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

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Bastion 2011-9-11

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评论： Walter Roberson 2015-6-11

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?

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Basma M 2015-6-10

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

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回答（5 个）

Walter Roberson 2011-9-11

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

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Bastion 2011-9-11

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

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Bastion 2011-9-11

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在 MATLAB Online 中打开

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.

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Walter Roberson 2011-9-12

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 2011-9-12

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

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Dmitry Borovoy 2011-9-12

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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

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Bastion 2011-9-12

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.

Walter Roberson 2011-9-12

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)

Walter Roberson 2011-9-12

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 2011-9-13

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 2011-9-13

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 2011-9-20

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 2011-9-22

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.

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