Find indices in cell array
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Hello everybody,
I have some problems to get the indices of a certain value, in this case -1, within a cell array. The structure of the cell array is like this: A is a 27x1 cell where each of the 27 rows has 500000 cell entries (numeric values between -1 and 999). What I want are the indices where the value -1 occurs, since I have to replace them afterwards. For example A{1,1}{4,1} or A{24,1}{95423,1} are containing the value -1. First of all I tried index=cellfun(@(x) x==-1, A{1,1}, 'Uniform', false) and then I used index2=cellfun(@(x) x==-1, A{2,1}, 'Uniform', false) and so on (. So I could get the information separately but I think there should exist a solution where I don't have to define 27 index cell arrays. Hopefully someone can help and recommend a more meaningful solution. I used the search function but none of the solutions in similar topics did work for my case (using cellfun(@(x) find(x==-1), A{1,1}, 'UniformOutput',false) leads to the same approach like I did it already for example).
Thanks in advance :)
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Respuestas (3)
Andrei Bobrov
el 25 de Oct. de 2017
Aw = cellfun(@(x)[x{:}]',A,'un',0);
Aw = [Aw{:}];
[ii,jj] = find(Aw == -1) ;
out = [jj,ii];
7 comentarios
Stephen23
el 27 de Oct. de 2017
Editada: Stephen23
el 27 de Oct. de 2017
@Timo: always store data in the simplest array-class possible. It is up to you to decide where that boundary is. A good rule of thumb is to start with simple numeric arrays, and only change to a more complex data class when it proves necessary to do so.
Jos (10584)
el 25 de Oct. de 2017
A = {[-1 0 0 -1],1:4,[1 -1 1 -1 1 -1]} % example array
fhFind = @(x) find(x == -1) ;
IDX = cellfun(fhFind ,A,'un',0)
0 comentarios
Jan
el 25 de Oct. de 2017
Editada: Jan
el 25 de Oct. de 2017
After the data are converted to a numerical array, the processing is very efficient. Consider, if it would be better to store the data as numerical array directly instead of the complicated nested cell.
If you really need the cells, and want to replace the -1 only, it might be better to avoid the conversion to one huge numerical array:
for iA = 1:numel(A)
B = A{iA};
for iB = 1:numel(B)
C = B{iB};
C(C == -1) = 42;
B{iB} = C;
end
end
This is not nice, but it is worth to try if it is faster.
7 comentarios
Jan
el 28 de Oct. de 2017
@Tima: You have set all values of the output to {-99} in this line already:
union{i,1}(:,2)={-99};
and later crop of all other information:
union2{i,1}=union{i,1}(:,2);
Therefore it is not clear to me, what you want to achieve.
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