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Find the index of zero in cell and put it as empty

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Na
Na on 7 Jan 2020
Commented: Na on 8 Jan 2020
A={[45;101],[2;7],[5;8],0,0};
A(cell2mat(A)==0)={[]}; % I want to find zero and put in empty
A should be
A={[45;101],[2;7],[5;8],[],[]};

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Guillaume
Guillaume on 7 Jan 2020
Edited: Guillaume on 7 Jan 2020
A(cellfun(@(x) isequal(x, 0), A)) = {[]}; %replace any cell whose content is the scalar 0 by empty
is one way.
edit: fixed code.

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Stephen Cobeldick
Stephen Cobeldick on 7 Jan 2020
Actually the original idea is fine, it just needs a scalar cell array on the RHS:
>> A = {[45;101],[2;7],[5;8],0,0}
A =
[2x1 double] [2x1 double] [2x1 double] [0] [0]
>> A(cellfun(@(x)isequal(x,0), A)) = {[]} % SCALAR CELL ARRAY
A =
[2x1 double] [2x1 double] [2x1 double] [] []
When allocating to the LHS the RHS can always be a scalar, and this applies to cell array just like any other array type. Instead of making this more complex, all that is required is the scalar cell array {[]} rather than the empty cell array {}.
Guillaume
Guillaume on 7 Jan 2020
Do'h! I knew this was simple. Thanks, Stephen. For some reason, I had it in my head that {} and {[]} were the same thing.

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