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PCOLOR command in Matlab App Designer

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Michal Cerny
Michal Cerny el 2 de En. de 2022
Comentada: Image Analyst el 2 de En. de 2022
I have achieved working and running code in Matlab, but I have problem converting it for use in the App Designer. I am mostly struggling with the syntax. For the graph I am using UIAxes in app designer.
the code:
E=[1 1 0 0 1 0;1 1 0 0 1 0;1 1 0 0 1 0;1 1 0 0 1 0;1 1 0 0 1 0;1 1 0 0 1 0;]
grid=pcolor(padarray(E,[1 1],'replicate','post'))
grid.EdgeColor=[0 0 0]
grid.LineWidth=3
colormap(jet(4))
set(gca,'ydir','reverse')
axis equal
Thank you very much for your help
Michal Cerny

Respuesta aceptada

Walter Roberson
Walter Roberson el 2 de En. de 2022
Instead of passing the uiaxes as the first parameter, use the 'Parent' name/value pair.
grid = pcolor(padarray(E,[1 1],'replicate','post'), ...
'Parent', app.UIAxes, 'EdgeColor', [0 0 0], 'LineWidth', 3);
colormap(app.UIAxes, jet(4))
app.UIAxes.YDir = 'reverse';
axis(app.UIAxes, 'equal')
  2 comentarios
Michal Cerny
Michal Cerny el 2 de En. de 2022
Error using pcolor (line 50)
Too many input arguments.
--------------------------------
Thank you for help. For some reason it still gives the error above. There is no other "pcolor" command used anywhere in the code.
--------------------------------
Image Analyst
Image Analyst el 2 de En. de 2022
Try this:
E = rand(5,5); % Just for demo - you should delete this line.
app.UIAxes = axes(); % Just for demo - you should delete this line.
paddedE = padarray(E,[1 1],'replicate','post')
paddedE = 6×6
0.1886 0.4694 0.1405 0.3392 0.2175 0.2175 0.8035 0.1835 0.3030 0.5094 0.1928 0.1928 0.6563 0.1289 0.4795 0.6464 0.4215 0.4215 0.6175 0.4941 0.3259 0.2673 0.0400 0.0400 0.2382 0.3621 0.1431 0.6461 0.7562 0.7562 0.2382 0.3621 0.1431 0.6461 0.7562 0.7562
hp = pcolor(paddedE, 'Parent', app.UIAxes);
hp.EdgeColor = [0 0 0];
hp.LineWidth = 3;
colormap(app.UIAxes, jet(4))
app.UIAxes.YDir = 'reverse';
axis(app.UIAxes, 'equal')

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