Make structures from a cell and store them in a new cell

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Asim Ismail 2017-8-16

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I have a function yal which gives a square cell of size nxn, each cell containing a vector of two columns and multiple rows. I want to make structure for each cell, the structures will have:

 a.fc=[:,1];  % First column of structure a will store first column of vector
 a.sc=[:,2]; % Second column of structure a will store second column of vector
 a.addition=[:,1]+[:,2]; % Third column is the sum of both columns of vector
 (There can be more fields added to the structure a)

At the end these structures need to be stored in a new cell.

Can anyone please help me out..

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Stephen23 2017-8-16

" I think storing them in a cell will be more convenient."

Splitting up a perfectly good non-scalar structure into the cells of a cell array will be inefficient, very inconvenient, and most likely a total waste of time.

Asim Ismail 2017-8-16

编辑：Asim Ismail 2017-8-16

Ok lets ignore the final part (storing them in a cell) and just get structures for each vector of the cell.

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Guillaume 2017-8-16

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As per Stephen's comment using a non-scalar structure as the final result would make a lot more sense (and be a lot easier).

Anyway, to do your conversion the easiest way would be to rearrange your cell array so that the columns of the matrices are actually split into cells of the cell array in the 3rd dimension. A call to cell2struct then do the conversion to structure:

C = arrayfun(@(x) randi([1 25], x, 2), magic(5), 'UniformOutput', false);  %generated demo data that fits the bill (nx2 matrix in each cell)
temp = cellfun(@(c) cat(3, {c(:, 1)}, {c(:, 2)}, {c(:, 1) + c(:, 2)}), C, 'UniformOutput', false);
temp = reshape([temp{:}], [size(C), 3]);   %the 3 is the number of field
resultstruct = cell2struct(temp, {'fc', 'sc', 'addition'}, 3);  %the 3 means use 3rd dimension for fields