Insert rows in a matrix

2 vues (au cours des 30 derniers jours)
Ionut  Anghel
Ionut Anghel le 24 Juin 2015
Commenté : David Verrelli le 27 Mar 2018
Hi, I have the following problem: matrix
AA=[NaN 1 2 3 4 5
NaN 10 20 30 40 50
...............
NaN 1E8 2E8 3E8 4E8 5E8 ];
There is possible to insert every 10 row the following row:
[NaN NaN 99 Nan 77 NaN];
without a loop?
Thank you

Réponse acceptée

Guillaume
Guillaume le 24 Juin 2015
One way of doing this:
%work out how to split AA
rowdist = ones(1, ceil(size(AA, 1) / 10)) * 10;
rowdist(end) = 10 + mod(size(AA, 1), -10);
%do the splitting and transpose into a row
splitAA = mat2cell(AA, rowdist, size(AA, 2))';
%add another row with the data to insert
splitAA(2, :) = {[NaN NaN 99 NaN 77 NaN]};
%avoid insertion of the new row at the end if the height
%of the matrix is not a multiple of ten
if mod(size(AA, 1), 10)
splitAA{2, end} = [];
end
%join it all together
newAA = vertcat(splitAA{:})
  1 commentaire
Ionut  Anghel
Ionut Anghel le 24 Juin 2015
Very nice method! Thank you.

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Plus de réponses (1)

Anthony Poulin
Anthony Poulin le 24 Juin 2015
Modifié(e) : Anthony Poulin le 24 Juin 2015
You can try something like this (with B = [NaN NaN 99 Nan 77 NaN]):
for i=10:10:100
AA = [AA(1:i, 1:end); B; AA(i+1:end,1:end)];
end
(100 is an arbitrary value)
  4 commentaires
Anthony Poulin
Anthony Poulin le 24 Juin 2015
Or just replacing "i" by "i + i/10 -1".
David Verrelli
David Verrelli le 27 Mar 2018
Or run the loop 'backwards', as in for i=100:-10:10 (although 100 might not be the correct terminal value for your case, as already noted). Even though the matrix is still resized at each step, at least this way the subsequent changes aren't affected by previous steps.
BTW, "1:end" can just be replaced with ":" alone.

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