[SciPy-User] scipy.interpolate.rbf sensitive to input noise ?
Fri Feb 19 10:26:41 CST 2010
On Feb 18, 4:48 pm, Robert Kern <robert.k...@gmail.com> wrote:
> On Thu, Feb 18, 2010 at 09:19, denis <denis-bz...@t-online.de> wrote:
> > Running rbf on 100 1d random.uniform input points
> > (after changing the linalg.solve in rbf.py to lstsq)
> Why would you do this? This does not magically turn Rbf interpolation
> into a smoothing approximation. Use the "smooth" keyword to specify a
> smoothing parameter.
i'm interpolating y = np.sin(x) + np.random.normal( 0, .1 ), not
The idea was to look at gaussian vs thin-plate;
true, that 1d snippet doesn't say much,
but my 2d plots were so noisy that I went down to 1d.
Use "smooth" ? rbf.py just does
self.A = self._function(r) - eye(self.N)*self.smooth
and you don't know A .
Bytheway (googling), http://www.farfieldtechnology.com/products/toolbox
have an O(N lnN) FastRBF TM in matlab, $
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