# scipy.interpolate.rbf sensitive to input noise ?

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## scipy.interpolate.rbf sensitive to input noise ?

 A followup to the 15feb thread "creating a 3D surface plot from collected data": the doc for scipy.interpolate.rbf says that 3 of the 7 radial functions use a scale paramater `epsilon`, 4 do not. Odd -- how can some be scale-free ? Running rbf on 100 1d random.uniform input points (after changing the linalg.solve in rbf.py to lstsq) shows that all but linear and gaussian are noisy, giving interpolants way outside the input range:     N: 100  ngrid: 50  sigma: 0.1     # min  max  max |delta|  Rbf      -1.0  1.0   0.5  gaussian      -1.0  1.2   0.3  linear      -1.4  2.1   2.5  thin-plate      -1.0  2.2   3.0  inverse multiquadric      -1.0  2.4   3.2  multiquadric      -2.1 12.8   7.4  quintic     -10.6  7.0  11.7  cubic Should Rbf be moved off to noisymethods/...  until experts can look it over, or have I done something stupid ? (A back-of-the-envelope method for choosing epsilon aka r0 would be nice, and a rough noise amplification model would be nice too.) Also, the doc should have a link to http://www.scipy.org/Cookbook/Matplotlib/Gridding_irregularly_spaced_dataand should say that the initial rbf() is N^2. cheers   -- denis
# from http://scipy.org/Cookbook/RadialBasisFunctions    import sys     import numpy as np     from rbf import Rbf  # lstsq     # from scipy.interpolate import Rbf     Rbfuncs = ( 'gaussian', 'linear', 'thin-plate', 'inverse multiquadric',         'multiquadric', 'quintic', 'cubic', )     N = 100     ngrid = 50     sigma = .1     exec "\n".join( sys.argv[1:] )  # N=     np.random.seed(1)     np.set_printoptions( 2, threshold=50, edgeitems=5, suppress=True )  # .2f     x = np.sort( np.random.uniform( 0, 10, N+1 ))     y = np.sin(x) + np.random.normal( 0, sigma, x.shape )     xi = np.linspace( 0, 10, ngrid+1 )     print "N: %d  ngrid: %d  sigma: %.2g" % (N, ngrid, sigma)     print "# min  max  max |delta|  Rbf"     for func in Rbfuncs:         rbf = Rbf( x, y, function=func )         fi = rbf(xi)         maxdiff = np.amax( np.abs( np.diff( fi )))         print "%5.1f %4.1f  %4.1f  %s" % (             fi.min(), fi.max(), maxdiff, func )
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Thu, Feb 18, 2010 at 09:19, denis <[hidden email]> wrote: > A followup to the 15feb thread "creating a 3D surface plot from > collected data": > the doc for scipy.interpolate.rbf says that 3 of the 7 radial > functions > use a scale paramater `epsilon`, 4 do not. > Odd -- how can some be scale-free ? > 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. > shows that all but linear and gaussian are noisy, > giving interpolants way outside the input range: > >    N: 100  ngrid: 50  sigma: 0.1 >    # min  max  max |delta|  Rbf >     -1.0  1.0   0.5  gaussian >     -1.0  1.2   0.3  linear >     -1.4  2.1   2.5  thin-plate >     -1.0  2.2   3.0  inverse multiquadric >     -1.0  2.4   3.2  multiquadric >     -2.1 12.8   7.4  quintic >    -10.6  7.0  11.7  cubic > > Should Rbf be moved off to noisymethods/...  until experts can look it > over, > or have I done something stupid ? I'm not sure what else you would expect from interpolating uniform random noise. Many interpolation methods tend to give such oscillations on such data. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Feb 18, 4:48 pm, Robert Kern <[hidden email]> wrote: > On Thu, Feb 18, 2010 at 09:19, denis <[hidden email]> 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. Robert,   i'm interpolating  y = np.sin(x) + np.random.normal( 0,  .1 ), not random noise. 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, \$ cheers   -- denis _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Fri, Feb 19, 2010 at 10:26, denis <[hidden email]> wrote: > On Feb 18, 4:48 pm, Robert Kern <[hidden email]> wrote: >> On Thu, Feb 18, 2010 at 09:19, denis <[hidden email]> 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. > > Robert, >  i'm interpolating  y = np.sin(x) + np.random.normal( 0,  .1 ), not > random noise. Okay. You said otherwise. It would help if you simply attached the code that you are using. > 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 . I have no idea what you mean by the last part of your sentence. Did you actually try using the smooth keyword? What were your results? > Bytheway (googling), http://www.farfieldtechnology.com/products/toolbox> ... > have an O(N lnN) FastRBF TM in matlab, \$ Okay. And? -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 In reply to this post by denis-bz-gg On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: > On Feb 18, 4:48 pm, Robert Kern <[hidden email]> wrote: >> On Thu, Feb 18, 2010 at 09:19, denis <[hidden email]> 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. > > Robert, >  i'm interpolating  y = np.sin(x) + np.random.normal( 0,  .1 ), not > random noise. > 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 . rbf is a global interpolator, and as Robert said for some cases you get very unstable results. with a large number of points self._function(r)  is not very well behaved and you need the penalization to make it smoother. In a variation of your example, I printed out the eigenvalues for self.A and they don't look nice without penalization. There are still some strange things that I don't understand with the behavior rbf, but when I looked at it in the past, I got better results by using only local information, i.e. fit rbf only to a neighborhood of the points, although I think thinplate did pretty well also with a larger number of points. Josef > > Bytheway (googling), http://www.farfieldtechnology.com/products/toolbox> ... > have an O(N lnN) FastRBF TM in matlab, \$ > > cheers >  -- denis > _______________________________________________ > SciPy-User mailing list > [hidden email] > http://mail.scipy.org/mailman/listinfo/scipy-user> _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Feb 19, 5:41 pm, [hidden email] wrote: > On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: > > Use "smooth" ? rbf.py just does > >    self.A = self._function(r) - eye(self.N)*self.smooth > > and you don't know A . That's a line from scipy/interpolate/rbf.py: it solves     (A - smooth*I)x = b  instead of     Ax = b Looks to me like a hack for A singular, plus the caller doesn't know A anyway. I had looked at the eigenvalues too (100 points, 2d, like test_rbf.py): gauss     : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1.2e-10 ... 44 linear    : -1.8e+02 ... -0.027 5.6e+02 thinplate : -4.7e+03 ... -5.4e+02 0.0032 So gauss is singular => don't do that then. (Odd that linalg.solve didn't LinAlgError though.) > rbf is a global interpolator, and as Robert said for some cases you > get very unstable results. > > with a large number of points self._function(r)  is not very well > behaved and you need the penalization to make it smoother. In a > variation of your example, I printed out the eigenvalues for self.A > and they don't look nice without penalization. > There are still some strange things that I don't understand with the > behavior rbf, but when I looked at it in the past, I got better > results by using only local information, i.e. fit rbf only to a > neighborhood of the points, although I think thinplate did pretty well > also with a larger number of points. Yes -- but you don't know which points are local without some k-nearest-neighbor algorithm ? in 2d, might as well triangulate. Gaussian weights nearer points automatically BUT rbf gauss looks to me singular / unusable. cheers   -- denis _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Mon, Feb 22, 2010 at 7:35 AM, denis <[hidden email]> wrote: > On Feb 19, 5:41 pm, [hidden email] wrote: >> On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: > >> > Use "smooth" ? rbf.py just does >> >    self.A = self._function(r) - eye(self.N)*self.smooth >> > and you don't know A . > > That's a line from scipy/interpolate/rbf.py: it solves >    (A - smooth*I)x = b  instead of >    Ax = b > Looks to me like a hack for A singular, plus the caller doesn't know A > anyway. It's not a hack it's a requirement, ill-posed inverse problems need penalization, this is just Ridge or Tychonov with a kernel matrix. A is (nobs,nobs) and the number of features is always the same as the number of observations that are used. (I was looking at "Kernel Ridge Regression" and "Gaussian Process" before I realized that rbf is essentially the same, at least for 'gauss') I don't know anything about thinplate. I still don't understand what you mean with "the caller doesn't know A".  A is the internally calculated kernel matrix (if I remember correctly.) > > I had looked at the eigenvalues too (100 points, 2d, like > test_rbf.py): > gauss     : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > 1.2e-10 ... 44 > linear    : -1.8e+02 ... -0.027 5.6e+02 > thinplate : -4.7e+03 ... -5.4e+02 0.0032 > > So gauss is singular => don't do that then. > (Odd that linalg.solve didn't LinAlgError though.) > >> rbf is a global interpolator, and as Robert said for some cases you >> get very unstable results. >> >> with a large number of points self._function(r)  is not very well >> behaved and you need the penalization to make it smoother. In a >> variation of your example, I printed out the eigenvalues for self.A >> and they don't look nice without penalization. >> There are still some strange things that I don't understand with the >> behavior rbf, but when I looked at it in the past, I got better >> results by using only local information, i.e. fit rbf only to a >> neighborhood of the points, although I think thinplate did pretty well >> also with a larger number of points. > > Yes -- but you don't know which points are local > without some k-nearest-neighbor algorithm ? in 2d, might as well > triangulate. I used scipy.spatial for this, or if there are not too many points use the full dense distance matrix setting far away points to zero (kind of). My example script should be on the mailing list. rbf works for nd not just 1d or 2d, I think only the number of observations is important, not the number of features. > Gaussian weights nearer points automatically BUT rbf gauss looks to me > singular / unusable. unusable without large enough smooth>0 , e.g. I tried smooth=0.4 in bad cases. Cheers, Josef > > cheers >  -- denis > _______________________________________________ > SciPy-User mailing list > [hidden email] > http://mail.scipy.org/mailman/listinfo/scipy-user> _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Feb 22, 3:08 pm, [hidden email] wrote: > On Mon, Feb 22, 2010 at 7:35 AM, denis <[hidden email]> wrote: > > On Feb 19, 5:41 pm, [hidden email] wrote: > >> On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: > > >> > Use "smooth" ? rbf.py just does > >> >    self.A = self._function(r) - eye(self.N)*self.smooth > >> > and you don't know A . > > > That's a line from scipy/interpolate/rbf.py: it solves > >    (A - smooth*I)x = b  instead of > >    Ax = b > > Looks to me like a hack for A singular, plus the caller doesn't know A > > anyway. > It's not a hack it's a requirement, ill-posed inverse problems need OK, I must be wrong; but (sorry, I'm ignorant) how can (A - smooth) penalize ? For gauss the eigenvalues are >= 0, many 0, so we're shifting them negative ?? Or is it a simple sign error, A + smooth ? > penalization, this is just Ridge or Tychonov with a kernel matrix. A > is (nobs,nobs) and the number of features is always the same as the > number of observations that are used. (I was looking at "Kernel Ridge > Regression" and "Gaussian Process" before I realized that rbf is > essentially the same, at least for 'gauss') > I don't know anything about thinplate. > > I still don't understand what you mean with "the caller doesn't know > A".  A is the internally calculated kernel matrix (if I remember > correctly.) Yes that's right; how can the caller of Rbf() give a reasonable value of "smooth" to solve (A - smoothI) inside Rbf, without knowing A ?  A is wildly different for gauss, linear ... too. Or do you just shut your eyes and try 1e-6  ? Thanks Josef, cheers   -- denis _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Mon, Feb 22, 2010 at 10:14, denis <[hidden email]> wrote: > On Feb 22, 3:08 pm, [hidden email] wrote: >> I still don't understand what you mean with "the caller doesn't know >> A".  A is the internally calculated kernel matrix (if I remember >> correctly.) > > Yes that's right; how can the caller of Rbf() give a reasonable value > of "smooth" > to solve (A - smoothI) inside Rbf, without knowing A ?  A is wildly > different for gauss, linear ... too. Ah! *That's* what you meant. > Or do you just shut your eyes and try 1e-6  ? Basically, yes. You can try different values while watching the residuals if you want a more rigorous approach. But it really does work. Try using 'multiquadric' and smooth=0.1. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 In reply to this post by denis-bz-gg On Mon, Feb 22, 2010 at 11:14 AM, denis <[hidden email]> wrote: > On Feb 22, 3:08 pm, [hidden email] wrote: >> On Mon, Feb 22, 2010 at 7:35 AM, denis <[hidden email]> wrote: >> > On Feb 19, 5:41 pm, [hidden email] wrote: >> >> On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: >> >> >> > Use "smooth" ? rbf.py just does >> >> >    self.A = self._function(r) - eye(self.N)*self.smooth >> >> > and you don't know A . >> >> > That's a line from scipy/interpolate/rbf.py: it solves >> >    (A - smooth*I)x = b  instead of >> >    Ax = b >> > Looks to me like a hack for A singular, plus the caller doesn't know A >> > anyway. > >> It's not a hack it's a requirement, ill-posed inverse problems need > > OK, I must be wrong; but (sorry, I'm ignorant) how can (A - smooth) > penalize ? > For gauss the eigenvalues are >= 0, many 0, so we're shifting them > negative ?? > Or is it a simple sign error, A + smooth ? ouch, standard Ridge is A + smooth * identity_matrix to make it positive definite. I don't know why there is a minus. When I checked the eigenvalues, I found it strange that there were some large *negative* eigenvalues of A, but I didn't have time to figure this out. Generically, A - smooth*eye would still make it invertible, although not positive definite I haven't looked at the sign convention in rbf, but if you figure out what's going on, I'm very interested in an answer. I just briefly ran an example with a negative smooth (-0.5 versus 0.5), rbf with gauss seems better, but multiquadric seems worse. If smooth is small 1e-6, then there is not much difference. Even with negative smooth, all except for gauss still have negative eigenvalues. I have no idea, I only looked at the theory for gaussian process and don't know how the other ones differ. > >> penalization, this is just Ridge or Tychonov with a kernel matrix. A >> is (nobs,nobs) and the number of features is always the same as the >> number of observations that are used. (I was looking at "Kernel Ridge >> Regression" and "Gaussian Process" before I realized that rbf is >> essentially the same, at least for 'gauss') >> I don't know anything about thinplate. >> >> I still don't understand what you mean with "the caller doesn't know >> A".  A is the internally calculated kernel matrix (if I remember >> correctly.) > > Yes that's right; how can the caller of Rbf() give a reasonable value > of "smooth" > to solve (A - smoothI) inside Rbf, without knowing A ?  A is wildly > different for gauss, linear ... too. > Or do you just shut your eyes and try 1e-6  ? That's the usual problem of bandwidth selection for non-parametric estimation, visual inspection, cross-validation, plug-in, ... I don't know what's recommended for rbf. Cheers, Josef > > Thanks Josef, > cheers >  -- denis > _______________________________________________ > SciPy-User mailing list > [hidden email] > http://mail.scipy.org/mailman/listinfo/scipy-user> _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Mon, Feb 22, 2010 at 11:45 AM,  <[hidden email]> wrote: > On Mon, Feb 22, 2010 at 11:14 AM, denis <[hidden email]> wrote: >> On Feb 22, 3:08 pm, [hidden email] wrote: >>> On Mon, Feb 22, 2010 at 7:35 AM, denis <[hidden email]> wrote: >>> > On Feb 19, 5:41 pm, [hidden email] wrote: >>> >> On Fri, Feb 19, 2010 at 11:26 AM, denis <[hidden email]> wrote: >>> >>> >> > Use "smooth" ? rbf.py just does >>> >> >    self.A = self._function(r) - eye(self.N)*self.smooth >>> >> > and you don't know A . >>> >>> > That's a line from scipy/interpolate/rbf.py: it solves >>> >    (A - smooth*I)x = b  instead of >>> >    Ax = b >>> > Looks to me like a hack for A singular, plus the caller doesn't know A >>> > anyway. >> >>> It's not a hack it's a requirement, ill-posed inverse problems need >> >> OK, I must be wrong; but (sorry, I'm ignorant) how can (A - smooth) >> penalize ? >> For gauss the eigenvalues are >= 0, many 0, so we're shifting them >> negative ?? >> Or is it a simple sign error, A + smooth ? > > ouch, standard Ridge is A + smooth * identity_matrix to make it > positive definite. > > I don't know why there is a minus. When I checked the eigenvalues, I > found it strange that there were some large *negative* eigenvalues of > A, but I didn't have time to figure this out. > Generically, A - smooth*eye would still make it invertible, although > not positive definite > > I haven't looked at the sign convention in rbf, but if you figure out > what's going on, I'm very interested in an answer. > > I just briefly ran an example with a negative smooth (-0.5 versus > 0.5), rbf with gauss seems better, but multiquadric seems worse. If > smooth is small 1e-6, then there is not much difference. > > Even with negative smooth, all except for gauss still have negative > eigenvalues. I have no idea, I only looked at the theory for gaussian > process and don't know how the other ones differ. My intuition for gaussan might not be correct for the other rbfs, gauss is decreasing in distance, all others are increasing in distance >>> for func in Rbfuncs: print func, Rbf( x, y, function=func ,smooth=smooth)._function(np.arange(5)) gaussian [ 1.      0.9272  0.739   0.5063  0.2982] linear [0 1 2 3 4] thin-plate [  0.       0.       2.7726   9.8875  22.1807] multiquadric [ 1.      1.0371  1.1413  1.2964  1.4866] quintic [   0    1   32  243 1024] cubic [ 0  1  8 27 64] and the distance matrix itself has negative eigenvalues Josef > > >> >>> penalization, this is just Ridge or Tychonov with a kernel matrix. A >>> is (nobs,nobs) and the number of features is always the same as the >>> number of observations that are used. (I was looking at "Kernel Ridge >>> Regression" and "Gaussian Process" before I realized that rbf is >>> essentially the same, at least for 'gauss') >>> I don't know anything about thinplate. >>> >>> I still don't understand what you mean with "the caller doesn't know >>> A".  A is the internally calculated kernel matrix (if I remember >>> correctly.) >> >> Yes that's right; how can the caller of Rbf() give a reasonable value >> of "smooth" >> to solve (A - smoothI) inside Rbf, without knowing A ?  A is wildly >> different for gauss, linear ... too. >> Or do you just shut your eyes and try 1e-6  ? > > That's the usual problem of bandwidth selection for non-parametric > estimation, visual inspection, cross-validation, plug-in, ... I don't > know what's recommended for rbf. > > Cheers, > > Josef > >> >> Thanks Josef, >> cheers >>  -- denis >> _______________________________________________ >> SciPy-User mailing list >> [hidden email] >> http://mail.scipy.org/mailman/listinfo/scipy-user>> > _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 Robert, Josef,   thanks much for taking the time to look at RBF some more. Summary, correct me:     A - smooth*I in rbf.py is a sign error (ticket ?)     for gauss, start with A + 1e-6*I  to move eigenvalues away from 0     others have pos/neg eigenvalues, don't need smooth. Looking at Wikipedia Tikhonov (thanks Josef) reminded me of lstsq_near on advice.mechanicalkern: minimize |Ax-b| and w|x| together, i.e. Tikhonov with Gammamatrix = I. But for RBF, why minimize |x| ?  don't we really want to minimize     |Ax-b|2 + w2 xRx where R is a roughness penalty ? On a regular grid Rij ~ Laplacian smoother, not much like I. Here I'm over my head; any ideas for a q+d roughness matrix for scattered data ? (just for fun -- for RBF we've reached the point of diminishing returns). cheers   -- denis _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Tue, Feb 23, 2010 at 11:43, denis <[hidden email]> wrote: > Robert, Josef, >  thanks much for taking the time to look at RBF some more. > Summary, correct me: >    A - smooth*I in rbf.py is a sign error (ticket ?) Not necessarily. It seems to work well in at least some cases. Find a reference that says otherwise if you want it changed. >    for gauss, start with A + 1e-6*I  to move eigenvalues away from 0 >    others have pos/neg eigenvalues, don't need smooth. No. If you need a smoothed approximation to noisy data rather than exact interpolation than you use the smooth keyword. Otherwise, you don't. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Tue, Feb 23, 2010 at 12:46 PM, Robert Kern <[hidden email]> wrote: > On Tue, Feb 23, 2010 at 11:43, denis <[hidden email]> wrote: >> Robert, Josef, >>  thanks much for taking the time to look at RBF some more. >> Summary, correct me: >>    A - smooth*I in rbf.py is a sign error (ticket ?) > > Not necessarily. It seems to work well in at least some cases. Find a > reference that says otherwise if you want it changed. chapter 2 page 16, for gaussian process. As I said I don't know about the other methods http://docs.google.com/viewer?a=v&q=cache:qs8AaAxO6nkJ:www.gaussianprocess.org/gpml/chapters/RW2.pdf+gaussian+process+noise+Ridge&hl=en&gl=ca&pid=bl&srcid=ADGEESj4j8osT6cOIc65r3OaeAtQO_dzgZD4YxSAEkFTeRZajBcROJpJJ9zTlMSrD2OaK1iOJYgy8QqH_Nr0rNxf41faNihCdIzWyVOYxtCFIR7H8mdQZAKFoeaRkFamQlCKhp_s1FOI&sig=www.gaussianprocess.org/gpml/chapters/RW2.pdf Josef > >>    for gauss, start with A + 1e-6*I  to move eigenvalues away from 0 >>    others have pos/neg eigenvalues, don't need smooth. > > No. If you need a smoothed approximation to noisy data rather than > exact interpolation than you use the smooth keyword. Otherwise, you > don't. > > -- > Robert Kern > > "I have come to believe that the whole world is an enigma, a harmless > enigma that is made terrible by our own mad attempt to interpret it as > though it had an underlying truth." >  -- Umberto Eco > _______________________________________________ > SciPy-User mailing list > [hidden email] > http://mail.scipy.org/mailman/listinfo/scipy-user> _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Tue, Feb 23, 2010 at 12:57 PM,  <[hidden email]> wrote: > On Tue, Feb 23, 2010 at 12:46 PM, Robert Kern <[hidden email]> wrote: >> On Tue, Feb 23, 2010 at 11:43, denis <[hidden email]> wrote: >>> Robert, Josef, >>>  thanks much for taking the time to look at RBF some more. >>> Summary, correct me: >>>    A - smooth*I in rbf.py is a sign error (ticket ?) >> >> Not necessarily. It seems to work well in at least some cases. Find a >> reference that says otherwise if you want it changed. > > chapter 2 page 16, for gaussian process. As I said I don't know about > the other methods > http://docs.google.com/viewer?a=v&q=cache:qs8AaAxO6nkJ:www.gaussianprocess.org/gpml/chapters/RW2.pdf+gaussian+process+noise+Ridge&hl=en&gl=ca&pid=bl&srcid=ADGEESj4j8osT6cOIc65r3OaeAtQO_dzgZD4YxSAEkFTeRZajBcROJpJJ9zTlMSrD2OaK1iOJYgy8QqH_Nr0rNxf41faNihCdIzWyVOYxtCFIR7H8mdQZAKFoeaRkFamQlCKhp_s1FOI&sig=AHIEtbQK35MLfnZAySw3lF-dR_mNcSaP3wgoogle links are very short, missed a part > > www.gaussianprocess.org/gpml/chapters/RW2.pdf > > Josef >> >>>    for gauss, start with A + 1e-6*I  to move eigenvalues away from 0 >>>    others have pos/neg eigenvalues, don't need smooth. >> >> No. If you need a smoothed approximation to noisy data rather than >> exact interpolation than you use the smooth keyword. Otherwise, you >> don't. >> >> -- >> Robert Kern >> >> "I have come to believe that the whole world is an enigma, a harmless >> enigma that is made terrible by our own mad attempt to interpret it as >> though it had an underlying truth." >>  -- Umberto Eco >> _______________________________________________ >> SciPy-User mailing list >> [hidden email] >> http://mail.scipy.org/mailman/listinfo/scipy-user>> > _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user
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## Re: scipy.interpolate.rbf sensitive to input noise ?

 On Tue, Feb 23, 2010 at 11:59,  <[hidden email]> wrote: > On Tue, Feb 23, 2010 at 12:57 PM,  <[hidden email]> wrote: >> On Tue, Feb 23, 2010 at 12:46 PM, Robert Kern <[hidden email]> wrote: >>> On Tue, Feb 23, 2010 at 11:43, denis <[hidden email]> wrote: >>>> Robert, Josef, >>>>  thanks much for taking the time to look at RBF some more. >>>> Summary, correct me: >>>>    A - smooth*I in rbf.py is a sign error (ticket ?) >>> >>> Not necessarily. It seems to work well in at least some cases. Find a >>> reference that says otherwise if you want it changed. >> >> chapter 2 page 16, for gaussian process. As I said I don't know about >> the other methods >> > > http://docs.google.com/viewer?a=v&q=cache:qs8AaAxO6nkJ:www.gaussianprocess.org/gpml/chapters/RW2.pdf+gaussian+process+noise+Ridge&hl=en&gl=ca&pid=bl&srcid=ADGEESj4j8osT6cOIc65r3OaeAtQO_dzgZD4YxSAEkFTeRZajBcROJpJJ9zTlMSrD2OaK1iOJYgy8QqH_Nr0rNxf41faNihCdIzWyVOYxtCFIR7H8mdQZAKFoeaRkFamQlCKhp_s1FOI&sig=AHIEtbQK35MLfnZAySw3lF-dR_mNcSaP3w> > google links are very short, missed a part I've found a couple of things on RBFs specifically that agree. However, the original source on which our implementation is based does subtract. He may have a source that uses a negative sign. http://www.mathworks.co.uk/matlabcentral/fileexchange/10056-scattered-data-interpolation-and-approximation-using-radial-base-functionsPlaying around, it seems to me that some of the radial functions create smoother approximations with large positive values (with the current implementation) while others create smoother approximations with large negative values. It's possible that it's just a convention as to which sign you use. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth."   -- Umberto Eco _______________________________________________ SciPy-User mailing list [hidden email] http://mail.scipy.org/mailman/listinfo/scipy-user