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About preallocating for speed

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Ender Rencuzogullari
Ender Rencuzogullari el 2 de Dic. de 2015
Comentada: Stephen23 el 17 de Mayo de 2021
Dear Contributers,
There is a " for loop" in my program and Matlab gives me a suggestion to consider "Preallocating" for speed. I want to learn and do that. However, I don't know where to start to learn it. Of course, I looked the documents in mathworks but I found a few documents in there. I can read all of them but I don't want to rush headlong into this topic. Please, I am asking your ideas where I should start from those documents;
By the way, My code is;
for k1 =1: length(X)
for k2 =1: length(X_inv)
DE(k1,k2)= hypot(X(k1)-X_inv(k2), Y(k1)-Y_inv(k2));
end
end
And Matlab suggested me to preallocate the DE
I am not sure whether this question is appropriate for here or not. If it is not okay, accept my apologies.

Respuesta aceptada

John D'Errico
John D'Errico el 2 de Dic. de 2015
Editada: John D'Errico el 3 de Dic. de 2015
When you grow an array incrementally as you are doing with DE, MATLAB is forced to reallocate the entire array at EVERY iteration, copying over the entire mess just to add ONE element. This will cause the operation to get slower, and the time required will grow quadratically. SO it will get SLOW.
You know in the end EXACTLY how large DE will be. So just add ONE extra line before the loop.
DE = zeros(length(X),length(X_inv));
This essentially creates the array in advance, filling it with zeros. It need no longer reallocated at each iteration, because it will no longer need to change size at every step.
Much of the time, preallocation is not needed. MATLAB is not a language where you need to initialize/allocate every array. Variables that are scalars, and will remain so are never an issue. It is only when an array is dynamically grown that preallocation is important. A problem arises when you don't know the final size of your array. Even there, there are tricks that one can do. I posted a submission to the FEX long ago that allows the user to store information in a form that can be more efficiently grown, then at the end, you can unpack the object into a regular array.
  4 comentarios
SINDU GOKULAPATI
SINDU GOKULAPATI el 17 de Mayo de 2021
Editada: Stephen23 el 17 de Mayo de 2021
hey john im facing a similar issue , for context below is my code
a=csvread('D:\sem6\PROJECT\hygdata.csv',1,5,[1,5,5068,7]); %reading x,y,z
n=5068;
//r = zeros(1,n) %error:Unable to perform assignment because brace indexing is not supported for variables of this type.
for i=1:n
r{i} = transpose(a(i,:)); %mztrix [x y z]
end
what is the alternative here?
Stephen23
Stephen23 el 17 de Mayo de 2021
"what is the alternative here?"
Preallocate the array using the correct class:
r = cell(1,n)

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Más respuestas (1)

Madison Schossow
Madison Schossow el 8 de Mzo. de 2018
try doing length(X):-1:1

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