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How can I make the layout in the attached image with tiledlayout

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AP
AP 2024 年 5 月 19 日
編集済み: Matt J 2024 年 5 月 20 日
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.

採用された回答

Star Strider
Star Strider 2024 年 5 月 19 日
編集済み: Star Strider 2024 年 5 月 19 日
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.
.

その他の回答 (2 件)

Matt J
Matt J 2024 年 5 月 19 日
編集済み: Matt J 2024 年 5 月 20 日
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

Image Analyst
Image Analyst 2024 年 5 月 20 日
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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