How to store a string of n elements as one element of a matrix

I have a function wich give me an array of elements like Nff=[5 0 5 0] and I want to insert Nff into a matrix N(i,j)=Nff. It can be numeric or string. It`s only important to me to store the array in the same row of the others elements 'i' and 'j' as one element of the matrix.

Answer 1

[Peng Li]

You may want to try a cell array instead? N{i, j} = Nff;

[Gustavo]

Hi. Thank for your replay. It would work if I only have Nff but later in my code I insert Nff into a big matrix and it doesn`t work for cells what I do. I`ll try to show.

for....

pat(c,:)=[ptt(a,:) froN(pit(b,a),:) Nt(b,a) Nit(b,a) Npit(b,a) Nff(b,a) Npt1(b,a) tt(b,a)]

end

pat( all(~pat,2), : ) = [];%rows

I want to eliminate the zeros and with pat matrix became a cell matrix so I can not do it

[Gustavo]

I need to keep the numeric numbers in the main matrix. I want to do something like

Nff(i,j)=Nff which change at each iteraction, then I want to insert then into a numeric matrix but I can consider Nff is number, cell, string, whatever

pat=[1 2 3 Nff 4 5 6; 2 1 4 Nff 5 6 1; .....]

pat=[1 2 3 '[5 0 5 0]' 4 5 6; 2 1 4 '[11 12 11 12 11 12]' 5 6 1; .....]

[Peng Li]

Sorry I don't quite understand your question. It's tricky to extend a matrix this way in a loop as you add a block during one loop process to one dimension, the size changes and it conflicts with other dimension. For example, in your example above, you add four element in the first row, it's gonna stop because in this matrix pat, there are only 6 columns in the second row while you force it to have 10 elements in the first row. It's not allowed, and there is no way that you add 6 elements to the second row. They have to be consistent.