I need help preallocating array?

I need help making this array preallocated.
p = 10;
newArray = [];
array = [];
for i = 1:p
a= variable1;
b = variable2;
c = variable3;
d = variable4;
e = variable5;
array = [a,b,c,d,e]
newArray= [newArray array];
end

 Respuesta aceptada

pfb
pfb el 12 de Abr. de 2015
Look at the function "zeros".
array = zeros(1,5);
newArray = zeros(1,5*p);
and, in the loop
newArray((1+(i-1)*5):(i*5))=array;
This way all you need is allocated before the loop.
I assume that the assignments of a through e come from function that have a different value for different i's.
If they are the same, it's just a tiling problem. You need "repmat".

5 comentarios

ME
ME el 12 de Abr. de 2015
Thank you soooo much thats really helpful. could you please explain this newArray((1+(i-1)*5):(i*5))=array; so i am able to manipulate it to suit my program. because I want this to create rows = p which in my case in 10.
pfb
pfb el 12 de Abr. de 2015
well, it's simply what you want to do, I guess.
At the ith iteration you put the content of "array" into the elements of newArray going from 1+(i-1)*5 to (i*5).
That is:
- when i = 1 the elements go from 1 to 5 - when i = 2 the elements go from 6 to 10 - and so on
This avoids concatenating array at the end of newArray, which in general slows you down.
But now you talk of rows.. What exactly do you want to attain here? Should newArray be a very long row vector or a p by 5 matrix?
ME
ME el 12 de Abr. de 2015
I want new array to be 5 columns and 'p' rows.
ME
ME el 12 de Abr. de 2015
in my original loop the reason i have
myArray = [myArray array];
was so that in each iteration array in copied in new row.
pfb
pfb el 12 de Abr. de 2015
Editada: pfb el 12 de Abr. de 2015
not really in a new row.
In your code "array" is copied at the end of the row vector newArray. No new rows.
Use the semicolon operator to add new rows. You should have written
myArray = [myArray; array];
I think you should read some documentation.
Anyway, to allocate your matrix (not vector)
newArray = zeros(p,5);
Filling it, now is much simpler. This fills the ith row of newArray with the current values in array.
newArray(i,:) = array;
You should state your problem better. Apparently, what you needed was filling the rows of a matrix, all the time.
And perhaps take a look into the basics of matlab.

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Preguntada:

ME
el 12 de Abr. de 2015

Editada:

pfb
el 12 de Abr. de 2015

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