如何实现从Numba调用的C以与nquad高效集成? [英] How can you implement a C callable from Numba for efficient integration with nquad?

问题描述

我需要在python中的6D中进行数值积分.由于scipy.integrate.nquad函数运行缓慢，因此我目前正在尝试通过将被积分数定义为scipy.LowLevelCallable以及Numba来加快处理速度.

I need to do a numerical integration in 6D in python. Because the scipy.integrate.nquad function is slow I am currently trying to speed things up by defining the integrand as a scipy.LowLevelCallable with Numba.

通过复制给定

10000次循环，最好为3:每个循环128 µs

10000 loops, best of 3: 128 µs per loop

# integration with compiled function
%timeit integrate.quad(nb_integrand.ctypes, 1, np.inf)

100000个循环，每个循环中最好为3个:7.08 µs

100000 loops, best of 3: 7.08 µs per loop

当我现在想使用nquad进行此操作时，nquad文档说:

When I want to do this now with nquad, the nquad documentation says:

如果用户希望改善集成性能，则f可能是 scipy.LowLevelCallable，并带有以下签名之一:

If the user desires improved integration performance, then f may be a scipy.LowLevelCallable with one of the signatures:

double func(int n, double *xx)
double func(int n, double *xx, void *user_data)

其中n是额外参数的数量，而args是一个数组 附加参数的两倍，则xx数组包含 坐标. user_data是包含在 scipy.LowLevelCallable.

where n is the number of extra parameters and args is an array of doubles of the additional parameters, the xx array contains the coordinates. The user_data is the data contained in the scipy.LowLevelCallable.

但是以下代码给了我一个错误:

But the following code gives me an error:

from numba import cfunc
import ctypes

def func(n_arg,x):
    xe = x[0]
    xh = x[1]
    return np.sin(2*np.pi*xe)*np.sin(2*np.pi*xh)

nb_func = cfunc("float64(int64,CPointer(float64))")(func)

integrate.nquad(nb_func.ctypes, [[0,1],[0,1]], full_output=True)

错误:四元组:第一个参数是签名不正确的ctypes函数指针

error: quad: first argument is a ctypes function pointer with incorrect signature

是否可以使用numba编译可直接在代码中与nquad一起使用而无需在外部文件中定义该函数的函数?

Is it possible to compile a function with numba that can be used with nquad directly in the code and without defining the function in an external file?

非常感谢您！

推荐答案

将函数包装在 查看全文