：

x = np.linspace(1,10)
r1 = cython_sphj2(4,x)
r2 = python_sphj(4,x)
assert np.allclose(r1,r2)
%timeit cython_sphj2(4,x)
%timeit python_sphj(4,x)

结果：

10000 loops,best of 3: 21.5 µs per loop
10000 loops,best of 3: 28.1 µs per loop

以下是100000个元素的结果：

x = np.linspace(1,100000)
r1 = cython_sphj2(4,x)

结果：

10 loops,best of 3: 44.7 ms per loop
1 loops,best of 3: 231 ms per loop

总结

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