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# Is it possible to have 3 or more variable inputs for a 3D contour plot in GUI?

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##### 1 Comment

Basma M
on 10 Jun 2015

### Answers (5)

Walter Roberson
on 11 Sep 2011

In what manner were you hoping to display your 4 dimensional data?

##### 1 Comment

Bastion
on 11 Sep 2011

Just to clarify:

I want to make a 3D CONTOUR plot in Matlab GUI. I have an equation for Concentration (C) which is a function of (X,Y,Z,t). I'm making t as a constant. But I want the user to input the data for values of X, Y and Z.

In the Matlab help files I got this for 3D contour:

[X,Y] = meshgrid([-2:.25:2]);

Z = X.*exp(-X.^2-Y.^2);

contour3(X,Y,Z,30)

This will only allow me to input the variables for X and Y and get Z as an output. So I'm wondering if it is possible for me to let X, Y and Z be inputs and C as my output.

##### 5 Comments

Walter Roberson
on 12 Sep 2011

According to current theory, Yes, it is possible, if you have a spinning negatively charged black hole and you can enter its outer "naked singularity" without touching its event horizon. However, theory has not yet determined whether it would ever be possible to re-enter *this* Universe afterwards, partly because the concept of "afterwards" would have little meaning once you had managed to complete the trade of time for another spacial dimension in order to be able to construct the graph with 4 spacial dimensions.

See http://www.mathworks.com/matlabcentral/answers/11433-plotting-3d-of-output

Bastion
on 12 Sep 2011

Walter Roberson
on 12 Sep 2011

You might also consider isosurface() or slice().

Also, some people have indicated that the scatter3() solution I proposed in the linked question has been fine for their needs. Representing a contour in that method might be a bit tricky, though.

Bastion
on 12 Sep 2011

yeah I'm looking at isosurface but can't seem to place equation, I'll check it out

Dmitry Borovoy
on 12 Sep 2011

If you really need to use function just create your own shell-function.

function MyContour3(X,Y,Z,t)

% add logic for assigning t in your equation

contour3(X,Y,Z)

end

##### 26 Comments

Dmitry Borovoy
on 12 Sep 2011

Walter Roberson
on 12 Sep 2011

Bastion
on 12 Sep 2011

If I input all the values for X,Y,Z and I join them together wouldn't that give me a contour?

Walter Roberson
on 12 Sep 2011

Bastion
on 12 Sep 2011

It's alright I've decided to just try with C(X,Y) etc.

but can you help me with this code?

------------------------------------------------------------------------

X = str2double(get(handles.Input_3D_x,'String'));%%%%% user inputs X

Y = str2double(get(handles.Input_3D_y,'String'));%%%%% user inputs Y

Z = str2double(get(handles.Input_3D_z,'String'));%%%%% user inputs Z

t = str2double(get(handles.Input_t,'String'));%%%%% user inputs t

[X,Y] = meshgrid(-10:0.25:10); %%%%% this part I'm not too sure of

C = X.*Y.*Z.*t %%%%% The equation that operates the user inputs

contour3(X,Y,C,100) %%%%% ploting a contour of C in the Z-axis with X and Y in their respective axis

------------------------------------------------------------------------

I get a squiggly orange line for X and Y in the first 2 lines, and it says: "The value assigned to varaible 'X' might be unused." ditto for Y.

If I input any variables for X or Y the contour doesn't change. I'm thinking I'm made an error somewhere.

Bastion
on 12 Sep 2011

If I put "C = X.*Y.*Z.*t" above "[X,Y] = meshgrid(-10:0.25:10);" then the equation doesn't work

Walter Roberson
on 12 Sep 2011

What format are you expecting the user to have entered data in to Input_3D_x and so on? Are they to be the X, Y, and Z ranges? If so are you expecting a pair of numbers or are you expecting a single number that you will use as the positive and negative bounds?

Try

Xmax = str2double(get(handles.Input_3D_x,'String'));%%%%% user inputs X

Ymax = str2double(get(handles.Input_3D_y,'String'));%%%%% user inputs Y

Zmax = str2double(get(handles.Input_3D_z,'String'));%%%%% user inputs Z

t = str2double(get(handles.Input_t,'String'));%%%%% user inputs t

[X,Y,Z] = ndgrid(-Xmax:0.25:Xmax, -Ymax:0.25:Ymax, -Zmax:0.25:Zmax);

C = X.*Y.*Z.*t %%%%% The equation that operates the user inputs

But now you have a problem, in that X, Y, and Z are now 3 dimensional arrays. When you talk about contour3(X,Y,C,100) you are implicitly wanting to project along an axes, but C will have multiple values along that axes for each (x,y) position. You need to resolve this.

One resolution that _sometimes_ makes sense is to take the maximum (or minimum) value along the axes. So, for example,

contour3(X(:,:,1), Y(:,:,1), max(C,[],3), 100)

Bastion
on 12 Sep 2011

Hey thanx for the reply, all of the user inputs are one value;

I was searching here and came across this:

http://www.mathworks.com.au/matlabcentral/answers/15558-plot-a-6x6-matrix-in-3d

"N=6;

[X,Y]=meshgrid(1:N,1:N);

surf(X,Y,A);"

Like the range is 1:X, so the users only have to input one number. Like a matrix thing, I tried that in my code but it doesn't work.

I'll try your one now and I'll get back to ya.

Bastion
on 12 Sep 2011

Walter Roberson
on 12 Sep 2011

At the command window, put in the command

dbstop if error

and then run the program. It will stop at the line that is causing the problem and tell you the kind of problem it is encountering. You can examine the variable sizes and classes and values to try to determine the source of the problem.

It is difficult for us to debug problems without knowing which line was the problem and what the error message was.

Walter Roberson
on 12 Sep 2011

I used Xmax and Ymax and Zmax all equal, with code I showed above that ends in the contour3 call. It seemed to work fine considering the limitations of having to project C down to 3D.

I also used

scatter3(X(:),Y(:),Z(:),10,C(:))

colormap(hsv)

and that produced something somewhat intelligible, though not contoured.

The biggest problems with the two tests above was that the mesh was too coarse to produce a good-looking plot over the range -5:+5 .

I refined the mesh to steps of 1/64 . That slowed things down a fair bit, matrix sizes 641 x 641 x 641

I then

Cmin = min(C(:));

Cmax = max(C(:));

[n,bin] = histc(C(:),linspace(Cmin,Cmax,11));

scatter3(X(:),Y(:),Z(:),10,bin)

Plotting with those array sizes was notably slow. The above is akin to doing a contour plot with (11-1) = 10 levels (the 11th level will exist in there but will be only the 4 peaks of the symmetric C calculation.) Did I mention this was notably slow? :( A few hundred million points to plot... Guess I might as well go sign in the after-hours log while I am waiting for it to finish...

Walter Roberson
on 13 Sep 2011

The above was just too slow with the grid step of 1/64 over -5:+5 .

Referring to the above variables, I then used

u = unique(round(rand(1,floor(length(bin)/1000))*length(bin)));

scatter3(X(u),Y(u),Z(u),10,bin(u))

That was still somewhat slow but at least it could be worked with. It did not produce surfaces as would be hoped for contours, but it was still useful in showing the different value levels (several simultaneously)

I also used

p = patch(isosurface(X,Y,Z,C,100));

set(p, 'FaceColor', 'red', 'EdgeColor', 'none');

daspect([1 1 1])

view(3)

camlight; lighting phong

That was a fairly reasonable time-frame and produced smooth surfaces for the one value level (100) that had been specified.

The examples in isosurface() show calls to isonormal() . That took at notable time and then failed trying to interp3. My speculation is that it ran out of memory or something similar, though the complaint it gave was that the coordinates had not been produced by meshgrid (producing them by ndgrid should have done fine.)

Bastion
on 13 Sep 2011

My computer won't handle it if there's too many points to plot

Yeah I'm going to try the:

u = unique(round(rand(1,floor(length(bin)/1000))*length(bin)));

scatter3(X(u),Y(u),Z(u),10,bin(u))

p = patch(isosurface(X,Y,Z,C,100));

set(p, 'FaceColor', 'red', 'EdgeColor', 'none');

daspect([1 1 1])

view(3)

camlight; lighting phong

Thanx for the answers, I'll let you know how it goes

Walter Roberson
on 13 Sep 2011

I'm not sure you would want to use both scatter3() and patch(), but it might make sense to do so if you had a particular level you needed emphasized (via the patch) and wanted to get an idea of how the others were distributed (via the scatter3).

The scatter3() part was much slower than the construction of the isosurface, so if the isosurface gives you what you need, you will probably want to skip the scatter3()

Hmmm, you might want to add a max() of the floor(length(bin)/1000) and some minimum number of points, in case the initial size of the mesh is very small -- you don't want to end up with u being empty or only having 1 point for example.

Walter Roberson
on 13 Sep 2011

Bastion
on 13 Sep 2011

Sorry I'm confused at where you mentioned isosurface? I did not see this part

I'm not very versed in Matlab GUI at all, let alone Matlab so some of your statements are not really registering.

Bastion
on 13 Sep 2011

I went back to the first equation that you suggested:

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

M = str2double(get(handles.Input_M,'String'));

n = str2double(get(handles.Input_n,'String'));

Vx = str2double(get(handles.Input_Vx,'String'));

Dx = str2double(get(handles.Input_Dx,'String'));

Dy = str2double(get(handles.Input_Dy,'String'));

Dz = str2double(get(handles.Input_Dz,'String'));

Xmax = str2double(get(handles.Input_3D_x,'String'));

Ymax = str2double(get(handles.Input_3D_y,'String'));

Zmax = str2double(get(handles.Input_3D_z,'String'));

t = str2double(get(handles.Input_t,'String'));

[X,Y,Z] = ndgrid(-Xmax:0.25:Xmax, -Ymax:0.25:Ymax, -Zmax:0.25:Zmax);

C = ((M./n)./(8).*((3.14.*t).^1.5).*sqrt(Dx.*Dy.*Dz)).*exp((-((X-Vx.*t).^2)./4.*Dx.*t)-(Y.^2/4.*Dy.*t)-(Z.^2./4.*Dz.*t));

contour3(X(:,:,1), Y(:,:,1), max(C,[],3), 100)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

This works the best and produces something similar and allows the user to set boundaries. Do you think it's reasonable? Or explore isosurface?

Walter Roberson
on 16 Sep 2011

Looks to me like you would get a divide by zero error if any of the Dx, Dy, or Dz were zero, and that you would get a complex result if an odd number of them were negative or if t is negative.

Question: does 3.14 represent Pi ?

Bastion
on 17 Sep 2011

Yeah 3.14 is pi, not sure of the notation in Matlab yet.

Ummmm If i change any of the variable inputs it will change the contour HOWEVER this is not true for Xmax and Ymax as both does not change the contour no matter what value I input

Walter Roberson
on 20 Sep 2011

The MATLAB notation for pi is pi

Test values would help so that I do not end up looking at a different oddity than you are looking at.

Bastion
on 22 Sep 2011

Thanx for helping,

M=10

n=1

Vx=1

Dx=1

Dy=1

Dz=1

Xmax=2

Ymax=2

Zmax=2

t=10

would be fine

at the moment the graph doesn't change when I change the Xmax and Ymax input when I plot contour3(X,Y,C). But if I change any of the other variables the graph changes.

Sean de Wolski
on 12 Sep 2011

An isosurface with different colored and shaded patch es can be used to visualize 4d data.

doc isosurface

doc patch

##### 1 Comment

Basma M
on 10 Jun 2015

##### 1 Comment

Walter Roberson
on 11 Jun 2015

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