CODEWITHSUNDEEP
matlabmatrix
student beginner
student beginner

Reputation: 29

matlab the index of the next smallest element in a matrix

I could get the minimum value and its index like in here matlab how to get min value and its index in a matrix.

From a matrix A

A=[1 3 6 2 0 4
   6 8 9 5 1 3
   7 2 7 8 9 2]

To get the minimal value MinVal(where the row is given (r) and the column is in an interval ([c.. c+x]) and the index ind (number of column of it)). I have to do

[MinVal,I]=min(A(r,c:c+x))
 ind= c-1+I;

Example

[MinVal, ind]=min(A(2,3:3+2))
will give me

 % MinVal= 1
 % ind =5

Then I have 

B.state=[ 0
          0
          1
          0
          1]

So here I can't take ind=5 because B(5).state==1, I need to move to the next MinVal= 5 and ind = 4. Here, it is ok, I can stop but if B(4).state ==1, then I need to move the next smallest and so on

But then the problem is that I have another structure B where I am going to check if B(ind).state== 1 then I have to move to the next smallest element and get its index and so on until I find the first empty one.

If I try like this

        MinD = A(r,c:c+x);
        [MinVal,Ind]=min(MinD);
        ind= nbrT+Ind;

        MinD2 = sort(MinD(:));
        p=2;
        while (B(ind).state == 1)
            MinVal= MinD2(p);
            %need to get the new index 
            %something like this
            ind=find (A == MinVal) ;
            p=p+1;
        end

The problem is that I can get the next minimum value but the index I will get can be of more than one value if MinVal appears more than once so how can I get the one with state == 0 I don't want to use unique either because even if I have two different elements with the same minimum, they refer to two different places and I have to keep both (I can use the second one if the first one is full). Modified code to

MinD = A(r,c:c+x); 
[MinVal,Ind]=min(MinD); 
ind= nbrT+Ind; 
[MinD2, IndMinD2] = sort(MinD(:)); 
p=2; 
while (B(ind).state == 1) 
    MinVal=MinD2(p); 
    Ind=IndMind2(p); 
    p=p+1; 
end
Ind= c-1+Ind;

So how can I do it?

Upvotes: 0

Views: 202

Answers (1)

shamalaia
shamalaia

Reputation: 2347

I think this should work:

           MinD2 = MinD(:));

           for ii=1:numel(MinD2)

              [MinVal,Ind]=min(MinD2);

              %do you stuff with the index 
              %and at the end do this:
              MinD2(Ind)=Inf;

           end

Upvotes: 0

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