How to cleanup/fix my mapping of cell-array MATLAB code so I do not hit a recursive limit?

I have to solve a problem where I have to take a vector for ex [1 2 3] and map it using a table such that 1 = [1 1], 2 = [4 6 8], 3 = [6 9 12 15] (a shorter example of actual problem) So my function ideally would be convert([1 2 3]) = [1 1 4 6 8 6 9 12 15]

My thinking process is that I have to make the vector a cell array first so that I can replace the values to others that different dimensions, and then convert cell array back to vector/matrix. This is what I have so far

y = num2cell(x);

for n = 1:length(x)
    if y{n} == 0
        y{n} = [0 2];
    elseif y{n} == 1
        y{n}= [1 1];
    elseif y{n} == 2
        y{n} = [4 6 8];
    elseif y{n} == 3
        y{n} = [6 9 12 15];

      elseif y{n} < 0
          y{n} = 1 - convert(-(x+1));
      end
      output = cell2mat(y);
  end
  end

Everything works fine if my input has the positive values of 0, 1 , 2, or 3 in the initial vector. However, I need to have this condition where if the vector has a negative entry (x), the negative entry gets entered such that y{n} = 1 - convert(-(x+1)) However, when I do this, I get an error of hitting recursive limit. Is there a way to fix my code such that it will work with negative values and not produce an error? Also, is there a way to do it without using a for or while loop? Thanks

PS: Maybe to clarify, I'll give an example of input vector with a negative entry of -1 convert([1 2 3 -1]) = [1 1 4 6 8 6 9 12 15 1 -1]

Basically because the -1 entry will be computed in the formula 1 - convert(-(-1+1)), which then becomes 1 - convert(0) which becomes 1- [0 2], finally to [1 -1]

[Jan]

And convert is the name of the function, correct?

Answer 1

[Andrei Bobrov]

x = [1 2  3 -1];

d0 = 0:3;
d = {[0 2] , [1 1] , 4:2:8, 6:3:15};
tp = x < 0;
x(tp) = -(x(tp)+1);
out = cell(size(x));
[a,ii] = ismember(x,d0);
out = d(ii(a));
out(tp) = cellfun(@(x)1-x,out(tp),'un',0);
out = [out{:}];

Answer 2

[Jan]

You do not want:

y{n} = 1 - convert(-(x+1))

but

y{n} = 1 - convert(-(y{n}+1))

Calling the function with the complete input x must cause an infinite recursion, when x contains positive elements also.

[Joe]

Thanks, however when I do this I get the error Error using convert(line 9) Input must be a column or row vector

Error in convert(line 46) y{n} = 1 - convert(-(y{n}+1));

One of my conditions is that it needs to be a row or column vector

============================ Hmm, if I get rid of the error check, it works. Maybe there is someting wrong with my error check? Here it is [row col] = size(x); if row == col error('Input must be a column or row vector'); end

[Jonathan Sullivan]

The more correct way to check if something is a row vector is as follows:

sz = size(x);         % Get the size of the array
sz(2) = [];           % Remove the second dimension
isRow = all(sz == 1); % Check that all other dimensions have length == 1

[Jan]

@Joe: "row == col" is true for scalars, 2x2 matrixes and [6x3x2] arrays also.

To check for a vector shape, the best idea is obviously isvector(). Alternatively I'd use:

if ndims(x) == 2 && any(size(x) == 1)