Index 2D matrix along line rotated about the matrix midpoint

2 次查看（过去 30 天）
Roberto Brenes
Roberto Brenes 2019-10-9
回答： Daniel M 2019-10-20
Hi,
I have an N x N matrix that I would like to sequentially index along a line that slices the middle of the matrix and is then rotated by 180/(2(N-1)) degrees until it completes half a full rotation. In the end, we want to end up with 2(N-1) distinct lines. Picture attached for a 5x5 case for clarity, since it's easier to describe by a picture than by words, I think. If N were even, then the "equivalent" center point of the matrix could be arbitrarily decided by ceil(N/2).
So for example, if the matrix is:
a = [1 2 3 4 5; 6 7 8 9 10; 11 12 13 14 15; 16 17 18 19 20]
Then the elements returned for each iteration would be the middle row, middle column, diagonals as well as:
[2 13 24]
[4 13 22]
[6 13 20]
[10 13 16]
The only way I can think of implementing this is to rotate the 2D matrix by the angle 180/(2(N-1)) and then always slice the middle column of the matrix but I'm not sure this is the most optimal way to do so for large matrices. Is there any other, more efficient way to achieve this?
  3 个评论
显示 1更早的评论
Roberto Brenes
Roberto Brenes 2019-10-9
编辑：Roberto Brenes 2019-10-9
Hopefully the new updated post makes sense. In the end, we would want 2(N-1) lines in total but making sure that every element in the matrix appears at least once. In that case, it might be best to think of the lines in between the diagonals and the mid row or column as a staircase or the linear interpolation between the start pixel and end pixel due to the discreteness of the matrix.
Daniel M
Daniel M 2019-10-9
Sorry I can't help because what you're asking for is still unclear. You have not been able to specify what you want in the general case. My post above still applies regarding if N = 7 (let alone if N = 99 or higher), or if N is even. What you want is unknown, so how can I help?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（2 个）

Daniel M
Daniel M 2019-10-9
编辑：Daniel M 2019-10-9
360/(N-1) doesn't make sense. What if N = 8? I think you just mean 180. But it's irrelevant. It would not be efficient to recreate the matrix by shifting the elements around.
It seems all you want to do is get the middle column, middle row, and the diagonals. Well that is simple enough for odd N (you can provide more clarity for even-numbered N if you wish).
A = magic(5);
sz = size(A,1);
mid = ceil(sz/2);
midcol = A(:,mid);
midrow = A(mid,:);
maindiagonal = diag(A);
oppositediagonal = A(sz:sz-1:end-1);
Other than that I'm not sure what order or arrangement you want the values it, but it should be pretty straightforward from here. Use commands like flip and transpose.
  3 个评论
显示 1更早的评论
Daniel M
Daniel M 2019-10-9
Those are an odd pairing of indices for sure. I hope when you rewrite your question, you pose it in a very clear way that generalizes it for a matrix of size N.
Roberto Brenes
Roberto Brenes 2019-10-9
Just edited the question above with a picture, hopefully that explains it. Also, disregard the first set of elements that I replied with the first time, I have no idea what I wrote and they didn't make any sense. I have edited them again to hopefully clear up the confusion.

请先登录，再进行评论。

Daniel M
Daniel M 2019-10-20
Here is another idea, perhaps you can play with it. You'll have to implement this in a loop.
[cx,cy,c] = improfile(I,xi,yi,n);
xi are the line endpoints x values. yi are the endpoints y values. n is the number of samples along the line that you want to extract. c are those image values, and cx,cy are the exact points where the measurements were made.
For example, the green line in your image:
x = magic(5);
[cx,cy,c] = improfile(x,[2,4],[1 5],3)

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

产品


版本

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by