How can I make the layout in the attached image with tiledlayout
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I am able to get plot 1 and plot 2 in but not 3, 4 and 5. I can also get 3, 4 and 5 in without plot 1 and 2 but that is not what I want, per the attached image.
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Accepted Answer
Star Strider
on 19 May 2024
Edited: Star Strider
on 19 May 2024
It took a few experiments to get this to work.
Try this —
x = 0:0.1:31;
y = sin((1:5).'*x*2*pi/32);
figure
tiledlayout(4,4)
nexttile([1 2])
plot(x,y(1,:))
grid
title('Plot 1')
nexttile([2 2])
plot(x,y(3,:))
grid
title('Plot 3')
nexttile([1 2])
plot(x, y(2,:))
grid
title('Plot 2')
nexttile([2 2])
plot(x, y(4,:))
grid
title('Plot 4')
nexttile([2 2])
plot(x, y(5,:))
grid
title('Plot 5')
EDIT — Corrected typographical errors.
.
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More Answers (2)
Matt J
on 19 May 2024
Edited: Matt J
on 20 May 2024
If you download nestedLayouts,
then you can do,
x = 0:0.1:31;
y = sin((1:5).'*x*2*pi/32);
[ax,t]=nestedLayouts([2,2],[2,1]);
delete(ax([4,6,8]));
ax=ax([1,2,3,5,7]);
for k=1:numel(ax)
plot(ax(k), x,y(k,:));
title(ax(k), "Plot "+k);
if k>2, ax(k).Layout.TileSpan=[2,1]; end
end
for i=1:numel(t), ylabel(t(i),"YLabel "+i); end
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Image Analyst
on 20 May 2024
If you understand how subplots work, it's easy. The bottom and right plots just use a 2,2 layout while the upper left two use a 4,2 layout:
subplot(4, 2, 1);
title('Plot 1');
subplot(4, 2, 3);
title('Plot 2');
subplot(2, 2, 2);
title('Plot 3');
subplot(2, 2, 3);
title('Plot 4');
subplot(2, 2, 4);
title('Plot 5');
Many people don't realize it but whatever layout you use for your first subplot(s) does not need to be used for ALL subplots on the figure, so you can switch from 4 rows, 2 columns to 2 rows and 2 columns like I did above.
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