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

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AP
AP am 19 Mai 2024 um 20:43
Bearbeitet: Matt J am 20 Mai 2024 um 11:03
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.

Akzeptierte Antwort

Star Strider
Star Strider am 19 Mai 2024 um 21:51
Bearbeitet: Star Strider am 19 Mai 2024 um 21:53
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.
.

Weitere Antworten (2)

Matt J
Matt J am 19 Mai 2024 um 23:02
Bearbeitet: Matt J am 20 Mai 2024 um 11:03
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 am 20 Mai 2024 um 3:16
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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