# [SciPy-user] SciPy cubic interpolation coefficients

josef.pktd@gmai... josef.pktd@gmai...
Sun Jun 21 16:28:15 CDT 2009

```On Sun, Jun 21, 2009 at 4:13 PM, Celvin<read.beyond.data@gmx.net> wrote:
> josef.pktd@gmail.com @ Jun, 21, 2009 09:23PM
>
>> Did you try
>> UnivariateSpline.get_coeffs()
>
>> UnivariateSpline.get_knots()
>
> >From my experiments with interpolate splines, I would think this
>> provides what you want. But the documentation is still a bit sparse.
> Yes, I did try using UnivariateSpline. Apart from being "more off"
> than using splrep with a smoothing factor of 0, get_coeffs() also
> returns an 1-d array, with far too few coefficients.
>
> Assuming a signal with about 390 data points, I also expect about
> 390 coefficients (which is consistent with what I get using
> signal.cspline1d for example, but I need 4 arrays, not just one).
>
> I would expect coefficients to be a list of shape
>
> coeff = [ [a0, b0, c0, d0],
>          [a1, b1, c1, d1],
>          [a2, b1, c2, d2],
>          ...
>          ...
>          [an, bn, cn, dn],
>        ]
>
> but I am only able to obtain 1-d array of coefficients, no matter what
> function or module I use.
>
> http://www.physics.utah.edu/~detar/phys6720/handouts/cubic_spline/cubic_spline/node1.html
>
> ...is basically what I'm looking for. I used to do the matrix
> calculations for Si(x) myself using C++, but I'd prefer to somehow get the
> coefficients using numpy/scipy and not use an extension.
>
> Any further ideas?

interpolate.spalde   gives you the derivatives of as a 1 by 4 array
for each point
given it's a cubic function, it should be possible to recover the coefficients

something like

coeff = np.vstack(interpolate.spalde(x,tck))*[1,1,0.5,1/3.]

I have no idea yet, whether this actually does what you want.
If you find out how it works, it would be good to get an example since
figuring out the splines with the current documentation is pretty
difficult.

Josef
```