[SciPy-User] B-spline basis functions?

Nathaniel Smith njs@pobox....
Wed Aug 1 11:53:14 CDT 2012


On Tue, Jul 31, 2012 at 4:46 PM, Charles R Harris
<charlesr.harris@gmail.com> wrote:
> On Tue, Jul 31, 2012 at 9:37 AM, Nathaniel Smith <njs@pobox.com> wrote:
>>
>> Hi all,
>>
>> I'd like to be able to do spline regression in patsy[1], which means
>> that I need to be able to compute b-spline basis functions. I am not
>> an initiate into the mysteries of practical spline computations, but I
>> *think* the stuff in scipy.signal is not quite usable as is, because
>> it's focused on doing interpolation directly rather than exposing the
>> basis functions themselves?
>>
>> Specifically, to achieve feature parity with R [2], I need to be able to
>> take
>>   - an arbitrary order
>>   - an arbitrary collection of knot positions (which may be irregularly
>> spaced)
>>   - a vector x of points at which to evaluate the basis functions
>> and spit out the value of each spline basis function evaluated at each
>> point in the x vector.
>>
>> It looks like scipy.signal.bspline *might* be useful, but I can't
>> quite tell? Or alternatively someone might have some code lying around
>> to do this already?
>>
>> Basically I have a copy of Schumaker here and I'm hoping someone will
>> save me from having to read it :-).
>>
>
> I have this floating around
[...]
>         v[i] = spl.splev(x, (knots, d[i], deg))

This looks fabulous, thank you! From a quick look it seems to be
producing numerically identical results to R's spline.des().

The bit with the coefficient vectors like [0, 0, 0, 1, 0] worries me a
bit -- do you know if it's producing the entire spline basis
internally on every call, and then throwing away most of them when
forming the inner product? If so then in practice it's probably not a
show-stopper, but it would be a pointless factor-of-len(knots)
increase in running time for no reason.

-n


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