我注意到scipy.specialn阶和参数x的Bessel函数在x中jv(n,x)被向量化：
In [14]: import scipy.special as sp In [16]: sp.jv(1, range(3)) # n=1, [x=0,1,2] Out[16]: array([ 0., 0.44005059, 0.57672481])
但是，球形贝塞尔函数没有相应的矢量化形式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，球形贝塞尔函数可一遍计算N的所有阶数。因此，如果我想将n=5Bessel函数用作参数x=10，它将返回n = 1,2,3,4,5。实际上，它一次返回了jn及其派生变量：
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]))
为什么在API中存在这种不对称性，并且有人知道一个将返回球形Bessel函数矢量化或至少更快（即在cython中）的库吗？
您可以编写一个cython函数来加快计算速度，首先要做的就是获取fortran函数的地址SPHJ，这是在Python中的实现方法：
from scipy import special as sp
sphj = sp.specfun.sphj
import ctypes
addr = ctypes.pythonapi.PyCObject_AsVoidPtr(ctypes.py_object(sphj._cpointer))
然后，您可以直接在Cython中调用fortran函数，请注意，我prange()使用多核来加快计算速度：
%%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, &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
PyMem_Free(sj)
PyMem_Free(dj)
PyMem_Free(mn)
return res.base
为了进行比较，下面是sphj()在forloop中调用的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
这是10个元素的％timit结果：
x = np.linspace(1, 2, 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
这是100000个元素的结果：
x = np.linspace(1, 2, 100000)
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：
10 loops, best of 3: 44.7 ms per loop
1 loops, best of 3: 231 ms per loop
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