python - python中的矢量化球贝塞尔函数？

Question

我注意到 scipy.special n 阶贝塞尔函数和参数 x jv(n,x) 在 x 中向量化:

在 [14] 中:将 scipy.special 导入为 sp 在 [16] 中:sp.jv(1, range(3)) # n=1, [x=0,1,2] 输出 [16]: 数组 ([ 0., 0.44005059, 0.57672481])

但是球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，球形贝塞尔函数一次计算 N 的所有阶数。因此，如果我想要参数 x=10 的 n=5 Bessel 函数，它会返回 n=1,2,3,4,5。它实际上一次性返回 jn 及其导数:

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，有没有人知道一个库可以返回向量化的球贝塞尔函数，或者至少更快(即在 cython 中)？

Answer 1

您可以编写一个 cython 函数来加速计算，您必须做的第一件事是获取 fortran 函数的地址 SPHJ，以下是在 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))

然后就可以在Cython中直接调用fortran函数了，注意我用prange()来使用多核加速计算:

%%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, &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

为了比较，这里是在 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

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