How can you implement a C callable from Numba for efficient integration with nquad?

问题

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.

I was able to do this in 1D with the scipy.integrate.quad by replicating the example given here:

import numpy as np
from numba import cfunc
from scipy import integrate

def integrand(t):
    return np.exp(-t) / t**2

nb_integrand = cfunc("float64(float64)")(integrand)

# regular integration
%timeit integrate.quad(integrand, 1, np.inf)

10000 loops, best of 3: 128 µs per loop

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

100000 loops, best of 3: 7.08 µs per loop

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

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)
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.

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)

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

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?

Thank you very much in advance!

回答1:

Wrapping the function in a scipy.LowLevelCallable makes nquad happy:

si.nquad(sp.LowLevelCallable(nb_func.ctypes), [[0,1],[0,1]], full_output=True)
# (-2.3958561404687756e-19, 7.002641250699693e-15, {'neval': 1323})

回答2:

The signature of the function you pass to nquad should be double func(int n, double *xx). You can create a decorator for your function func like so:

import numpy as np
import scipy.integrate as si
import numba
from numba import cfunc
from numba.types import intc, CPointer, float64
from scipy import LowLevelCallable


def jit_integrand_function(integrand_function):
    jitted_function = numba.jit(integrand_function, nopython=True)

    @cfunc(float64(intc, CPointer(float64)))
    def wrapped(n, xx):
        return jitted_function(xx[0], xx[1])
    return LowLevelCallable(wrapped.ctypes)

@jit_integrand_function
def func(xe, xh):
    return np.sin(2*np.pi*xe)*np.sin(2*np.pi*xh)

print(si.nquad(func, [[0,1],[0,1]], full_output=True))
>>>(-2.3958561404687756e-19, 7.002641250699693e-15, {'neval': 1323})

来源：https://stackoverflow.com/questions/45823212/how-can-you-implement-a-c-callable-from-numba-for-efficient-integration-with-nqu

标签

python

numpy

scipy

numba