Assigning row to array using cellfun
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Konrad Warner
am 11 Nov. 2019
Kommentiert: Konrad Warner
am 12 Nov. 2019
Hello,
I have a 1x4 cell array, each cell contains a 1000x8 zeros array:
A = arrayfun(@(~) zeros(1000,8),(1:4),'un',0);
Now I want to assign every same row (definded by a background counter) with a new array same length, e.g B = [1 1 1 1 1 1 1 1]
The following loop discribes it quite well...
counter = randi(1000) % any row
for i = 1:4
A{i}(counter,:) = B;
end
I was just wondering if there is a way doing it with cellfun?
1 Kommentar
JESUS DAVID ARIZA ROYETH
am 11 Nov. 2019
maybe this code can help you
A=zeros(1000,1);
counter = randi(1000); % any row
A(counter)=1;
A=repmat({repmat(A,1,8)},1,4)
Akzeptierte Antwort
Adam Danz
am 11 Nov. 2019
Bearbeitet: Adam Danz
am 11 Nov. 2019
Here's your loop, within cellfun().
A = cellfun(@(x)[x(1:counter-1,:);B;x(counter+1:end,:)], A, 'UniformOutput', false);
3 Kommentare
Adam Danz
am 11 Nov. 2019
Really? I just timed the loop method from your question and the cellfun() method from my answer. Each method was repeated 100,000 times and each iteration was timed using tic/toc. When comparing the median speeds, the loop method was ~22 times faster than the cellfun method (I repeated that process twice and got nearly the same results).
The cellfun() method is (arguably) cleaner and reduces the number of lines of code but often times loops are faster than cellfun(), arrayfun() etc. I wonder if your timing test involved additional computations.
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