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

I'm trying to make a 3D contour plot with an equation that has X,Y,Z and t as a variable input by the user in the GUI. Is it possible to achieve?

[Basma M]

hi Bastion, i was wondering if you ever knew the correct code for this problem. i am very much interested in the solution you came up with cuz i need to solve the same equation but having difficulties doing so. I would really appreciate your feedback on this, it is somewhat urgent and i've already spent too much time and effort trying to work it out. THANKS

Answer 1

[Walter Roberson]

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

[Bastion]

On a axes plot in matlab GUI. I wish to make a 3D contour

Answer 2

[Bastion]

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.

[Bastion]

Anyone? Is it possible?

[Walter Roberson]

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]

Ok so it is practically impossible, thanx for the replies, I have to use contour. I guess I will just make 3 different contours, C(X,Y) C(X,Z) and C(Y,Z)

[Walter Roberson]

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]

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

Answer 3

[Dmitry Borovoy]

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

[Bastion]

what do you mean? I'm doing this in GUI

[Dmitry Borovoy]

I don't know what change when GUI is used. I understood that you need 3D contour. But for some reasons you need function that uses 4 input variable where first 3 variables is a simple coordinats and fourth is time that is constant for current moment. Is it correct? So just create your function as I show before

[Walter Roberson]

Dmitry, you missed that Bastion needs X, Y, and Z to calculate C (concentration), and thus is trying to do 4 dimensional display: a distance along X to represent the X coordinate, a distance along Y to represent the Y coordinate, a distance along Z to represent the Z coordinate -- but then how do you do the distance along C to represent the resulting value?

[Dmitry Borovoy]

sorry, my bad

[Bastion]

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

[Walter Roberson]

Do you mean all of the values that give you a specific concentration? If so that would be an isosurface rather than a contour. I suppose you could do multiple isosurface plots in the same axes, but I am not at all sure that you would be able to make any sense out of the result.

[Bastion]

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]

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

[Walter Roberson]

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]

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]

Hey I tried it, it works, but I think there is some error, I keep hearing the 'ding' sound. Also I'm not quite sure what you did there, an explanation would be good.

[Walter Roberson]

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]

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]

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]

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

[Bastion]

I tried the codes but it is not working, maybe I didn't put it in right...

[Walter Roberson]

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]

The 1 in 1000 sampling I did in constructing u was related to when I was using 1/64 as the step size instead of the 1/4 that you originally coded; depending on the input ranges you specified, the 1 in 1000 random sampling might have been too small to be useful.

[Bastion]

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]

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]

Could you suggests test values for each of the parameters?

[Bastion]

any values is fine at the moment

[Walter Roberson]

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]

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]

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]

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.

Answer 4

[Sean de Wolski]

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

doc isosurface
doc patch

[Bastion]

I'm still reading on that

Answer 5

[Basma M]

hi Bastion, i was wondering if you ever knew the correct code for this problem. i am very much interested in the solution you came up with cuz i need to solve the same equation but having difficulties doing so. I would really appreciate your feedback on this, it is somewhat urgent and i've already spent too much time and effort trying to work it out. THANKS

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency