How to index this matrix ?

Hi all,

I have a problem while indexing a matrix with the result of the "min" function. My problem can be resumed as this :

B is a 5D matrix. (size = [ 2 2 3 3 240])

I have to know the position of the minimums of B accoarding to the 5th dimension.

Knowinw the position of this minimum, I want to transform M, another matrix with the same size (5D), into a 4D matrix keeping the values of the 5th dimension corresponding to the minimum values of B.

Now, I use this :

[U,Xm] = min(B,[],5)

Xm is a 4D matrix as expected.

The problems is that Xm values are the index in the dim.6, and not the indexing of the matrix. So I can not use : M(Xm).

Do you have an idea for that ?

Thank you a lot

Answer 1

[Image Analyst]

Maybe try

minValue = min(X(:));
linearIndexes = find(X(:) == minValue);

then use ind2sub(size(X), linearIndex) to get the sets of 5 indexes.

[Ameye Simon]

Like this ?

vals = min(B,[],5);
linearIndexes = find(B == vals);
[a1,a2,a3,a4,a5] = ind2sub(size(B), linearIndexes);

The problem is that I get this size for a1,a2... :

size(a1) =   234     1

So I can not get the right values inside M using something like :

M(a1,a2,a3,a4)

Because I want M to be the size :

size = [ 2     2     3     3   1]

Do you know how to have the result with the right size ?

[Image Analyst]

How many locations does the max appear in?

Attach B in a .mat file.

Answer 2

[Stephen23]

Several years ago I wrote an efficient function that returned the linear indices corresponding to any ND indices returned by min or max. But there was a resounding lack of interest from other users of FEX, so I deleted it again.

The straightforward way (but not the fastest) is to generate all of the subscript indices yourself, substitute the ones returned by min / max and then use those to generate an array of linear indices:

szb = size(B);
szx = size(Xm);
szx(end+1:numel(szb)) = 1;
tmp = arrayfun(@(n)1:n,szx,'uni',0);
[tmp{:}] = ndgrid(tmp{:});
tmp{szx~=szb} = Xm;
idx = sub2ind(szb,tmp{:});

And then all you need is:

M(idx)

Answer 3

[Abdallah]

I know it's too late but might help someone else

Use five for loops as follows:

B_min=min(B)

for i=1:2

for j=1:2

for u=1:3

for v=1:3

for w=1:240

if B(i,j,u,v,w)==B_min

ii=i;

jj=j;

uu=u;

vv=v;

ww=w;

end

end

end

end

end

end