creating squares of images using imagesc.

The idea is to produce 2D image but is not working. I want each square to correspond to different values of rho(1x101) and beta(1x101) for JCpBest(101x101). please help me

close all
clear all 
% general parameters; will be fixed for all runs
L=1;
T=100;
r=0.03;
I1=0.5;
 
% we will be changing P and epsilon to generate grous of graphs
p=0.0005;
epsilon=0.3;
 
% individual policy plots will be generated by a double loop 
% over beta and rho
beta=0.1;
rho=6000;
 
Mybeta=0:0.003:0.3;
Myrho=0:60:6000;
MycpBest=zeros(numel(Myrho),1);
JCpBest=zeros(numel(Myrho),1);
 
for j = 1:numel(Myrho)
    rho=Myrho(j);
 
for ii = 1:numel(Mybeta)
    beta=Mybeta(ii);
 
% the inner loop will be over Mycp, to find its 'best' value which
% minimmises the loss; here this value is fixed to see what happens
% create an index vector
Mycp = 0:10:100; 
n = zeros(numel(Mycp),1 );
n2 = zeros(numel(Mycp),1 );
n3 = zeros(numel(Mycp),1 );
Jcp = zeros(numel(Mycp),1 );
 
for i = 1:numel(Mycp )
    MycpCurrent=Mycp(i);
delta = 1-MycpCurrent/100 ;
tau = (1/(beta*(L+delta*p)))*log((L*(I1+delta*p))/(delta*p*(L-I1 )));
t05 =(1/(beta*(L+delta*p)))*log((L*(0.05*L+delta*p))/(delta*p*(L-0.05*L )));
I2= @(t)(L*delta*p*(exp (beta*(L+delta*p)*t)-1)) ./ (L + delta*p* exp(beta*(L+delta*p)*t ));
I3= @(t)(L*(I1+delta*p)*exp((epsilon*beta)*(L+delta*p)*(t-tau))-...
    delta*p*(L -I1))./(L-I1+(I1+delta*p)*exp(epsilon*beta*(L+delta*p)*(t-tau)));
 
 
fun = @(t,MycpCurrent) MycpCurrent*L*exp(-r*t);
fun2=@(t)rho*I2(t).*exp(-r*t);
fun3=@(t)rho*I3(t).*exp(-r*t);
 
n(i) = integral(@(t)fun(t,MycpCurrent),0,100, 'ArrayValued',1);
n2(i)= integral(fun2,t05,tau);
n3(i)= integral(fun3,tau,100);
JCp(i)= n(i)+n2(i)+n3(i);
 
end

 
MycpBest(j)=Mycp(JCp==min(JCp));
JCpBest(j)=min(JCp);
end
end
 
 
imagesc(Mybeta,Myrho,JCpBest)
colormap jet
ylim([0 6000]);
xlim([0 0.3]);
ylabel('Loss,\rho_1');
xlabel('Secondary disease transmission rate \beta ');
title('P=0.0005')

Answer 1

[Walter Roberson]

imagesc(Mybeta,Myrho,JCpBest)

You have invoked the form

imagesc(x, y, C)

where x and y give information about where the image is to be placed on the screen, and C is the image to be drawn.

Your Mybeta is Mybeta=0:0.003:0.3 . When used in the first argument, that would be treated the same as if you had passed in [0 0.3] in the first argument.

Your Myrho is Myrho=0:60:6000 . When used in the second argument, that would be treated the same as if you had passed in [0 6000] in the second argument.

These two parameters together mean that the center of the lower left pixel is to be placed at data units x=0 y=0, and that the center of the upper right pixel is to be placed at data units x=0.3 y=6000

Your JCpBest is JCpBest=zeros(numel(Myrho),1) which is a column vector.

You are therefore asking that the one column of data be drawn from data units x=0 to x=0.3, and y=0 to y=6000 . The result of that is going to be a vertical band, not a 2D image.

None of the three parameters you pass to imagesc() are 2D, so it is not just a matter of having accidentally used the wrong order of parameters.

[JACINTA ONWUKA]

Thanks so much for your reply, it just an eye opener. Please advice me on how to go about it?. I dont Know create a matrix from this column vectors.

The second loop(beta), stopped the code from working. If i end it after the first end, it will work but i dont know if is correct.