0 个投票

For 2D arrays, the operation that I want to execute is:
a = rand(3,3);
for i = 1:numel(a)
for j = 1:numel(a)
out(i,j) = (a(i) - a(j)) / (a(i) + a(j));
end
end
Which will turn the out into a skew-symmetric matrix. Now I want to perform the same operation instead the matrix is in 3D, how can I do that? The speed is important since my original matrix is (100,100,726). Thanks!

请先登录，再回答此问题。

 采纳的回答

Dyuman Joshi
Dyuman Joshi 2022-5-31
编辑：Dyuman Joshi 2022-5-31
Ran in:

2 个投票

Ran in:
a = rand(3,3,4);
for k=1:size(a,3)
y=a(:,:,k);
for i=1:size(y,1)
for j=1:size(y,2)
out(i,j,k)=(y(i) - y(j))/(y(i) + y(j));
end
end
end
out
out =
out(:,:,1) = 0 0.6498 -0.1418 -0.6498 0 -0.7248 0.1418 0.7248 0 out(:,:,2) = 0 -0.6166 -0.5396 0.6166 0 0.1154 0.5396 -0.1154 0 out(:,:,3) = 0 0.1018 0.1400 -0.1018 0 0.0387 -0.1400 -0.0387 0 out(:,:,4) = 0 0.2123 0.1857 -0.2123 0 -0.0277 -0.1857 0.0277 0

2 个评论
显示 无 隐藏 无

Dyuman Joshi
Dyuman Joshi 2022-5-31
It will be skew-symmetric matrix. I have made an edit, take a look at it again.
MarshallSc
MarshallSc 2022-5-31
Thanks, yeah, changing the i and j indices from 1:size(y,1) to numel(y) would do the trick. Thanks again!

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by