Replacing elements of multilevel cell arrays

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

Guillaume
Guillaume le 10 Avr 2018
Why eps(1)? If it's just an arbitrary small value, eps(1) is not that small (compared to realmin)
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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 Réponse acceptée

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.

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

Sukru Yavuz
Sukru Yavuz le 10 Avr 2018
What does x stand for? If they stand for meancalc, then I am getting this error: Undefined function 'isinf' for input arguments of type 'cell'.
Birdman
Birdman le 10 Avr 2018
Hmm, which version of MATLAB are you using?
Guillaume
Guillaume le 10 Avr 2018
Modifié(e) : Guillaume le 10 Avr 2018
No cellfun won't help because:
a) you want to do an assignment which is not supported by anonymous functions. You'd need a separate function to do the assignment, at which point you're better off using an explicit loop (in my opinion)
b) the content of the cell array is itself a cell array, so you'd need at least a double cellfun but... see a)
Birdman
Birdman le 10 Avr 2018
Actually this approach worked for me.
Birdman
Birdman le 10 Avr 2018
Modifié(e) : Birdman le 10 Avr 2018
Oh, it says multiple cell arrays. Then,
idx=cellfun(@(x) isinf(x{:}),meancalc)
this one works.
isinf(x{:})
will return Error using isinf. Too many input arguments if the cell array x contains more than one element. e,g.
x = {1 2 3}
isinf(x{:})
x{:} is expanded in the comma-separated list 1,2,3, so the above is equivalent to
isinf(1, 2, 3)
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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