Switching values around in a matrix
显示 更早的评论
Hi,
Say I have a 5 by 2 matrix in the form:
A = [2 5; 9 7; 10 2; 3 2; 1 9]
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]
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
1 个评论
Stephen23
2025-8-2
采纳的回答
A = [2 5; 9 7; 10 2; 3 2; 1 9];
B([1:2:10,2:2:10],1)=A(:)
3 个评论
Good illustration of MATLAB indexing operations...more of the same
A = [2 5; 9 7; 10 2; 3 2; 1 9];
B=cell2mat(arrayfun(@(r)cat(1,A(r,:).'),1:height(A),'UniformOutput',false).')
isOdd=@(i)mod(i,2)==1;
B=nan(numel(A),1);
B(isOdd(1:numel(B)))=A(:,1);
B(~isOdd(1:numel(B)))=A(:,2)
B=accumarray([1:2:numel(A) 2:2:numel(A)].',A(:))
<lol>
Scott Banks
2025-8-2
dpb
2025-8-2
@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?
更多回答（1 个）
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)
% option B
B=reshape(A.',[],1) % how to implement the above wanted but unallowable syntax...
The key is to recognize need to transpose to get in needed/wanted column order first...
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