filling the gaps in the sequence of dates

Question

antonet 2012-8-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/45475-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

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Answer 1

Andrei Bobrov 2012-8-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/45475-filling-the-gaps-in-the-sequence-of-dates#answer_55630

编辑：Andrei Bobrov 2012-8-8

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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antonet 2012-8-8

编辑：antonet 2012-8-8

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 2012-8-8