[SciPy-Dev] vectorized scipy.integrate.quad

[Nathan Woods]

The API part is simple. Something like, 
try: 
    func([xes])
except TypeError:
    func = lambda xlist: [func x for x in xlist]

A much smarter way would use a class instead of a lambda, but hopefully this conveys the idea. 

> On Sep 28, 2015, at 4:28 AM, Christoph Groth <christoph at grothesque.org> wrote:
> 
> Nathan Woods wrote:
> 
>> Unless I'm mistaken, though, integration of a python function seems
>> unlikely to take advantage of a vectorized C integrator, so the speed
>> ups would be limited to those calling quad with ctypes functions.
> 
> What I proposed is making it possible that the integrand gets called
> with a sequence of x-values instead of a single x-value at a time.  It
> would be then up to the user-specified integrand function to perform the
> computation.  I can think of at least three ways that can provide speed-ups:
> 
> • vectorized evaluation using vectorized numpy operations (it should be faster
>  to call np.sin(x) once with x being an array of 21 elements than to
>  call math.sin(x) 21 times).
> 
> • concurrent.futures
> 
> • MPI
> 
> Is there a problem that I fail to see?
> 
>> Also, I think breaking the API would be a bad idea, but you could, for
>> instance, allow quad to accept both numbers and sequences, and then
>> wrap the number in an array before sending it on to the
>> integrator. This would preserve the API, while allowing the desired
>> vectorization.
> 
> Of course breaking the API is a bad idea.  That’s why I have in mind
> keeping scipy.integrate.quad as it is, and adding a sibling that could
> be called scipy.integrate.vectorized_quad.  If the algorithm of
> vectorized_quad remains the same (as planned), quad could be eventually
> implemented as a thin wrapper around vectorized_quad.  This would allow
> to remove the unomodified QUADPACK from scipy.
> 
> I am sorry, I do not understand your proposal for extending the API in a
> backwards-compatible way without introducing a new function.  Could you
> please explain it in more detail?
> 
> Christoph
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