Search multiple matrices for values at common index positions

3 次查看（过去 30 天）
AndyT
AndyT 2015-6-23
回答： Martin Brown 2015-6-24
I am trying to analyse several tens of 90x90 matrices and I want to find a summary matrix that only shows a logical 1 if an element has any non zero value that is common to all the matrices at the same index position. In the example below the desired result is to show that there is a value in common between all matrices in position [2,2] and [4,4]. I do not wish to show the values themselves or compare their values. I've tried working with ismember , find , intersect , a==b for just two matrices but nothing seems to do the trick. I can't seem to find a similar query in the archives either. Does anyone know of a suitable function or do I need to write an element by element search routine across all the matrices . All matrices are double and all identically sized .
a =
1 0 0 0 0
0 3 0 0 0
0 0 5 0 0
0 0 0 7 8
>> d = [ 0 0 0 12 0; 0 3 0 4 0; 0 0 0 0 0 ; 34 0 0 9 0]
d =
0 0 0 12 0
0 3 0 4 0
0 0 0 0 0
34 0 0 9 0
>> e = [ 0 0 0 0 0; 0 15 0 0 0; 0 0 0 2 0 ; 0 0 0 5 0]
e =
0 0 0 0 0
0 15 0 0 0
0 0 0 2 0
0 0 0 5 0
>> Result=[ 0 0 0 0 0; 0 1 0 0 0; 0 0 0 0 0 ; 0 0 0 1 0]
Result =
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 1 0
  6 个评论
显示 4更早的评论
AndyT
AndyT 2015-6-24
Thanks for the good advice Stephen, I am now coming to understand the power of matlab when you properly use dimensional properties of arrays and you have stopped me learning bad practices early on.
Stephen23
Stephen23 2015-6-24
编辑：Stephen23 2015-6-24
Learn to use the dimensions of arrays and MATLAB will fly...
MATLAB is a high-level language, and it works best when data is kept together as much as possible. Avoid nesting, multiple variables and loops (unless you really have to). Then you can use vectorized code to really get the best out of MATLAB: neat and fast operations.
You should read some other tips for writing efficient MATLAB code.
Good luck and have fun!

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

James Tursa
James Tursa 2015-6-23
编辑：James Tursa 2015-6-23
If you are just looking for when there are non-zero entries in the same index in all matrices, e.g.,
Result = a & d & e;
To process "several tens" of these matrices, you may want to write a function. How the function is written will depend on how you have your matrices stored (individual variables, members of a cell array or struct, etc). E.g., if they are individual variables, then a function like this might be a good start:
function g = allnz(varargin)
n = nargin;
if( n )
g = logical(varargin{1});
for k=2:n
g = g & varargin{k};
end
else
g = [];
end
Then you could call it like this:
Result = allnz(a,d,e);
  2 个评论
Guillaume
Guillaume 2015-6-24
In my opinion, a better implementation (no loop) of your allnz would be:
function g = allnz(varargin)
dimplus1 = ndims(varargin{1}) + 1;
g = all(cat(dimplus1, varargin{:}), dimplus1);
end
AndyT
AndyT 2015-6-24
Thanks James and Guillaume , I think this gets to the heart of my problem and combines the good advice to concatenate all the matrices into one larger array. Thanks to all for taking the time to assist me .

请先登录，再进行评论。

更多回答（4 个）

Andrei Bobrov
Andrei Bobrov 2015-6-24
a.*d.*e > 0;
  1 个评论
AndyT
AndyT 2015-6-24
Thanks Andrei , very elegant though I take the points above that I may need to reformat all my matrices into one larger stacked array for the work I need to do,

请先登录，再进行评论。

Guillaume
Guillaume 2015-6-23
编辑：Guillaume 2015-6-24
As given in the example:
result = all(cat(3, a, d, e), 3)
Note that if you tens of these matrices, rather than putting them in individual variables, you're better off putting them in a cell array:
allmatrices = {a, b, c, d, e, f, ...};
In which case the above is simply:
result = all(cat(3, allmatrices{:}), 3);
Or even better since they all have the same size, just stack them all up in the third dimension as I have done as a first step, so:
allmatrices = cat(3, a, b, c, d, e, f, ...);
In which case the code is simply:
result = all(allmatrices, 3);
  1 个评论
AndyT
AndyT 2015-6-24
Thanks Guillaume, an excellent way of using the dim=3 parameter to stack the all and cat functions. I'll remember this.

请先登录，再进行评论。

Azzi Abdelmalek
Azzi Abdelmalek 2015-6-23
编辑：Azzi Abdelmalek 2015-6-23
a=[ 1 0 0 0 0 ; 0 3 0 0 0 ; 0 0 5 0 0 ; 0 0 0 7 8]
d=[ 0 0 0 12 0; 0 3 0 4 0; 0 0 0 0 0 ; 34 0 0 9 0]
e=[ 0 0 0 0 0; 0 15 0 0 0; 0 0 0 2 0 ; 0 0 0 5 0]
aa=~~a
dd=~~d
ee=~~e
aa&dd&ee
Martin Brown
Martin Brown 2015-6-24
As above, reformat your code to put the matrices into a single 3D matrix, something like:
A(:,:,1) = [1 0; 0 1];
A(:,:,2) = [2 0; 0 0];
find(sum(A(:,:,1:end)~=0,3)==2);

类别

在 Help Center 和 File Exchange 中查找有关 Logical 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