Switching values around in a matrix

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Scott Banks
Scott Banks 2025 年 8 月 1 日
コメント済み: dpb 2025 年 8 月 2 日
Hi,
Say I have a 5 by 2 matrix in the form:
A = [2 5; 9 7; 10 2; 3 2; 1 9]
A = 5×2
2 5 9 7 10 2 3 2 1 9
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And I want to make a 10 by 1 matrix from these values so that I get:
B = [2;5;9;7;10;2;3;2;1;9]
B = 10×1
2 5 9 7 10 2 3 2 1 9
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How would I do this?
I know there is a probably a simple fix, but I haven't been able to do it.
Many thanks,
Scott

採用された回答

Matt J
Matt J 2025 年 8 月 1 日
A = [2 5; 9 7; 10 2; 3 2; 1 9];
B([1:2:10,2:2:10],1)=A(:)
B = 10×1
2 5 9 7 10 2 3 2 1 9
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  3 件のコメント
Scott Banks
Scott Banks 2025 年 8 月 2 日
Thanks, guys
dpb
dpb 2025 年 8 月 2 日
@Scott Banks, you're free to choose the solution you wish, but I'm curious why you chose the direct indexing solution over the generic one?
As I noted, it does illustrate using vectors as indices in MATLAB which is a powerful tool/feature and is sometimes invaluable, but the other code is general for any size array rather than only working for the specific case and more difficult to code generically (as my other tongue-in-cheek examples illustrate).
Again, just wondering what was your thinking here? Did my response need more amplification beyond the comments?

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その他の回答 (1 件)

dpb
dpb 2025 年 8 月 1 日
編集済み: dpb 2025 年 8 月 1 日
A = [2 5; 9 7; 10 2; 3 2; 1 9];
% option A
B=A.'; B=B(:) % B=A.'(:); is invalid MATLAB syntax, unfortunately. (Octave allows this in at least some contexts)
B = 10×1
2 5 9 7 10 2 3 2 1 9
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% option B
B=reshape(A.',[],1) % how to implement the above wanted but unallowable syntax...
B = 10×1
2 5 9 7 10 2 3 2 1 9
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The key is to recognize need to transpose to get in needed/wanted column order first...

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