3D plot help

Hi,

I have a row data composed of independent variables x, y, and z and a dependent variable D, and I would like to 3d plot the geometry. Each independent and dependent variable is a 1D array. However, all 3d plotting functions that I am aware of such as isosurface require 3D arrays. My question is how I can transfom the different arrays into 3d arrays, given that they are not uniform.

[Walter Roberson]

griddata() .

Or possibly you could

ux = unique(x);
uy = unique(y);
uz = unique(z);
D3 = reshape(D, length(ux), length(uy), length(uz));

and then isosurface() on D3

[Diab Abueidda]

Thanks, Walter for the reply.

I did as you suggested:

ux = unique(x);
uy = unique(y);
uz = unique(z);
D3 = reshape(D, length(ux), length(uy), length(uz));

and received the following error

Error using reshape

To RESHAPE the number of elements must not change.

Error in IsoSurface (line 12)

D3 = reshape(D, length(ux), length(uy), length(uz));

Also, I don't think unique function is what I need. In my case, x, y, and z are three spatial coordinates. Hence, some data points will share one or two coordinates, but the remaining coordinates will be different. In this sense, I don't have a problem of nonuniqueness.

[Walter Roberson]

Your original line about "Each independent and dependent variable is a 1D array." was not clear. If each one is a 1D array, then that would imply that the variables are independent and that you have all combinations of values available. If that is not the case, then see griddata() that I mentioned before.

[Diab Abueidda]

Thanks, Walter for the reply.

I think griddata is in the right direction, but still I am not getting what I am expecting. The code I am using is

D = xval;
minv= 0; maxv=10; inc=0.05;
[xq,yq,zq]=meshgrid(minv:inc:maxv,minv:inc:maxv,minv:inc:maxv);
Dq = griddata(x,y,z,D,xq,yq,zq,'linear');

margin = 0.1;
xb_index = [find(xq<=minv+margin) find(xq>=maxv-margin)];
yb_index = [find(yq<=minv+margin) find(yq>=maxv-margin)];
zb_index = [find(zq<=minv+margin) find(zq>=maxv-margin)];

cutoff = 0.5;
Dq(xb_index)=cutoff;
Dq(yb_index)=cutoff;
Dq(zb_index)=cutoff;

p1 = patch(isosurface(xq,yq,zq,Dq, cutoff),'FaceColor','blue','EdgeColor','none');
% p2 = patch(isocaps(xq,yq,zq,Dq, cutoff+0.1),'FaceColor','cyan','EdgeColor','none');
view(3);
axis tight
camlight
lighting gouraud
daspect([1,1,1])
isonormals(Dq,p1)

If I use 'nearest' for griddata I get this,

I get what I need as a closed geometry, but it is not smooth. I changed the method to natural. I started getting a bit smoother isosurface, but I could not close the sides of the geometry. Please see below

Any recommendation?

[Walter Roberson]

Please show

scatter3(x, y, z)

[Diab Abueidda]

Hi Walter,

Here it is

[Diab Abueidda]

Hi Walter,

Any thoughts?

Thanks a lot

[Walter Roberson]

I think I will need the data to play with. It is not obvious from the scatter plot that any semi-regular structure exists inside that at all.

[Diab Abueidda]

Hi Walter,

Please find the spatial coordinates attached.

Many thanks!

[Walter Roberson]

I will need the D as well.

I was correct in my original speculation that you have a regular 3D grid of data.

[Diab Abueidda]

Hi Walter,

Thanks for the reply. Please find the data attached.

The data I sent indeed has a regular 3D grid, but please note that this is the simplest case I have, and it is the only one with a regular grid. What I am trying to write is a code that is applicable for all cases, and I am still failing to obtain the geometry even with the simplest case.

Thanks for the support.

[Diab Abueidda]

Hi Walter,

Did you have the chance to check it out?

Many thanks!

[Walter Roberson]

I did have a look, but I couldn't figure out why I was getting the visuals that I was getting.

One thing I do not understand is why you assign 0.5 to locations in the margins, considering that the value distribution is "a lot of values near 0.1" and "a lot of values near 0.9" together with an equal but much smaller distribution of values in-between. It would seem more natural to assign 0 or nan to the margin.

[Diab Abueidda]

Hi Walter,

Thanks for having a look. The values are obtained mathematically by using an optimization framework. Physically, the densities should be binary (zero or one), but we relax this condition and apply penalization to push the density either to zero or one. For numerical stability purposes, we put a minimum value for densities to be 0.001.

Regarding the assigned threshold, any value from 0.3 to 0.7 should be physically acceptable. Feel free to change it.

I hope this gives you an overview of the problem.

[Walter Roberson]

When I look at the data with a 3D visualizer, I can see that there is something that is roughly X shaped, and I do not see anything special happening in the margins. I do not understand why a higher cutoff is being assigned than the values.

If you let H be high values near 0.9, and L be low values near 0.01 then you get a shape that near the top is roughly

HHLLLLLLLLHH
LHHLLLLLLHHL
LLHHLLLLHHLL
LLLHHLLHHLLL
LLLLHHHHLLLL

and I do not understand why you want to modify that to

MMMMMMMMMMMM
MHHLLLLLLHHM
MLHHLLLLHHLM
MLLHHLLHHLLM
MLLLHHHHLLLM

This introduces a "flange" into the visuals that I do not see a purpose for.

[Diab Abueidda]

I added this part to close the isosurface at the boundaries. Otherwise, I would get two isosurfaces as shown in the second figure I shared. I believe that this part I added closed the geometry as portrayed in Figure 1. Is there a better way to close the geometry? I tried to use isocaps as shown in the shared code, but still I had the same problem.

[Diab Abueidda]

Hi Walter,

Regardless of the code I wrote, how would you do it? Basically, I need to get something similar to figure 1 but with smoothed isosurface.

Thanks

[Diab Abueidda]

Hi Walter,

I hope you have an idea of what I am trying to do. Please let me know if you have any ideas/suggestions. I have been stuck with this issue for a while, and your help is highly appreciated.

Thanks

[Walter Roberson]

Sorry, nothing yet, I have been a bit busy.

[Diab Abueidda]

No worries, take your time.

Many thanks

[Diab Abueidda]

Hi Walter,

Any thoughts?

Thanks

Answer 1

[Thomas Satterly]

If you're expecting the independant variables X, Y, and Z to make some nice looking 3D surface, you can use Delaunay Triangulation to make a best guess at what that surface would look like and set the corresponding vertex colors to a colormap scaled by dependant variable D. However, you have to be careful about extrapolating data between points using this approach, as the edges generated by triangulation are purely ficticious.

If there is no expected surface between the independant variables, I'd recommend using a 3D scatterplot of X, Y, and Z with the point colors or size determined by dependant variable D. This might be harder to visualizes than a surface, since there's not as strong of relational queues to nearby points, but it does eliminate some misleading extrapolation that generating a surface would cause.

[Diab Abueidda]

Thanks a lot, Thomas!

It worked.