[Numpy-discussion] How to make scipy.interpolate give a an extrapolated result beyond the input range?

[josef.pktd at gmail.com]

On Fri, Apr 30, 2010 at 12:18 PM, Salim, Fadhley (CA-CIB)
<fadhley.salim at ca-cib.com> wrote:
> I'm trying to port a program which currently uses a hand-rolled C++
> interpolator (developed by a mathematician colleage) over to use the
> interpolators provided by scipy. I'd like to use or wrap the scipy
> interpolator so that it's behavior is as close as possible behavior to
> our old interpolator.
>
> A key difference between the two functions is that in our original
> interpolator - if the input value is above or below the input range, our
> original interpolator will extrapolate the result. If you try this with
> the scipy interpolator it raises a ValueError.  Consider this program as
> an example:
>
> # EXAMPLE
> import numpy as np
> from scipy import interpolate
>
> x = np.arange(0,10)
> y = np.exp(-x/3.0)
> f = interpolate.interp1d(x, y)
>
> print f(9)
> print f(11) ##### Causes ValueError, because it's greater than max(x) #
> END OF EXAMPLE
>
> In the example above, I'd like the last line not to raise a ValueError,
> but to return a value calculated from the gradient of the line between
> f(x[-2]) and f(x[-1]).
>
> Is there a sensible way to make it so that instead of crashing, the
> final line will simply do a linear extrapolate, continuing the gradients
> defined by the first and last pairs of input data-points to infinity?
>
> I know that this is a simple enough function to write myself, however
> I'd rather not re-invent the wheel, especially as if I wanted to
> introduce new basic math functions into our library they would need to
> be validated by a number of gate-keepers before they were permitted into
> our library!
>
> I'm on Python 2.4, scipy 0.7 on Windows XP, 32bit
>
> Incidentally, I have seen this tutorial which has a "left" and "right"
> argument on the interpolator. This does not seem to exist on any version
> of the interp1d function which I can use on Windows Python 2.4 - can
> anybody speculate which version of Scipy this tutorial is intended for?
> http://projects.scipy.org/scipy/browser/branches/Interpolate1D/docs/tuto
> rial.rst?rev=4591

Interesting question, after renaming, the branch is at
http://projects.scipy.org/scipy/browser/branches/interpolate?rev=

Does anyone know what the purpose and status of this branch is?

Josef


>
> Sal
>
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