Indexing arrays with matrices

2 次查看（过去 30 天）
Andrew Newell
Andrew Newell 2011-3-31
Matt's recent Hump-day challenger has led me ponder a strange form of indexing used in repmat. Consider a 3d array like the following:
A = reshape(1:8,[2 2 2])
A(:,:,1) =
1 3
2 4
A(:,:,2) =
5 7
6 8
Now run repmat and stop at the second last line:
B = A(subs{:});
Putting in the values for subs, this is the same as
B = A([1 1; 2 2],[1; 2],[1 1; 2 2]);
How is MATLAB interpreting indexing like this?

请先登录，再回答此问题。

采纳的回答

Matt Fig
Matt Fig 2011-3-31
Now that is the kind of wondering that lead me to the project which lead me to the challenger!
I think we can see what is going on by taking it one at a time:
A([1 1; 2 2],:,:)
So this produces the same thing as this:
A([1 2 1 2],:,:) % Read the indexing array as (:).
Now it makes sense: Take - Row one, then Row two, then Row one, then Row two for all higher dimensions (i.e., for each column and page).
Similarly with the other indexing arrays.
REPMAT and BSXFUN are deep funtions in terms of really understanding how they (and consequently, MATLAB indexing in general) work.
  1 个评论
Andrew Newell
Andrew Newell 2011-3-31
I see the light! Just to add to this explanation, if you do the following series of commands
B = A([1 2 1 2],:,:);
C = B(:,[1; 2],:);
D = C(:,:,[1 2 1 2]);
you get the same as
E = A(subs{:});
isequal(D,E)
ans =
1
Thank you!

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by