how preallocate structure for better memory
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Salvatore Mazzarino
am 22 Sep. 2012
Kommentiert: Dyuman Joshi
am 27 Mär. 2024
I had created a structure made so:
head.number = 3;
head.pck_rcv = [1 0 0];
heads(2).number = 5;
head(2).pck_rcv = [1 1 0];
and so on.
How can I preallocate a structure?
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Akzeptierte Antwort
Jan
am 22 Sep. 2012
Bearbeitet: Jan
am 2 Okt. 2017
for k = n:-1:1 % Backwards!
head(k).number = 3;
head(k).pck_rcv = [1 0 0];
end
Now the final size of the struct array is created in the first iteration.
[EDITED] Alternatively:
head = struct('number', cell(1, 10), 'pck_rv', cell(1, 10));
Now head is a [1 x 10] struct array withe the fields 'number' and 'pck_rv'. Pre-allocating the contents of the fields is another job and you need a loop to do this. But now it can run in forward direction also.
5 Kommentare
Igor Gitelman
am 20 Mai 2022
thanks! that
head = struct('number', cell(1, 10), 'pck_rv', cell(1, 10));
work fine!
Dyuman Joshi
am 27 Mär. 2024
I am accepting Jan's answer as it provides a robust solution to the question posted.
Weitere Antworten (2)
Azzi Abdelmalek
am 22 Sep. 2012
Bearbeitet: Azzi Abdelmalek
am 22 Sep. 2012
heads=struct('numbers',zeros(10,1), 'pck_rcv',zeros(10,3))
%then
for k=1:n
heads.numbers(k)=2
heads.pck_rcv(k,:)=[1 2 3]
end
3 Kommentare
Jan
am 2 Okt. 2017
@Alexandra: I do not agree. Salvatore asked for a struct array: "head(2).numbers and so on". Azzi's suggestion creates a scalar struct only.
Alexandra Simpson
am 2 Okt. 2017
True, I just tried it out and realised it wasn't what I wanted either! Thanks for the response.
Walter Roberson
am 13 Dez. 2012
head = struct('number', {3, 5}, 'pck_rcv', {[1 0 0], [1 0 1]})
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