Contour plot of cubic interpolation with set number of contour levels

Question

Joseph Cunningham 2024-9-5

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https://ww2.mathworks.cn/matlabcentral/answers/2150464-contour-plot-of-cubic-interpolation-with-set-number-of-contour-levels

编辑： dpb 2024-9-6

I have 2 sets of raw experimental data arranged in a grid of X- and Y- points with Z-values at each point representing intensity. Xvec, Yvec, and Zvec are vectors of the same length (24) where corresponding positions give the X- and Y- positions and Z values. I have used a cubic interpolation to fit a surface to this data, and have been able to plot it with a contour plot function. My code is as below, with an example Xvec, Yvec, and Zvec provided.

Xvec= [-1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750]';
Yvec=[1.6250    1.6250    1.6250    0.5000    0.5000    0.5000   -0.6250   -0.6250   -0.6250   -1.8750   -1.8750   -1.8750   -1.8750   -1.8750   -1.8750   -0.6250   -0.6250   -0.6250    0.5000    0.5000    0.5000    1.6250    1.6250    1.6250]';
Zvec= [0.5516    0.7743    1.0936    1.1194    2.1194    2.6033    1.0840    2.1033    2.7903    0.4421    1.0936    0.7645    0.7420    0.4292    0.4292    2.6228   1.7775    0.9130    2.5453    1.7389    0.8743    1.0743    0.8358    0.6712]';

--Minimal coding example:

Zfit1=fit([Xvec Yvec],Zvec,'cubicinterp');
figure (1)
plot(Zfit1,[Xvec Yvec],Zvec,'Style','Contour')
c=colorbar;
cc=get(c,'Limits');
set(c,'Ticks',linspace(cc(1),cc(2),10))
set(gcf,'color','w');

--End example

Yields the following figure:

I would like to have both the contour of the fit and the X-Y data points on the chart, as above, but I would like to have 12 instead of 6 color/contour levels.

I would appreciate any assistance that can be provided by the community.

Thank you in advance.

JC

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Answer 1

John D'Errico 2024-9-5

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https://ww2.mathworks.cn/matlabcentral/answers/2150464-contour-plot-of-cubic-interpolation-with-set-number-of-contour-levels#answer_1511604

编辑：John D'Errico 2024-9-5

在 MATLAB Online 中打开

Simple enough. You should have looked at the handle returned by plot. Though I guess it helps if you know what to look for. ;-)

Xvec= [-1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750   -1.8750   -1.1250   -0.3750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750    0.3750    1.1250    1.8750]';
Yvec=[1.6250    1.6250    1.6250    0.5000    0.5000    0.5000   -0.6250   -0.6250   -0.6250   -1.8750   -1.8750   -1.8750   -1.8750   -1.8750   -1.8750   -0.6250   -0.6250   -0.6250    0.5000    0.5000    0.5000    1.6250    1.6250    1.6250]';
Zvec= [0.5516    0.7743    1.0936    1.1194    2.1194    2.6033    1.0840    2.1033    2.7903    0.4421    1.0936    0.7645    0.7420    0.4292    0.4292    2.6228   1.7775    0.9130    2.5453    1.7389    0.8743    1.0743    0.8358    0.6712]';
Zfit1=fit([Xvec Yvec],Zvec,'cubicinterp');
figure (1)
H = plot(Zfit1,[Xvec Yvec],Zvec,'Style','Contour')
H = 
  2x1 graphics array:

  Contour    (curvefit.gui.FunctionContour)
  Line

Note that H is an array of two handles. The first one is what we need.

c=colorbar;
cc=get(c,'Limits');
set(c,'Ticks',linspace(cc(1),cc(2),10))
set(gcf,'color','w');

If we take a look at what is stuffed in there,

get(H(1))
            Annotation: [1x1 matlab.graphics.eventdata.Annotation]
          BeingDeleted: off
            BusyAction: 'queue'
         ButtonDownFcn: ''
              Children: [0x0 GraphicsPlaceholder]
              Clipping: on
           ContextMenu: [0x0 GraphicsPlaceholder]
         ContourMatrix: [2x560 double]
             CreateFcn: ''
       DataTipTemplate: [1x1 matlab.graphics.datatip.DataTipTemplate]
             DeleteFcn: ''
           DisplayName: ''
             EdgeAlpha: 1
             EdgeColor: [0.0706 0.0706 0.0706]
             FaceAlpha: 1
             FaceColor: 'flat'
      HandleVisibility: 'on'
               HitTest: off
         Interruptible: on
            LabelColor: [0.1300 0.1300 0.1300]
           LabelFormat: @string
          LabelSpacing: 144
             LevelList: [0.3945 0.5000 1 1.5000 2 2.5000]
         LevelListMode: 'auto'
             LevelStep: 0.5000
         LevelStepMode: 'auto'
             LineStyle: '-'
             LineWidth: 0.5000
                Parent: [1x1 Axes]
         PickableParts: 'none'
              Selected: off
    SelectionHighlight: on
              ShowText: off
                   Tag: 'curvefit.gui.FunctionContour'
              TextList: [0.5000 1 1.5000 2 2.5000]
          TextListMode: 'auto'
              TextStep: 0.5000
          TextStepMode: 'auto'
                  Type: 'contour'
              UserData: []
               Visible: on
                 XData: [51x49 double]
             XDataMode: 'manual'
           XDataSource: ''
                 YData: [51x49 double]
             YDataMode: 'manual'
           YDataSource: ''
                 ZData: [51x49 double]
           ZDataSource: ''
             ZLocation: 0

Look at what you can set there. The set of levels originally chosen by the plot code were (in LevelList):

[0.3945 0.5000 1 1.5000 2 2.5000]

It looks like the min was probably at 0.3945. So I'll choose a different set now.

H(1).LevelList = .4:.1:2.5;

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Joseph Cunningham 2024-9-5

Thank you both for your suggestions and assistance. I'll play around with this a bit tomorrow, but it looks to be performing as I hoped! :-)

John D'Errico 2024-9-5

As I said, these things are always simpler when you know to look there. ;-) The good thing about Answers is we all learn something new everyday. I know I do.

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Answer 2

dpb 2024-9-5

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2150464-contour-plot-of-cubic-interpolation-with-set-number-of-contour-levels#answer_1511609

编辑：dpb 2024-9-6

在 MATLAB Online 中打开

Xvec= [-1.8750 -1.1250 -0.3750 -1.8750 -1.1250 -0.3750 -1.8750 -1.1250 -0.3750 -1.8750 -1.1250 -0.3750 0.3750 1.1250 1.8750 0.3750 1.1250 1.8750 0.3750 1.1250 1.8750 0.3750 1.1250 1.8750]';

Yvec=[1.6250 1.6250 1.6250 0.5000 0.5000 0.5000 -0.6250 -0.6250 -0.6250 -1.8750 -1.8750 -1.8750 -1.8750 -1.8750 -1.8750 -0.6250 -0.6250 -0.6250 0.5000 0.5000 0.5000 1.6250 1.6250 1.6250]';

Zvec= [0.5516 0.7743 1.0936 1.1194 2.1194 2.6033 1.0840 2.1033 2.7903 0.4421 1.0936 0.7645 0.7420 0.4292 0.4292 2.6228 1.7775 0.9130 2.5453 1.7389 0.8743 1.0743 0.8358 0.6712]';

Zfit1=fit([Xvec Yvec],Zvec,'cubicinterp');

[X,Y]=meshgrid(linspace(min(Xvec),max(Xvec)),linspace(min(Yvec),max(Yvec)));

Zpred=Zfit1(X,Y);

%whos X* Y* Z*

hC=contourf(X,Y,Zpred,12);

c=colorbar;

%cc=get(c,'Limits');

%set(c,'Ticks',linspace(cc(1),cc(2),10))

set(gcf,'color','w');

That seems strange that the builtin fit object contour option doesn't have the number of levels as a property, but doesn't seem to have.

It appears you are short of data to create a smooth plot and that there may be a case of the cubic rolling over, maybe???

figure

hC3=contour3(X,Y,Zpred,12);

figure

% repeat John's specific levels

LevelList = 0.4:0.1:2.5;

contourf(X,Y,Zpred,LevelList);

colorbar

ADDENDUM: Evaluating the fit at a finer grid smooths out the plot...

One more slight perturbation -- add a finer delta at the peak...

[min(Zpred,[],'all') max(Zpred,[],'all')]

ans = 1x2

0.3937 2.9458

figure

hC3=contour3(X,Y,Zpred,12);

figure

% augment John's specific levels to see shape of peak more fully

LevelList = [0.3:0.1:2.5 2.55:0.05:3.0];

contourf(X,Y,Zpred,LevelList,'ShowText',1);

colorbar

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Joseph Cunningham 2024-9-6

Thank you for the update!

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Contour plot of cubic interpolation with set number of contour levels

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Contour plot of cubic interpolation with set number of contour levels

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