Concatenate two cell colums with different length, in Variables window it is possible while horzcat gives "not consistent" error.

I have two celll columns.

a={'11';'21';'31'};
b={'12';'22'};

if I issue the

c=[a,b];

command it gives the error

Error using horzcat
Dimensions of arrays being concatenated are not consistent.

However, if I do a

c=a;

and then I go to the Variables window and open 'c' and 'b' variables and copy the first (and only) column of 'b', I can paste the content ANYWHERE in the 'c' correctly. All the other (not involved) cell values become empty. How to reproduce it programmatically???

Answer 1

[madhan ravi]

 ab = {a, b};
 [Val, which_one] = max([numel(a),numel(b)]);
 c = cell(Val, 2);
 c(:, 1) = a;
 c(1 : numel(b), 2) = b

 % or perhaps you want
 c = cell(5);
 c(1:numel(a), 1) = a;
 c(end - numel(b) + 1 : end, end) = b

Answer 2

[madhan ravi]

c = {a, b} %? if this doesn't do what you want , you need to illustrate with an expected result.

[Csaba]

No, it does not do!

c={a,b}

c =

  1×2 cell array

    {3×1 cell}    {2×1 cell}

It makes an 1x2 cell array, instead of e.g. 3x2 cell array. What I want is e.g.

    c =

  3×2 cell array

    {'11'}    {'21'      }
    {'12'}    {'22'      }
    {'13'}    {0×0 double}

or

c =

  3×2 cell array

    {'11'}    {0×0 double}
    {'12'}    {'21'      }
    {'13'}    {'22'      }
    

or, if I paste at - let's say (4,5) field of 'c'

c =

  5×5 cell array

    {'11'      }    {0×0 double}    {0×0 double}    {0×0 double}    {0×0 double}
    {'12'      }    {0×0 double}    {0×0 double}    {0×0 double}    {0×0 double}
    {'13'      }    {0×0 double}    {0×0 double}    {0×0 double}    {0×0 double}
    {0×0 double}    {0×0 double}    {0×0 double}    {0×0 double}    {'21'      }
    {0×0 double}    {0×0 double}    {0×0 double}    {0×0 double}    {'22'      }

[madhan ravi]

ab = {a, b};
[Val, which_one] = max([numel(a),numel(b)]);
c = cell(Val, 2);
c(:, 1) = a;
c(1 : numel(b), 2) = b

% or perhaps you want
c = cell(5);
c(1:numel(a), 1) = a;
c(end - numel(b) + 1 : end, end) = b

[Csaba]

Thanks, although it is a workaround (probably not possible to do it more elegantly) I would accept your answer if you would put it as an answer.