How to plot a legend with multiple confidence intervals?

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I have a plot with a line and 2 confidence intervals (90% and 68%). So, 5 lines in total. I would like to display only the legend for the central line. There are many answers around the web but I didn't find one that fixes my issue.
This is my code I tried without success:
X.label = [0 10 20 30];
x= xlsread('x.xlsx')
plotarea_b(x,X.label);
ylabel('%')
title('X1', 'FontWeight', 'Normal')
legend({'', '', '', '', 'X1'}, 'Location', 'southoutside')
Attached I provide the function I use to plot ('plotarea_b') and the dataset of the series I am plotting ('x.xlsx').
Can anyone help me?
Thanks!

Accepted Answer

Adam Danz
Adam Danz on 14 Nov 2020
Edited: Adam Danz on 14 Nov 2020
Use the graphics object handles to specify which objects you want in the legend.
See 4 "% ADDED" comments below
X.label = [0 10 20 30];
x= xlsread('x.xlsx');
[output,ha1,ha2,ha3] = plotarea_b(x,X.label); % ADDED
ylabel('%')
title('X1', 'FontWeight', 'Normal')
legend([ha1(2),ha2(2),output],{'error1','error2', 'X1'}, 'Location', 'southoutside') % ADDED
function [output,ha1,ha2,ha3] = plotarea_b(datamat,xlabel) % ADDED
ha1 = area([datamat(:,1) datamat(:,end)-datamat(:,1)],'linestyle','none');
hold on;
set(ha1(1), 'FaceColor', 'none');
set(ha1(2), 'FaceColor', [0.3,0.6,0.9], 'FaceAlpha', 0.5) ;
if size(datamat,2)>3
ha2 = area([datamat(:,2) datamat(:,end-1)-datamat(:,2)],'linestyle','none');
set(ha2(1), 'FaceColor', 'none');
set(ha2(2), 'FaceColor', [0.25,0.3,0.8], 'FaceAlpha', 0.2) ;
output = plot(datamat(:,3), 'Color', [0.25,0.3,0.8],'linewidth',1);
else
output = plot(datamat(:,2),'Color', [0.25,0.3,0.8],'linewidth',1);
end
ha3 = plot(zeros(size(datamat,1),1),'k','linewidth',0.5);% ADDED
xticks(xlabel)
xticklabels(cellstr(num2str(xlabel')))
end
  2 Comments
Adam Danz
Adam Danz on 15 Nov 2020
Edited: Adam Danz on 15 Nov 2020
Glad I could help. Lesson learned: specify object handles when creating the legend.
It's also a good idea to use the DisplayName property to define legend strings (see this answer).

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