plot different parts of vector
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Robin Strak
le 17 Mar 2020
Commenté : Sriram Tadavarty
le 17 Mar 2020
I´ve got a vector M(1x80) and would like to plot four different graphs (graph 1 equals values 1:20, graph 2 values 21:40, ...) and have a name of each of them in the legend (including a trendcurve):
I thougt of a loop like that:
for i = 1:20:80
plot(M3(i);
legend('Graph 1');
end
Can you tell me where the error is?
Thanks a lot for helping me!
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Réponse acceptée
Sriram Tadavarty
le 17 Mar 2020
Modifié(e) : Sriram Tadavarty
le 17 Mar 2020
Hi Robin,
The following code will provide what you are looking for
M3 = rand(1,80);
j = 1;
str = [];
for i = 1:20:80
hold on
plot(M3(i:19+i)); % Update this to have 20 values, which is fixed to 1 in your code
str = [str ["Graph " + num2str(j)]]; % To have different legend for each plot through string array
j = j+1;
end
legend(str)
Hope this helps.
Regards,
Sriram
4 commentaires
Sriram Tadavarty
le 17 Mar 2020
Hi Robin,
Please do accept the answer, if helped.
Here, col is the cell array and it need to be accessed through curly brackets {} rather than open bracket (k).
I am not sure what the variable time is.
Use col{k} instead of col(k), it solves the issue. Here is a partial code
col = {'g', 'y', 'c', 'm'};
for i = 1:20:80
plot(M_3(i:19+i), col{k});
str = [str ["Graph " + num2str(j)]];
c_rms = polyfit(1:20,M_3(i:19+i),1);
y_est = polyval(c_rms,1:20);
hold on
str = [str ["Polyfit " + num2str(j)]];
plot(y_est, col{k}, 'LineWidth',2)
j = j+1;
k = k+1;
end
legend(str)
This should provide some insights of how to solve this. I suggest you to start a new question, if there is any other problem.
Thanking you
Regards,
Sriram
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