How to get the size or position of the plot box within a figure window?
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After much effort, I am still struggling to find the command to get the size or position of the plot box (i.e. the white box) within a figure window. What I really need is the aspect ratio of this white box, but a size or position command would get me there. I've tried variations of 'Position', 'PlotBoxAspectRatio', 'pbaspect',and 'DataAspectRatio' without luck.
For example, if I run the following command,
plot(1:100,1:100,'linewidth',5)
I want to know about the physical size/shape of the white box ONLY, not the grey box:
.
Thanks!
Paul
1 件のコメント
Mr M.
2018 年 7 月 3 日
and how to get the size of the grey area? Because as far as I know, figure size consist the whole are with title bar also. But I need only the grey 'drawable' area.
採用された回答
Dishant Arora
2014 年 3 月 6 日
編集済み: Dishant Arora
2014 年 3 月 6 日
h =figure;
axesHandles = findall(h,'type','axes');
get(axesHandles,'position')
4 件のコメント
Eric
2017 年 4 月 12 日
Hi Paul,
I'm raising this thread from the grave to ask a question about your algorithm for determining the width and height.
PlotBoxWidth=(axesPosition(3)-axesPosition(1))*windowPosition(3);
PlotBoxHeight=(axesPosition(4)-axesPosition(2))*windowPosition(4);
Shouldn't these instead be:
PlotBoxWidth=axesPosition(3)*windowPosition(3);
PlotBoxHeight=axesPosition(4)*windowPosition(4);
For axesPosition in default normalized units and windowPosition in default pixel units, using your code with a square axis doesn't give an aspect ratio of 1, whereas the two lines I gave does.
I am wondering if there is a situation where your code is actually what is needed and my code is wrong. I can imagine that if axesPosition was not in normalized units, one would have to do something differently.
その他の回答 (1 件)
Hieu Nguyen
2018 年 2 月 9 日
You can use the subplot function:
pos = [0.3 0.6 0.2 0.4]; % [left bottom width height]
subplot('Position',pos);
Then, you can plot the figure that you need:
plot(1:100,1:100,'linewidth',5)
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