How to arrange one row vector into a row x column matrix?

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Miroslav Josic
Miroslav Josic 2017 年 2 月 7 日
コメント済み: Miroslav Josic 2017 年 2 月 7 日
Hi guys, I would appreciate your help on this problem. I have a row vector (M) of size 1x2108 which contains data for 62 countries and 34 industries for each country. So, column 1:34 is industry data for country 1, 35:68 country #2 etc. I just need to make a matrix which will be of a size 62x2108 and have data in following way: - row #1, columns 1:34 for country #1, elsewhere zeros - row #2, columns 35:68 for country #2, elsewhere zeros - row #3, columns 39:102 for country #3, elsewhere zeros etc.
So far, I used this code:
M_new=zeros(62,2108);
for i=1:62
for j=1:62
M_new(i,((j-1)*34+1):(j*34))=M(1,((j-1)*34+1):(j*34));
end
end
but I only get row vector copied 62 times in each row.
Best,
Miroslav

採用された回答

Guillaume
Guillaume 2017 年 2 月 7 日
M_new = [reshape(M, 34, 62); zeros(2108, 62)];
M_new = reshape(M_new(1:end-2108), 2108, [])'
No need for a loop.
However, I question the wisdom of using 62 times more memory to store the same information.
  1 件のコメント
Miroslav Josic
Miroslav Josic 2017 年 2 月 7 日
Hi Guillaume,
Thanks a lot for your quick reply, indeed it works perfectly. I was using loop for some other re-arrangements and it did not take much time so I tried to use it here again.
Best,
Miroslav

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その他の回答 (1 件)

KSSV
KSSV 2017 年 2 月 7 日
You should have a look into reshape. Read about reshape
  1 件のコメント
Miroslav Josic
Miroslav Josic 2017 年 2 月 7 日
Hi KSSV,
Thanks for your hint. Guillaume proposed the solution and it managed to give me the exact output I needed.
In any case, thanks for your help!
Miroslav

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