1. Passing extra arguments through scipy.integrate.quad
The quad
docs say:
If the user desires improved integration performance, then f
may be a scipy.LowLevelCallable
with one of the signatures:
double func(double x)
double func(double x, void *user_data)
double func(int n, double *xx)
double func(int n, double *xx, void *user_data)
The user_data
is the data contained in the scipy.LowLevelCallable
. In the call forms with xx
, n
is the length of the xx
array which contains xx[0] == x
and the rest of the items are numbers contained in the args
argument of quad
.
Therefore to pass an extra argument to integrand
through quad
, you are better of using the double func(int n, double *xx)
signature.
You can write a decorator to your integrand function to transform it to a LowLevelCallable
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 integrand(t, *args):
a = args[0]
return np.exp(-t/a) / t**2
def do_integrate(func, a):
"""
Integrate the given function from 1.0 to +inf with additional argument a.
"""
return si.quad(func, 1, np.inf, args=(a,))
print(do_integrate(integrand, 2.))
>>>(0.326643862324553, 1.936891932288535e-10)
Or if you don't want the decorator, create the LowLevelCallable
manually and pass it to quad
.
2. Wrapping the integrand function
I am not sure if the following would meet your requirements but you could also wrap your integrand
function to achieve the same result:
import numpy as np
from numba import cfunc
import numba.types
def get_integrand(*args):
a = args[0]
def integrand(t):
return np.exp(-t/a) / t**2
return integrand
nb_integrand = cfunc(numba.float64(numba.float64))(get_integrand(2.))
import scipy.integrate as si
def do_integrate(func):
"""
Integrate the given function from 1.0 to +inf.
"""
return si.quad(func, 1, np.inf)
print(do_integrate(get_integrand(2)))
>>>(0.326643862324553, 1.936891932288535e-10)
print(do_integrate(nb_integrand.ctypes))
>>>(0.326643862324553, 1.936891932288535e-10)
3. Casting from voidptr
to a python type
I don't think this is possible yet. From this discussion in 2016, it seems that voidptr
is only here to pass a context to a C callback.
The void * pointer case would be for APIs where foreign C code does not every try to dereference the pointer, but simply passes it back to the callback as way for the callback to retain state between calls. I don't think it is particularly important at the moment, but I wanted to raise the issue.
And trying the following:
numba.types.RawPointer('p').can_convert_to(
numba.typing.context.Context(), CPointer(numba.types.Any)))
>>>None
doesn't seem encouraging either!