Insert rows in a matrix
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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
采纳的回答
Guillaume
2015-6-24
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{:})
更多回答（1 个）
Anthony Poulin
2015-6-24
编辑：Anthony Poulin
2015-6-24
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 个评论
Guillaume
2015-6-24
That's not going to work as you resize AA on every step of the loop, yet still use the original range of rows. On step 2, you're inserting B on row 20, which is the original row 19 of AA, on step 3, you're inserting B on row 30, which is the original row 28 of AA, etc. Each step, you're more and more off the target.
Rather than an arbitrary value, you'd use size(AA, 1) for the end index.
Ionut Anghel
2015-6-24
Anthony Poulin
2015-6-24
Or just replacing "i" by "i + i/10 -1".
David Verrelli
2018-3-27
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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