Error about preallocating for speed

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Rohin
Rohin 2022-9-28
回答： dpb 2022-9-28
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")

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回答（3 个）

KSSV
KSSV 2022-9-28
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;
Steven Lord
Steven Lord 2022-9-28
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
dpb
dpb 2022-9-28
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.

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