Draw rectangle on existing graph

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deejt 2021-6-4

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https://ww2.mathworks.cn/matlabcentral/answers/848095-draw-rectangle-on-existing-graph

评论： Star Strider 2023-3-10

采纳的回答： Star Strider

在 MATLAB Online 中打开

How can I plot a rectangle over an existing graph using vectors and matrices, instead coordinates?

I plotted a graph from a matrix.

a = rand([-3.5,1.5],1300,3) % matrix to plot with each row representing 1 line in graph 
plot(times,a) % times is the vector with the time points, x axis 
ylim([1.5 -3.5]) % I prepare the axis of y so that negative is on top, and positive at the bottom 
set(gca,'YDir','reverse') % I need to plot the negative as upward going, therefore I reverse the axis 
hold on; 

Now I have two vectors and would like to plot the vectors as two rectangles over the existing graph. However, I found some information about having to rezize the graph before I can plot the rectangles and in the documentation I could not find the information to use a vector. Instead the documentation requires that the coordinates are typed in.

rect_1idx = [66,166] % the first rectangle should start at 66 on the x axis and end on 166 on the x axis
rect_2idx = [170,270] % the same as above 
% the hight of the rectangle should go  from -3.5 to 1.5 on the y axis 

This is what I have tried and it did not work out

rectangle('Position',[-3.5 66 100 1.5])

The goal is to get a figure similar to this one https://www.researchgate.net/publication/330810145/figure/fig3/AS:721682620772352@1549073953701/ERP-wave-from-O2-ERP-waveform-from-electrode-O2-displaying-the-P1-component-Cannabis_W640.jpg

Do you guys have an idea how I could solve this?

Thank you for you help!

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Adam Danz 2021-6-4

This is not a valid syntax: a = rand([-3.5,1.5],1300,3).

Perhaps you meant,

bounds = [-3.5,1.5];

a = rand(1300,3)*range(bounds)+bounds(1);

This is also no valid: ylim([1.5 -3.5]) since the bounds must be in ascending order.

ylim(bounds)

The rectangle you drew starts at x=-3.5 and y=66; you probably meant x=66 and y=-3.5

rectangle('Position',[66 -3.5 100 1.5])

Finally, if you want curvature like the rectangles shown in the link you shared,

rectangle('Position',[66 -3.5 100 1.5],'Curvature',0.3)

All of this is explained in the documentation: rectangle.

deejt 2021-6-5

编辑：deejt 2021-6-5

在 MATLAB Online 中打开

Thank you for your response.

Maybe my problem was not very well explained.

The first block of code you commented on, was never the problem, as I explained with several comments next to it and also the text above. The code line you commenetd on is my matrix that I am plotting and holding on. That section represents 3 lines representing my data, not the rectangle. And I also explained, why I reversed the y axis.

I only posted the code block to present my problem as detailed as possible and provide you guys with a dataset.

a = rand([-3.5,1.5],1300,3) % matrix to plot with each row representing 1 line in graph 
plot(times,a) % times is the vector with the time points, x axis 
set(gca,'YDir','reverse') % I need to plot the negative as upward going, therefore I reverse the axis 
hold on; 

I do see that I made a mistake in the first line, when I posted the question here ( but not in my actual code, so its just a transfer error).

% a = rand([-3.5,1.5],1300,3) % wrong 
a = randi([-3,1],3,1300) % this is the correct code line

And the link I sent was mearley meant as a visualization. i do not need the curvate, otherwise I would have asked for it, too (or looked up the doc).

My problem starts after this first code block, namely how can I use existing vector variables to create rectangles, instead of the location input arguments as presented in the rectangle documentation of MATLAB.

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