[SciPy-User] interpolate.interp1d - constructing a cubic interpolator is Slllooooooow.

[Eraldo Pomponi]

Dear Andrew,

On Thu, Jan 15, 2015 at 9:46 AM, Andrew Nelson <andyfaff at gmail.com> wrote:

> I'm intending to use interpolation in a curvefitting function.  So have
> been investigating the use of interpolate.interp1d.
> I'd prefer to use cubic interpolation but it seems to take ages:
>
> import numpy as np
>
> from scipy.interpolate import interp1d
>
> a = np.linspace(-2 * np.pi, 2 * np.pi, 1000)
>
> b = np.cos(a)
>
> %timeit interp1d(a, b)
>
> 10000 loops, best of 3: 71.6 µs per loop
>
> %timeit interp1d(a, b, kind='cubic')
>
> 1 loops, best of 3: 5.15 s per loop
>
>
> I'm wondering why it takes 5 orders of magnitude (x72000) longer to
> calculate a cubic interpolator than a linear interpolator?
>

I can reproduce your results but I cannot comment on the reason why there
exist a so big difference between the two cases. On the other hand,
following the documentation, I would go for the use of the more recent
UnivariateSpline (
http://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.interpolate.UnivariateSpline.html#scipy.interpolate.UnivariateSpline)
class that doesn't have this drawback. On my system, following your
example, I get:

%timeit spl = UnivariateSpline(a,b,k=3)

1000 loops, best of 3: 271 µs per loop


Cheers,

Eraldo
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