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Dear all,
I have a 2d matrix (2358*2) which corresponds to the points of a 2d grid (A: x-y data). These points do not define a rectangle. A second matrix (B: 2358*1) represents a specific variable corresponding to each of the gird points. What I want to show is the a 3D representation of these two matrices; so that I can get the surface of B over the grid points (A). The function 'surface' may work but I could not apply it as I do not have exactly a fully rectangle grid.
What I need to get as the end needs to look like a 3d topography.
Your great help would be appreciated, many thanks.
Reyhaneh

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Mike Garrity
Mike Garrity 2015-10-27

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You have a couple of options. We were just discussing them in this other thread. In that case they wanted a 2D color representation, but you can use the same techniques for a 3D surface. In that case, the "pcolor" becomes "surf", and leave the "view(2)" out of the trisurf example.

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Hi Mike,
Many thanks for your answer.
Unfortunately, I did not get what I wanted by applying the explained approach.
Actually, I don not want to do any kind of interpolation for extra points. What I have is already a grid-based data (x,y,z) where x and y are already representing grid points. In other words I would like just to show these specific points of the grid meshgrid(x,y) with a height (as third dimension) which corresponds to z, not all the grid points with interpolated values. By plot3(x,y,z), I get the scatter representation. but I would like to get the continuous 3D surface built solely on these points, since afterwards I need to get the volume of this 3D object.
Your help would be highly appreciated, many thanks.
Reyhaneh
Mike Garrity
Mike Garrity 2015-12-31
If you have X,Y,Z triplets for all of the locations in a 2D X,Y domain, then it's just a matter of getting your arrays into the right shape. I can't tell you how to do that without knowing more about the shape they're in, but it's likely to involve the reshape function.
As an example, lets say I had this data:
[x1,y1] = meshgrid(linspace(-pi,pi,10), ...
linspace(-pi,pi,7));
z1 = sin(x1).*sin(y1);
pts = [x1(:), y1(:), z1(:)];
clear x1 y1 z1
The array pts has 70 rows and 3 columns and looks something like this:
-3.1416 -3.1416 0.0000
-3.1416 -2.0944 0.0000
-3.1416 -1.0472 0.0000
-3.1416 0 0
-3.1416 1.0472 -0.0000
-3.1416 2.0944 -0.0000
-3.1416 3.1416 -0.0000
-2.4435 -3.1416 0.0000
-2.4435 -2.0944 0.5567
-2.4435 -1.0472 0.5567
...
To use it with surf, I need to get it back into the 7x10 shape of the original arrays x1,y1,z1. I can do that with reshape like so:
m = 7;
n = 10;
x2 = reshape(pts(:,1),[m n])
y2 = reshape(pts(:,2),[m n])
z2 = reshape(pts(:,3),[m n])
surf(x2,y2,z2)
But that's just an example. The details are going to depend on the order of the values. In some cases it will be more complicated than that example.

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Image Analyst
Image Analyst 2015-12-31

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If your x,y locations are not complete (some coordinates are missing), or not in a perfect grid, then you'll have to get them into a grid using scatteredInterpolant(). After you use scatteredInterpolant(), then you'll have a perfect rectangular gridded image and then you can use surf to map the value into a height above a plane, or you can use imshow() to display it as an image.

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