Elementwise shift of two matrices

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I have two 1xn matrices each of which with different index for their minima.
For instance the minimum of matrix A is the 3rd index:
A = [9 8 7 8 9];
[i idx_A] = min(A); idx_A
idx_A = 3
whereas the minimum of matrix B is the 4th index:
B = [8 8 7 6 9];
[i idx_B] = min(B); idx_B
idx_B = 4
How can I shift these two such that they both have their minima on the same index?
One way that comes to mind is cutting matrices from left and right. At the end of the day, I wan to get a matrix that looks like below:
A = [9 8 7 8];
[i idx_A] = min(A); idx_A
idx_A = 3
B = [8 7 6 9];
[i idx_B] = min(B); idx_B
idx_B = 3
How can this be done in a way that work for all matrices with the same size?

Accepted Answer

Bruno Luong
Bruno Luong on 15 Sep 2022
Edited: Bruno Luong on 15 Sep 2022
truncated matrices
[~, idx_A] = min(A);
[~, idx_B] = min(B);
di=idx_A-idx_B;
Ac=A(max(1,1+di):min(end,end+di))
Bc=B(max(1-di,1):min(end-di,end))

More Answers (1)

Walter Roberson
Walter Roberson on 15 Sep 2022
B = [8 8 7 6 9];
[i idx_B] = min(B);
circshift(B, 1-idx_B)
ans = 1×5
6 9 8 8 7
  1 Comment
Walter Roberson
Walter Roberson on 15 Sep 2022
You could examine the relative values of the indices and the length of the vectors to figure out which one would involve "less" work to move.. but it would probably involve more coding than the situation is worth.
Oh... I just had a thought: you might have a situation in which the minima for one of the vectors is beyond the length of the other vector. So if you have a hard rule about which one to shift, you could end up requiring that one of them be padded.
A = [9 8 7 8];
[~, idx_A] = min(A);
B = [8 8 7 6 9];
[~, idx_B] = min(B);
B = circshift(B, idx_A - idx_B);
A, B
A = 1×4
9 8 7 8
B = 1×5
8 7 6 9 8

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