MATLAB: Filling the gaps in the sequence of dates

dates

EDITED
Dear all,
I have A={
'kl'  '10/08'      [4.4840]    [4.1101]    [     0]
'kl'   '01/09'    [4.4840]    [4.1101]    [     0]
'kl'    '02/09'    [4.1101]    [4.0311]    [     0]
'kl'    '03/09'    [4.0311]    [3.9358]    [     0]
'kl'    '04/09'    [3.9358]    [3.9739]    [     0]
'kl'    '05/09'    [3.9739]    [3.9267]    [     0]
 'kl'   '07/09'    [3.9059]    [3.8655]    [     0]
 'kl'   '08/09'    [3.8655]    [3.8889]    [3.7498]
 'kl'   '10/09'    [3.7498]    [3.8857]    [     0]
 'kl'   '11/09'    [3.8857]    [4.4207]    [4.1647]
 'kl'   '01/10'    [4.1647]    [3.7704]    [     0]
 'kl'   '02/10'    [3.7495]    [3.7085]    [     0]
 'kl'   '04/10'    [3.7085]    [3.6800]    [     0]
 'kl'   '05/10'    [3.6800]    [3.7364]    [3.7867]
'kl'    '07/10'    [3.7867]    [3.7860]    [     0]
'kl'    '08/10'    [3.7860]    [3.7888]    [3.6435]
 'kl'   '10/10'    [3.6435]    [3.6149]    [     0]
'kl'    '11/10'    [4.2260]    [3.8786]    [     0]
 'kl'   '01/11'    [3.8786]    [3.5946]    [3.5765]
'kl'    '02/11'    [3.5765]    [3.5946]    [     0]
 'kl'   '04/11'    [3.5946]    [3.6493]    [     0]
'kl'    '05/11'    [3.6493]    [3.5918]    [3.6956]
'kl'    '07/11'    [3.6956]    [3.7282]    [     0]
'kl'    '08/11'    [3.7326]    [3.6308]    [     0]
 'kl'   '10/11'    [3.6308]    [3.6523]    [4.1421]
'kl'    '11/11'    [4.1421]    [2.0710]    [     0]}
The second column is month/year. Is it possible to fill the gaps in the sequence of the dates and for this additional row to set the rest of the elements equal to NaN? Specifically, the first date changes and is not fixed. Also the last date must be the date of the last row.
That is;
A={
 'kl'   '10/08'      [4.4840]    [4.1101]    [     0
   [NaN]     '11/08'      [NaN]           [NaN]      [NaN]
    [NaN]     '12/08'      [NaN]           [NaN]      [NaN]
 'kl'  '01/09'    [4.4840]    [4.1101]    [     0]
 'kl'     '02/09'    [4.1101]    [4.0311]    [     0]
    'kl'     '03/09'    [4.0311]    [3.9358]    [     0]
  'kl'    '04/09'    [3.9358]    [3.9739]    [     0]
    'kl'   '05/09'    [3.9739]    [3.9267]    [     0]
     [NaN]   '06/09'      [NaN]           [NaN]      [NaN]
   'kl'   '07/09'    [3.9059]    [3.8655]    [     0]
  'kl'    '08/09'    [3.8655]    [3.8889]    [3.7498]
   [NaN]   '09/09'      [NaN]           [NaN]      [NaN]
 'kl'   '10/09'    [3.7498]    [3.8857]    [     0]
  'kl'    '11/09'    [3.8857]    [4.4207]    [4.1647]
   [NaN]   '12/09'      [NaN]           [NaN]      [NaN]
And so forth . the last date must be
    '11/11'
Just to mentionthat the last date may change and is not fixed as it happens with the first date. SO the "last date" can be any date and the code must not create any new dates after the "last date"
Thanks in advance

Best Answer

d0 = datenum(A(:,1),'mm/yy');
k = diff(year(d0([1,end]))) + 1;
d1 = datenum(2009,(1:k*12)',1);
out = num2cell(nan(numel(d1),size(A,2)));
out(:,1) = cellstr(datestr(d1,'mm/yy'));
out(ismember(d1,d0),2:end) = A(:,2:end);
EDIT
[y,m] = datevec(A([1,end],2),'mm/yy');
mths = diff(y)*12+diff(m);
N = cellstr(datestr(datenum(y(1),(m(1)+(0:mths))',1),'mm/yy'));
out = repmat({nan},numel(N),size(A,2));
out(ismember(N,A(:,2)),[1,3:end]) = A(:,[1,3:end]);
out(:,2) = N;
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