find the location of minimum value

i have matrix with 4 variables, A=mintemp(a,b,c,d). i have find the minimum value especially at V=mintemp(:,:,1,1) dan the minimum value is out=min(V(:)). now, how to find the location / the coordinates the minimum value ?

if the matrix just two variables, i can the identify the coordinates with this :

A=magic(4);
out=min(A(:));
aa=1;
bb=1;
for xx=1:4
  for yy=1:4
      if(out==1)
          aa=xx;
          bb=yy;
     end
  end
end

i really have no idea if the matrix have 4 variables.

[Azzi Abdelmalek]

What is mintemp? and what do you mean by: the matrix have 4 variables.

Answer 1

[Azzi Abdelmalek]

Edit

A=rand(4,4,4,4) 
[ii1,ii2,ii3,ii4]=size(A)
count=0
id=zeros(ii3*ii4,4);
for k=1:ii3
for p=1:ii4
   count=count+1;
   v=A(:,:,k,p);
   [val,idx]=min(v(:));
   [id1,id2]=ind2sub(size(v),idx);
   minval(count)=val;
   idx1(count)=id1;
   idx2(count)=id2;
   idx3(count)=k;
   idx4(count)=p;
   id(count,:)=[id1 id2 k p]  % correspondant indices
end
end

[Internazionale]

if A=rand(4,4,4,4) and run the program the variable x is part like (:,:,1,1),(:,:,1,2) , ... until (:,:,4,4). After i find the minimum value every part, how to find the coordinates of minimum value every part ?

[Azzi Abdelmalek]

A=rand(4,4,4,4) 
[ii1,ii2,ii3,ii4]=size(A)
count=0
for k=1:ii3
  for p=1:ii4
     count=count+1;
     v=A(:,:,k,p);
     [val,idx]=min(v(:));
     [id1,id2]=ind2sub(size(v),idx);
     minval(count)=val;
     idx1(count)=id1;
     idx2(count)=id2;
     idx3(count)=k;
     idx4(count)=p;
  end
end

[Internazionale]

can we make every output the minimum value, the coordinates will go serially . example if minval=0,0087 so the output serailly (1,2,1,4).

[Internazionale]

i mean serially, sorry for the miswords

[Thorsten]

I changed my answer, minserial gives the desired output.

[Azzi Abdelmalek]

Internazionale. Have you tried the above code?

[Internazionale]

i have tried your last program. it works, but same with my last question, example, if we get minimum value at (:,:,1,2), the coordinates will get automatically with pattern (1,3,1,2). can we do that ?

[Azzi Abdelmalek]

That's what the code do! (previous comment) idx1,idx2,idx3,idx4 in the code correspond to 1,3,1,2

[Internazionale]

but, if you want to see the coordinates, we must open the workspace (idx1,idx2,idx3,idx4 separately). can we merge it all direct if we want to see the coordinates of minimum value ?

[Azzi Abdelmalek]

Look at the edited answer. It's the variable id

[Internazionale]

on the variable id [id1 id2 k p], next problem is, i want to make variable ei and fi.

ei = k-(id1*0.5) and fi = p-(id2*0.5). so the output is 64 value of ei and 64 value of fi.

Answer 2

[Thorsten]

That's easy

X = rand(4, 4, 4, 4);
[val ind] = min(X(:));

[Internazionale]

i mean the coordinates of the minimum value. so, after we find the minimum value, we must search the coordinates. i think the answer is, example mincoordinates=(1,2,1,4).

[Thorsten]

 [i j k l] = ind2sub(size(X), ind);

Answer 3

[Thorsten]

 R = rand(4,4,4,4);
 for i = 1:size(R, 3)
  for j = 1:size(R, 4)

    [min_val(i, j) min_ind(i, j)] = min(flatten(R(:,:, i, j)));
    [a b] = ind2sub(size(R), min_ind(i,j));
    minserial(end+1, :) = [i j a b];
  end
 end

With the convenience function flatten defined as

 function y = flatten(x)
 y = x(:);

[Internazionale]

matlab undefined variable minserial