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PCOLOR command in Matlab App Designer

조회 수: 11 (최근 30일)
Michal Cerny
Michal Cerny 2022년 1월 2일
댓글: Image Analyst 2022년 1월 2일
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

채택된 답변

Walter Roberson
Walter Roberson 2022년 1월 2일
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')
  댓글 수: 2
Michal Cerny
Michal Cerny 2022년 1월 2일
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 2022년 1월 2일
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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