Index 3D Array with 2D logical array

25 visualizaciones (últimos 30 días)
Florian Radack
Florian Radack el 25 de En. de 2020
Comentada: Walter Roberson el 26 de En. de 2020
I have an logical array B and an 3D array A. I wish to extract and manipulate values of A for specific k, whereby the values in the first two dimensions are given by B. I was hoping something like
A(B,[1 3 5])
would give me the values specified by B for the first, third and fifth slice of A, but MATLAB doesn't allow this kind of indexing.
In particular, I would like to switch values given by B for certain k (excuse the fauly syntax) akin to
A(B,[1 3 5]) = A(B,[2 4 6])
A minimum example of what I would like to achieve would be:
% coordinate mesh
[X,Y] = meshgrid(1:n,1:m);
% logical array B
B = (X-n/4).^2 + (Y-m/2).^2 < m;
% 3d array A
A = zeros(m,n,9);
A(:,:,1:4) = 1;
A(:,:,5:8) = 2;
A(:,:,9) = 3;
% switching of slices (in fauly syntax)
A(B,[1 2 3 ] = A(B,[5 6 7])
A(B,[5 6 7 ] = A(B,[1 2 3])
I need to do this switching of specific values in the 3. dimension for different slices of A and have not been able to come up with an efficient way of doing this. Any help would be greatly appreciated!

Respuestas (2)

Stephen23
Stephen23 el 25 de En. de 2020
Editada: Stephen23 el 25 de En. de 2020
P = size(A,3);
F = @(b,s)repmat(b,1,1,P)&reshape(ismember(1:P,s),1,1,P); % requires >=R2016b
A(F(B,[1,2,3])) = A(F(B,[5,6,7]));
A(F(B,[5,6,7])) = A(F(B,[1,2,3]));
For MATLAB versions earlier than R2016b replace the & operation with bsxfun.
  5 comentarios
Florian Radack
Florian Radack el 26 de En. de 2020
Thank you for your help so far. Unfortunately, running a basic test such as:
A = reshape(1:27,3,3,3);
B = false(3,3);
B(1,1) = true;
P = size(A,3);
F = @(b,s) bsxfun(@and, repmat(b,1,1,P), reshape(ismember(1:P,s),1,1,P));
A(F(B,[1 2 3])) = A(F(B,[3 2 1]));
yields no change in the array A.
Also, is there maybe an entirely different stratety that might be worth pursuing? The snippet of code will be placed in a (neccesary) loop to solve a time-dependent problem, and it seems calling the function F multiple times every loop might slow the code.
Walter Roberson
Walter Roberson el 26 de En. de 2020
Stephen's code, like mine, also assumed that the replacements have indices with increasing order. F(B,[1 2 3]) and F(B,[3 2 1]) are the same .
Every technique that uses logical matrices as indexing of the entire array is going to have the same problem of being insensitive to order of the panes. Logical indexing is always just "selected, or not selected" and so always works in linear indexing order, so trying to use logical indexing with [3 2 1] or other non-decreasing order is going to fail unless you take additional steps.

Iniciar sesión para comentar.


Walter Roberson
Walter Roberson el 25 de En. de 2020
You can do things like
mask1 = false(size(A));
mask2 = mask1;
B3 = repmat(B, [1 1 3]);
mask1(:,:,[1 2 3]) = B3;
mask2(:,:,[5 6 7]) = B3;
temp = A(mask1);
A(mask1) = A(mask2);
A(mask2) = temp;
This only works in the case where the pane indexes are increasing. It would not work if, for example, you were swapping A(B, [7 5 6])

Categorías

Más información sobre Matrix Indexing en Help Center y File Exchange.

Productos


Versión

R2016b

Community Treasure Hunt

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

Start Hunting!

Translated by