[sol]=aaa(@ptch);
[x1,x2]=ptch(sol)
plot(x1,x2,'r.-');hold on;
xlabel('x1');ylabel('x2')
legend('Modified');
pause('on')
fprintf('Press any key to continue to other algorithm')
pause
[Sol]=bbb(@ptch);
[x1 x2]=ptch(Sol)
plot(x1,x2,'k.-'); hold on;
xlabel('x1');ylabel('x2')
legend('pre modified');
toc
how to display the legend in such a case?

6 Kommentare

Azzi Abdelmalek
Azzi Abdelmalek am 30 Apr. 2016
What case?
JL555
JL555 am 30 Apr. 2016
Bearbeitet: JL555 am 30 Apr. 2016
That coding i posted...because only one legend comes and not 2
Walter Roberson
Walter Roberson am 30 Apr. 2016
You legend once for every iteration of your for loops. You need 42 legend entries.
JL555
JL555 am 30 Apr. 2016
Bearbeitet: JL555 am 30 Apr. 2016
My bad let me edit the code
Walter Roberson
Walter Roberson am 30 Apr. 2016
Are your x2 vectors or 2 dimensional arrays?
JL555
JL555 am 1 Mai 2016
Bearbeitet: JL555 am 2 Mai 2016
both x1 and x2 is 1x50 double

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 Akzeptierte Antwort

Walter Roberson
Walter Roberson am 2 Mai 2016

0 Stimmen

Do not call legend() twice. Call it once and pass in both entries.

1 Kommentar

JL555
JL555 am 2 Mai 2016
Bearbeitet: JL555 am 2 Mai 2016
yes walter i figured it out...by the way thanks anyway do you have any idea on pareto frontier for 2 objective functions?

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