App designer - display multiple ROI on the same image

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Teshan Rezel
Teshan Rezel 2021 年 3 月 10 日
コメント済み: Teshan Rezel 2021 年 3 月 10 日
Hi,
I want to display a crosshair and a circle right in the middle of my image. I wanted to do this by defining a function, as below.
I'm not sure how to proceed but whilst a similar approach works in the live editor, the app design function doesn't like this syntax. Is there any way to approach this please? I'm not familiar with the app design environment.
The image is always displayed in a UIFigure object. I've input the values [472, 323] as a position vector for testing. Ideally, they would point to the centre of the image.
Thanks in advance!
function DisplayImage(app)
app.Image.ImageSource = app.fullname;
app.crossHair = images.roi.Crosshair(gca, 'Position', [472, 323], 'Color', 'w');
app.circle = images.roi.Circle(gca, 'Centre', [472, 323], 'Radius', 100, 'Color', 'w');
end

採用された回答

Adam Danz
Adam Danz 2021 年 3 月 10 日
I assume "doesn't like this syntax" means that the crosshairs and circle are appearing on a different figure.
The reason that would happen is because of your use of gca instead of the actual axis handle.
Assuming your axis handle is app.UIAxes,
function DisplayImage(app)
app.Image.ImageSource = app.fullname; % ??? Not sure what this is
app.crossHair = images.roi.Crosshair(app.UIAxes, 'Position', [472, 323], 'Color', 'w');
% ^^^^^^^^^^
app.circle = images.roi.Circle(app.UIAxes, 'Center', [472, 323], 'Radius', 100, 'Color', 'w');
% ^^^^^^^^^^^
end
  3 件のコメント
Adam Danz
Adam Danz 2021 年 3 月 10 日
編集済み: Adam Danz 2021 年 3 月 10 日
Great question and it's not as straightforward as you may imagine.
The image dimensions and the axis dimensions do not necessarily agree. Consider this example where pixel (m,n) of the image is centered at coordinate (m,n) but the pixel size is 1x1 so each pixel extends +/-0.5 from the center. The far left side of the image starts at x=0.5 and the far right side ends at n+.5, same with the lower and upper edges.
I = imread('baby.jpg');
fig1 = figure();
ax = gca(fig1);
image(ax,I)
axis(ax,'equal')
axis(ax,'tight')
size(I); % [3600 2250 3]
xlim(); % [0.5 2250.5]
ylim(); % [0.5 3600.5]
Furthermore, some image plotting functions allows the user to specify the location of the image on the axes. Compare the axis ticks in the figure above to the one below.
fig2 = figure();
ax = gca(fig2);
image(ax, [1000,1100], [-1 -.5], I)
So, if the image consumes the axes, what you really want to find is the center of the axes.
axCnt = [ax.XLim(1)+range(ax.XLim)/2, ax.YLim(1)+range(ax.YLim)/2];
hold(ax,'on')
plot(ax, axCnt(1), axCnt(2), 'wx','MarkerSize',20)
Teshan Rezel
Teshan Rezel 2021 年 3 月 10 日
@Adam Danz this worked really well, thank you!

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