Replacing elements of multilevel cell arrays

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Sukru Yavuz
Sukru Yavuz le 10 Avr 2018
Modifié(e) : Guillaume le 10 Avr 2018
Hello, I have an multilevel cell arrays of meancalc. For example: To reach any of them, I have to write meancalc{1,1}{:,:}, meancalc{1,2}{:,:},meancalc{1,3}{:,:}. I want to reach the Inf values and change them with epsilon. But I couldn't reach them out with meancalc(meancalc{:,:}==Inf) = eps; Thanks in advance.
  3 commentaires
Sukru Yavuz
Sukru Yavuz le 10 Avr 2018
I am calculating the standard deviation of columns. And the values of some columns are identical. That's why the results are 0. After that step, I am using those zeros in a multiplication. That's why I am getting the Inf results. My advisor told me to replace them with epsilon to avoid Inf or NaN or 0s.
Guillaume
Guillaume le 10 Avr 2018
Modifié(e) : Guillaume le 10 Avr 2018
Well, you of course do whatever you want, but replacing an easily spotted erroneous value (maybe?) by an arbitrary constant value does not sound logical to me. Why eps(1) instead of exp(1) or sqrt(2) or eps(1e-10) or any other random arbitrary value that has no more significance than eps(1) to your algorithm?

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Guillaume
Guillaume le 10 Avr 2018
Assuming that all meancalc{t,u}{v,w} contain matrices, you'll have to use a double for loop:
for outer = 1:numel(meancalc)
for inner = 1:numel(meancalc{outer})
innermat = meancalc{outer}{inner};
innermat(isinf(innermat)) = eps(1);
meancalc{outer}{inner} = innermat;
end
end
Note: acellarray{:} will return a comma separated list, so your meancalc{:,:}==Inf would never work.
  1 commentaire
Sukru Yavuz
Sukru Yavuz le 10 Avr 2018
It works perfectly. Thank you!

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Plus de réponses (1)

Birdman
Birdman le 10 Avr 2018
Modifié(e) : Birdman le 10 Avr 2018
Usage of cellfun might help:
idx=cellfun(@(x) isinf(x),meancalc)
This will output logical array. Then,
meancalc(idx)={234};%Inf's are replaced with a random value, change this with eps
  8 commentaires
Birdman
Birdman le 10 Avr 2018
Modifié(e) : Birdman le 10 Avr 2018
Let me share the code that I worked so that we can clear the confusion:
%demo data
A(:,:,1)=[1 2;3 Inf];
A(:,:,2)=[Inf 2;5 6];
A(:,:,3)=[3 Inf;Inf 3];
meancalc=num2cell(A);
% what I understood by multilevel cell was each element was contained as a cell
meancalc=num2cell(meancalc)
%this found the indexes of Inf's
idx=cellfun(@(x) isinf(x{:}),meancalc)
Guillaume
Guillaume le 10 Avr 2018
Modifié(e) : Guillaume le 10 Avr 2018
I don't think that's what Sukry has (and a cell array where each element is a scalar would be a big waste of memory). A representative example would be
A{1} = {[1 2; 3 Inf], [1 2 3]; [Inf; -1; NaN], [3 Inf; Inf 3]}
A{2} = {[], [Inf 2; 5 6], magic(3)}
That is you have matrices inside cell arrays themselves inside a cell array. This would be consistent with the notation A{i}{j}

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