plot different parts of vector
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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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採用された回答
Sriram Tadavarty
2020 年 3 月 17 日
編集済み: Sriram Tadavarty
2020 年 3 月 17 日
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 件のコメント
Sriram Tadavarty
2020 年 3 月 17 日
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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