Using 3d matrix inmatlab

2013 5 29
1 个回答
回答已采纳
3 次查看（30 天）

0 个投票

A(:,:,1) =
1 1 0
0 1 1
1 1 1
A(:,:,2) =
1 0 0
0 0 1
0 0 1
I used [r,c,v] = ind2sub(size(A),find(A == 1)); to find r,c,v in matrix A that has a value 1. But I want to place a condition that while A == 1 then I(r,c) = v
for example I(1,1) = {1,2} since for v=1,2 A takes a value 1. Please help me out with a code.

请先登录，再回答此问题。

 采纳的回答

Matt J
Matt J 2013-5-29
编辑：Matt J 2013-5-29

0 个投票

[m,n,p]=size(A);
s=reshape(1:p,1,1,p);
A=bsxfun(@times,A,s);

4 个评论
显示 2更早的评论 隐藏 2更早的评论

jana
jana 2013-5-29
I wanted to put them as sets. example I(1,1) = {1,2} that is (1,1) takes value 1 at v ={1,2}. My matrix should look more like : [{1,2} 1 1; 0 1 {1,2}; 1 1 {1,2}]
Matt J
Matt J 2013-5-29
编辑：Matt J 2013-5-29
[m,n,p]=size(A);
s=reshape(1:p,1,1,p);
l=num2cell(bsxfun(@times,A,s),3);
l=cellfun(@(c) nonzeros(c).',l,'uni',0);
l(cellfun('isempty',l))={0}
jana
jana 2013-5-29
l=cellfun(@(c) nonzeros(c).',l,'uni',0); it shows an error message : Input # 2 expected to be a cell array, was double instead.
jana
jana 2013-5-29
Navermind I fixed it. Thankyou!

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 Operating on Diagonal Matrices 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