how to save calculating result ( form two variable) in 2-D matrix ?

let's me explaining my question :

for example in 1-D

using for loop:

z=[ ];     

for Eg=.5:.05:1.3;  
a = quad(@(x) (x.^2)./(exp(x/0.516)-1) , Eg , 100 );    
z = [z a]
end

then i use "arrayfun" saving result in 1-D matrix

z=[ ];

Eg=0.5:0.05:1.3; 
a = arrayfun(@(x1)quad(@(x)x.^2./(exp(x/0.516)-1),x1,100),Eg)

Now i want to this process with 2 variables change to 2-d matrix that Eg is on row and Ei is on column.

(here is my code)

z=[ ];     

for Eg=.5:.05:1.3; 
    for Ei=0:0.05:Eg-0.05
aa= quad( @(E)(E.^2) ./ (exp(E./0.5172)-1), Ei , Eg);
z = [z a]
    end
end

thanks :)

Answer 1

[Sargondjani]

the problem is that the length of Ei will differ for the values of Eg. To solve this with for loop i preallocate 'aa' with NaN everywhere

%with for loop
Eg=.5:.05:1.3; 
aa=zeros(17,26)+NaN;
for ig=1:length(Eg);
    Ei=0:0.05:Eg(ig)-0.05;
    for ii=1:length(Ei);
        aa(ig,ii)= quad( @(E)(E.^2) ./ (exp(E./0.5172)-1), Ei(ii) , Eg(ig));
    end
end
clear ig ii

without the for loop: i create a 'meshgrid' and then i cut of every value for Ei: if EI>=EG then i set EI=EG-0.05

%without for loop
Eg=.5:.05:1.3; 
Ei=0:0.05:Eg(end)-0.05;
[EI,EG]=meshgrid(Ei,Eg);
EI(EI>=EG)=EG(EI>=EG)-0.05;

bb = arrayfun(@(E1,E2)quad(@(x)x.^2./(exp(x/0.5172)-1),E1,E2),EI,EG);