Error about preallocating for speed

Hello, I keep getting an error in my code regarding my variable changes size on each iteration and I have to preallocate for speed, how do I fix it?

% Alpha Fe
a = [];
y = 1;
for T = 25:5:1000
    D = 0.0062*exp(-80000/(8.31*T));
    x(y) = 1/T; %error here
    a(y) = D; %error here
    y = y + 1;
end

% Gamma Fe
b = [];
y = 1;
for T = 25:5:1000
    D = 0.23*exp(-148000/(8.31*T));
    b(y) = D; %error here 
    y = y + 1;
end

% plot graph
subplot(195,195,100)
plot(x,a)
plot(x,b)
xlabel("Inverse Temperature (1/K)")
ylabel("Diffusion")
title("Diffusivity vs 1/T for Alpha Fe and Gamma Fe")

[Rohin]

I put "%error" on every line it gave me an error for

Answer 1

[KSSV]

You need to proceed likt his:

% Alpha Fe
a = [];
 T = 25:5:1000 ;
nT = length(T) ; 
a = zeros(nT,1) ; 
 for i = 1:length(T)
    D = 0.0062*exp(-80000/(8.31*T(i)));
    x = 1/T(i); %error here
    a(i) = D; %error here
 end

The above can also be achieved without loop.

 % Alpha Fe
 T = 25:5:1000 ;
 D = 0.0062*exp(-80000./(8.31*T));
 x = 1./T; 

Answer 2

[Steven Lord]

Those lines of code don't error. When Code Analyzer analyzes your code it issues a warning about those lines indicating that there may be room for improvement on those lines of code, but this is neither a run-time warning nor a run-time error.

See the code examples on this documentation page. The second one contains one line of code that is different from the first; that second example preallocates the vector x. Apply the same technique to your code.

That is, of course, if you're required (by the terms of a homework assignment) to use a for loop to construct your a, x, and b vectors. If you're not you can eliminate the loops entirely by using array operators to perform calculations element-wise on all the elements of an array at once.

% a = []; % No longer needed
% y = 1; % No longer needed
T = 25:5:1000; % Make a vector T
    D = 0.0062*exp(-80000./(8.31*T)); % Compute all the elements of D at once with ./
    x = 1./T; % Operate on all the elements of T at once with ./
    a = D; % Overwrite the variable a with the contents of D
%     y = y + 1; % No longer needed
% end % No longer needed

Answer 3

[dpb]

a = [];
y = 1;
for T = 25:5:1000
    D = 0.0062*exp(-80000/(8.31*T));
    x(y) = 1/T; %error here
    a(y) = D; %error here
    ...

You assigned an empty element to a and no reference at all to x so they're having to be reallocated dynamically every pass through the loop...

Do something like

N1=25; dN=5; N2=1000;       % use variables, don't bury data inside code unless it truly will NEVER change
N=numel(N1:dn:N2);          % compute the number elements going to have...
a=zeros(N,1);               % and preallocate...
x=a;
y = 1;
for T = N1:dn:N2
    D = 0.0062*exp(-80000/(8.31*T));
    x(y) = 1/T;
    a(y) = D;
    ...

However, "the MATLAB way" is to not use a loop in such instances at all, but take advantage of the vector operations -- that's what the "MATrix" part of MATLAB is all about..

N1=25; dN=5; N2=1000;   
T = 25:5:1000;
a=0.0062*exp(-80000./(8.31*T));
x=1./T;

and allocation comes along with the assignment of the RHS of the expression "for free".

NB the "dot" operator ./ to operate with the vector T on an element-wise basis instead of the matrix division operation without.