How do I set each legend on the curve in multiple plot?

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Dear friends!
I have multiple curves plot for which I need to specify the legends automatically. I use the following script in the code;
figure(2)
txt = ['\alpha=' num2str(p)];
plot(t, L, 'displayname', txt, 'linewidth', 2 );
xlabel('Time(Days)')
ylabel('L')
lgd = legend;
lgd.NumColumns = 2
title(lgd, 'Order of FDE, \alpha')
hold on
The point is, there are 10 legends in the same plot the problem is, each legend is associated with different colors not the shape of the curves when they reached more than 4 or 5 legends it repeat the color which create confusion in recognition of the curves. Now, please help me how can I set my legends on top of each curve like the legends shown in the picture below. It is a contour plot, I need exactly as it contain. While the problem is in the curves with legends associated with beta.
Thanks in advance!

Answers (1)

Chad Greene
Chad Greene on 10 Feb 2021
Perhaps the issue isn't the legend, but plotting the desired colors? How about specifying the colors of the plots explicitly, like this:
x = 1:100;
col = parula(10); % rgb values of 10 colors
hold on
for k = 1:10
h(k) = plot(x,k*x,'color',col(k,:),'linewidth',2)
end
legend(h,'location','northwest')
  3 Comments
Chad Greene
Chad Greene on 10 Feb 2021
Yep, you'll want to loop through each curve and plot it individually. So for 10 curves,
for k = 1:1
h(k) = plot(x,Y(:,k),'color',col(k,:));
end
Or perhaps it's Y(k,:), depending on how your data are arranged. Also be sure the index of the color is the loop index k, not txt as you have written.
Muhammad Sinan
Muhammad Sinan on 10 Feb 2021
Yes, I have used the txt because it is a vector of the values of p which represent the values of \alpha. That's why I used it, I also tried the p directly but showing this error
Index in position 1 is invalid. Array indices must be positive integers or logical values.
Error in Simulation (line 44)
plot(t, S,'displayname', txt, 'linewidth', 2, 'color', col(p, :) );
>>
I don't know how? While the plot(t, S) are already correct.

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