Maximum of 2D data interpolation

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Albert Zurita
Albert Zurita le 6 Fév 2023
Commenté : Albert Zurita le 7 Fév 2023
Hi, I have 2D data for which I know there is a single maximum, but the points spacing may not be ideal so the maximum may lie in between. Therefore I would like to use griddedInterpolant to find the maximum point. X and Y contain the points, Z the corresponding function value.
It should look something as:
F = griddedInterpolant(x_mat,y_mat,z_mat);
xmaxs = arrayfun(@(xx,yy)fminsearch(@(x,y)-F(x,y),xx,yy),[x_mat(ii) y_mat(ii)]);
Where I provide an initial guess ii. But this is not the right way of calling it for a 2D function I'm afraid...

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John D'Errico
John D'Errico le 6 Fév 2023
Modifié(e) : John D'Errico le 6 Fév 2023
You will NEED to use a spline based method of interpolation here, specifically spline or cubic. NOT makima, or pchip. Those interpolants will not generate functions with an extremum between the data points. And of course you certainly cannot use the default linear interpolant, as that will NEVER generate a solution that is not at a data point.
[X,Y] = ndgrid(0:0.5:6);
Z = cos(X+Y).*sin(X-Y);
G = griddedInterpolant(X,Y,Z,'spline');
fsurf(@(x,y) G(x,y),[0 6 0 6])
Warning: Function behaves unexpectedly on array inputs. To improve performance, properly vectorize your function to return an output with the same size and shape as the input arguments.
We can clearly see several peaks. Of course, since I used a trig function here, they will lie at transcendental locations, so never exactly at a grid point.
[xymin,fval] = fminsearch(@(xy) -G(xy(1),xy(2)),[1,1])
xymin = 1×2
0.7930 2.3606
fval = -0.9945
To within the tolerances used by fminsearch, this should be a solution.
xymin/pi
ans = 1×2
0.2524 0.7514
And that would clearly be the point at [pi/4,3*pi/4]. If you want a better solution you need a finer grid spacing, as the spline has limited approximative capabilities with that course grid.
  4 commentaires
Albert Zurita
Albert Zurita le 7 Fév 2023
I will investigate further, but I tested for a simple paraboloid and it seems to do things right, both for cubic and spline interpolations, as it should be. I will check what it's about my data then, perhaps there is some special pattern.
Brilliant answer, thank you!
Albert Zurita
Albert Zurita le 7 Fév 2023
I am thinking that given the fact that I'm looking at very precise maximum determination the problem is very likely related with the tolerance options to the fminsearch function. Changing the default value of 1e-4 to a lower value seems to provide better results!

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Plus de réponses (1)

Torsten
Torsten le 6 Fév 2023
Déplacé(e) : Torsten le 6 Fév 2023
That doesn't make sense.
F is only an approximating function between the grid points - you don't know if the "real" underlying function behaves like this.
If you cannot supply the "real" function for F, better just choose the z with maximum value and the corresponding x and y.
  1 commentaire
Albert Zurita
Albert Zurita le 6 Fév 2023
Sorry. I rely on the grid points and the corresponding z values, so that f(x,y) = z is known and well sampled. However even if I can sample denser and do max(z) I would like to find (xi,yi) interplated points which would correspond to a better approximation/interpolation of the maximum. I found some answers in 1D and thought it could be extrapolated to 2D.

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