Problems with the legend function.

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I'm in truble with the legend function.
I have a function plot_f.m which plots a function with its legend.
function plot_f
M=1000;
x=linspace(-11,15,M);
y=linspace(-7,11,M);
f1=0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2;
f2=0.1.*y.^3+12.*sin(y)+1.7.^(-y)-y.^2;
f=@(x)(0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2);
p1=plot(x,f1,'Color','b'); hold on;
p2=plot(y,f2,'Color',[0,1,0],'LineWidth',3);
p3=plot(-7.000,f(-7.000),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0]);
plot(-1.850,f(-1.850),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0]);
plot( 4.931,f( 4.931),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0]);
plot(10.007,f(10.007),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0]);
xlim([-15 15])
ylim([-70 80])
ax = gca;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
legend([p1,p2,p3],{'Función', 'Función en [-7,11]', 'Óptimos locales'})
box off
hold on
end
In the main program I make a call to this function and I want to add more information to the legend.
% Ejercicio 2:
% Uso de Matlab para la minimización de una funcion real
% con varios puntos de minimo local :
%
% min: f(x)=0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2
% restricciones: ninguna
clear all;close all;
options = optimset('Display','iter');%,'PlotFcns',@optimplotfval);
salida = fopen('output.txt','w');
global xx; xx=[];
figure; plot_f
hold on
x0=1.2;
% La convergencia del metodo numerico depende
% esencialmente del iterante inicial elegido.
% No obstante, nunca va a converger a -7
% converge a -7, que es 'a simple vista', el minimo
% global del problema con restricciones.
%Convergencias:
% x0 = -5; % xmin = -7.852.
% x0 = -3; % xmin = -1.850.
% x0 = 5; % xmin = 4.931.
% x0 = 10; % xmin = 10.007.
% Cambiar el solver local segun se
% desee: fminsearch o fminunc
f=@(x)f(x);
tic
[xmin,fmin,exitflag,outputs] = fminsearch('f',x0,options);
t1 = toc;
p4 = plot(x0,f(x0),'sq','MarkerSize',8,'MarkerFaceColor',[0 1 1]);
hold on
p5 = plot(xx(:,1),xx(:,2),'ro','MarkerSize',3);
hold on
p6 = plot(xmin,f(xmin),'o','MarkerSize',5,'MarkerFaceColor',[1 1 0]);
hold on
legend([p4,p5,p6],{'Iterante inicial','Iterantes','Optimo local'})
hold off
fprintf(salida,'\n');
fprintf(salida,'Solucion con optimizador local:');
fprintf(salida,'\n');
fprintf(salida,'Solucion optima: xmin = %.3f.' ,xmin);
fprintf(salida,'\n');
fprintf(salida,'Valor de f: fmin = %.3f.',fmin);
fprintf(salida,'\n');
fprintf(salida,'Evaluaciones de la funcion objetivo: k = %i.',outputs.funcCount);
fprintf(salida,'\n');
fprintf(salida,'Tiempo total: t1 = %.3f.',t1);
fprintf(salida,'\n');
fprintf(salida,'Iteraciones del metodo: n = %i',outputs.iterations);
fprintf(salida,'\n');
fprintf(salida,'\n');
function f=f(x)
global xx;
f = 0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2;
xy = [x,f];
xx = [xx; xy];
end
However, I'm just overwriting the data which is currently in the plot's legend. How can I solve this?

Accepted Answer

Sergey Kasyanov
Sergey Kasyanov on 22 Nov 2020
Hello!
The simplest way is to define 'DisplayName' parameter for curves. For example:
plot(-7.000,f(-7.000),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0], 'DisplayName', 'Función');
%...
%...
%...
p4 = plot(x0,f(x0),'sq','MarkerSize',8,'MarkerFaceColor',[0 1 1], 'DisplayName', 'Iterante inicial');
After that all you need to eval:
legend show

More Answers (1)

Alan Stevens
Alan Stevens on 22 Nov 2020
You might have to change the plot order a little. More like this perhaps:
M=1000;
x=linspace(-11,15,M);
y=linspace(-7,11,M);
f1=0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2;
f2=0.1.*y.^3+12.*sin(y)+1.7.^(-y)-y.^2;
f=@(x)(0.1.*x.^3+12.*sin(x)+1.7.^(-x)-x.^2);
x0 = 1.2;
[xmin,fmin,exitflag,outputs] = fminsearch(f,x0);
plot(x,f1,'Color','b'); hold on;
plot(y,f2,'Color',[0,1,0],'LineWidth',3);
plot(x0,f(x0),'o','MarkerSize',8,'MarkerFaceColor',[0 1 1]);
plot(-7.000,f(-7.000),'sq', ...
-1.850,f(-1.850),'sq', ...
4.931,f( 4.931),'sq', ...
10.007,f(10.007),'sq','MarkerSize',8,'MarkerFaceColor',[1 0 0]);
legend('Función', 'Función en [-7,11]','Iterante inicial','Óptimos locales')
xlim([-15 15])
ylim([-70 80])
ax = gca;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
box off

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