filling the gaps in the sequence of 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

Answer 1

[Andrei Bobrov]

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;

[Azzi Abdelmalek]

the code of andrei works fine; i suggest that you clear your variables, and then run his code

[Azzi Abdelmalek]

it's still andrei's code

[Azzi Abdelmalek]

you just modified his code, like i did with mine after your new observation.

[antonet]

ok Azzi. I left only the relevant parts. I hope this helps and sorry for causing any inconvenience. I am just struggling to find the solution and I can't. If you want please help me

thank you

[antonet]

Thanks Andrei!