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pythonscipyinterpolation
neo post modern
neo post modern

Reputation: 3404

How to specify boundary behavior for SciPy's interp1d

I'm trying to set the behavior at the boundaries for SciPy's interp1d function, which according to the documentation should be possible:

Behavior at the boundary can be specified at instantiation time.

But I have not found any further information on this. The interp1d documentation does not mention it all.

So: How can I define the behavior?

Side-curiosity: What is the default boundary behavior it uses?

Edit: Examples (Assuming data points between x=0 and x=n and cubic interpolation)

I know of at least three types of boundary behavior that I would like to be able to specify:

flat: I could demand the function to flat out, in other words derivations to be zero.

f'(0)  = f'(n)  = 0
f''(0) = f''(n) = 0

cyclic: Something among the lines of

f'(0)  = f'(n)
f''(0) = f''(n)

Thus the beginning and end have same "slope".

manual: Or I could manually provide the values for the derivatives...

f'(0)  = ...
f'(n)  = ...
f''(0) = ...
f''(n) = ...

Although probably not for all four of them at the same time.

Upvotes: 6

Views: 5578

Answers (1)

rth
rth

Reputation: 11201

I believe that sentence in the documentation is badly phrased, just ignore it.

The interp1d function, at present, does not allow to specify the boundary behaviour,

  • For linear interpolation, we simply don't have a choice about it.
  • For higher order interpolation interp1d uses the spline constructor scipy.interpolate.splmake with the default parameter kind='smoothest', so again no choice on the boundary behaviour there.

On the other hand, you might want to have a look at scipy.interpolate.PiecewisePolynomial which represents the curve with piecewise polynomials and has the ability to specify the derivatives (although for all knots and not only at the boundaries).

Upvotes: 3

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