Draw rectangle on existing graph

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!

[Adam Danz]

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]

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.

Answer 1

[Star Strider]

Create an anonymous function ‘rectplot’ (name it whatever you want) to do the calculations and plotting —

x = linspace(0, 300, 500);
y = 0.5*sin(2*pi*x.^0.35)-3;

rectplot = @(x1,x2) rectangle('Position',[x1 -3.5 (x2-x1) 1.5]);

figure
plot(x, y)
ylim([-5 -1])
rectplot(66,166)
rectplot(170,270)

.

[deejt]

Thank you very much! Your code solved my problem!!

if I wanted to specify that the two rectangles have different colours, i would modify this code, right?

rectplot = @(x1,x2) rectangle('Position',[x1 -3.5 (x2-x1) 1.5]); 

rectplot = @(x1,x2) rectangle('Position',[x1 -3.5 (x2-x1) 1.5], 'FaceColor','r');

How can I specify two different colours though?

Say, the first rectangle is supposed to be red and the second one blue?

[Star Strider]

My pleasure!

Add the colour to the argument list as the third argument —

rectplot = @(x1,x2,c) rectangle('Position',[x1 -3.5 (x2-x1) 1.5], 'FaceColor',c);

Note that it is now necessary to specify the colour each time the function is called.

x = linspace(0, 300, 500);
y = 0.5*sin(2*pi*x.^0.35)-2.8;

figure
rectplot(66,166,'b')
rectplot(170,270,'r')
hold on
plot(x, y)
hold off
ylim([-5 -1])

It is also necessary to plot the rectangles first, before plotting the waveform, so that the waveform will be visible, since there is no 'FaceAlpha' (transparency property) in rectangle objects.

If having the faces transparent is important, that would require coding a patch object. This is not difficult, however requires a different function. Also, some rectangle properties (such as 'Curvature') would not be available.

.

[deejt]

Thank you so much for your help! :)

[Star Strider]

As always, my pleasure!

[Elisheva Sherman]

Hi, This has been very hekpful for me too! How would I change the locations of the boxes with respect to the y axis? For example if I wanted to shift the blue box up and the red box down from there curret loction.

[Adam Danz]

Hint:

rectangle('Position',[x1 -3.5 (x2-x1) 1.5], 'FaceColor',c)
%                     x    y     w      h
% x = x coordinate of lower left corner
% y = y coordinate of lower left corner
% w = width of rectangle
% h = height of rectangle

[Elisheva Sherman]

Thank you so much!!

[Star Strider]

My pleasure!

A Vote would be appreciated!