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

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],[],[]};

Answer 1

[Guillaume]

A(cellfun(@(x) isequal(x, 0), A)) = {[]};  %replace any cell whose content is the scalar 0 by empty

is one way.

edit: fixed code.

[NA]

I have this error

Unable to perform assignment because the left and right sides have a different number of elements.

[Vladimir Sovkov]

Does not work due to the sizes of the left and right arrays can differ. The working version can be

A(cellfun(@(x) isequal(x, 0), A)) = cell(size(A(cellfun(@(x) isequal(x, 0), A))));

Can be optimized via precomputing cellfun...

[Guillaume]

Yes, I should have tested the code. Either

[A{cellfun(@(x) isequal(x, 0), A)}] = deal([]);  %replace any cell whose content is the scalar 0 by empty

which is a bit convoluted or

toreplace = cellfun(@(x) isequal(x, 0), A);
A(toreplace) = cell(1, nnz(toreplace));

[Stephen23]

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]

Do'h! I knew this was simple. Thanks, Stephen. For some reason, I had it in my head that {} and {[]} were the same thing.

[NA]

Thanks Stephen.