2d interpolation from a non-uniform grid
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oceanmod
on 24 Jun 2020
Commented: Akira Agata
on 26 Jun 2020
I have looked at the griddata function but am not clear if it applies to data on an irregular mesh. In my case, I have a 2d spatial dataset available on an [X,Y] grid where X and Y, which are the cartesian coordinates, are themselves 2d arrays. That is, both X and Y have sizes (m,n) but
X(:,1) ~= X(:,2) ~= X(:,3) and so on.
Similarly for the array Y. Visually, you can picturize the grid as follows, where X marks the location of the input data set:
X X X X
X X X X
X X X X
Along each row, the spacing between two Xs is uniform. But the spacing changes from row to row, leading to X(:,1), X(:,2), etc. all being different.
It looks like griddata will not accept such an input grid. Am I correct?
Thanks for your help.
4 Comments
Akira Agata
on 26 Jun 2020
Well, let me clarify my intention by the following small example.
I hope this will be somehow helpful for you to understand how these two functions works.
% Original Data (X and Y are non-uniform grid)
X = [...
1 3 5 7;...
2 4 6 8;...
1 3 5 7;...
2 4 6 8];
Y = [...
1 1 1 1;
2 2 2 2;
3 3 3 3;
4 4 4 4];
Z = X.^2 + Y.^2;
% Query points
[xGrid, yGrid] = meshgrid(1:0.2:8,1:0.2:4);
% [Solution 1]
F = scatteredInterpolant(X(:),Y(:),Z(:));
zGrid = F(xGrid,yGrid);
% [Solution 2]
X2 = X(1:2:end,:);
Y2 = Y(1:2:end,:);
Z2 = Z(1:2:end,:);
zGrid2 = interp2(X2,Y2,Z2,xGrid,yGrid);
Accepted Answer
Bjorn Gustavsson
on 25 Jun 2020
No you can use griddata and scatteredInterpolant. You can do something like this:
Zi = griddata(X(:),Y(:),Z(:),Xi,Yi);
And you do the same thing with scatteredInterpolant - the (:) construct just unwraps an array into a 1-D column array.
HTH
3 Comments
Bjorn Gustavsson
on 25 Jun 2020
Yes it does:
[X,Y] = meshgrid(1:12,1:11);
X(2:2:end,:) = X(2:2:end,:)+.5;
plot(X,Y,'+')
Z = peaks(12);
Z = Z(2:end,:);
subplot(1,2,1),
pcolor(X,Y,Z)
[Xi,Yi] = meshgrid(1:.1:12,1:.1:11);
Zi = griddata(X(:),Y(:),Z(:),Xi,Yi);
subplot(1,2,2)
pcolor(Xi,Yi,Zi),shading flat
You have to give the function the input it expects, for x, y and z. Then it demonstrably works.
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