How can i position the minimum value in the first cell for each column, without changing the sequence?

I have a 100 x 1000 array, with the minimum value at the different position for each column. How can I position the minimum value in the first cell for each column, without changing the sequence?

Answer 1

[Star Strider]

This should do what you want:

M = randi(9,6,4);                               % Create Matrix
for k1 = 1:size(M,2)
    [~,idx] = min(M(:,k1));                     % Index Of Minimum In Column ‘k1’
    Mr(:,k1) = circshift(M(:,k1), 1-idx, 1);    % Rotate To Put First Minimum In First Row
end

M =
     1     6     9     5
     6     4     3     3
     9     5     7     4
     3     9     3     2
     1     4     9     4
     4     7     3     4

Mr =
     1     4     3     2
     6     5     7     4
     9     9     3     4
     3     4     9     5
     1     7     3     3
     4     6     9     4

Here ‘M’ is the original matrix, ‘Mr’ is the ‘rotated’ matrix. The circshift function will do what you want.

Note that the min (and max) functions only return the index of the first value of the minimum in a vector, if there are duplicates.

[Utsav Vishal]

there is only one minimum in each column. but i am getting error. for k1 = 1:size(M,2) [~,idx] = min(M(:,k1)); % Index Of Minimum In Column ‘k1’ Mr(:,k1) = circshift(M(:,k1), 1-idx, 1); % Rotate To Put First Minimum In First Row end ---------- "Error using circshift Too many input arguments."

[Utsav Vishal]

The change of matrix M to Mr is what I want, but I get the error message. "Error using circshift Too many input arguments."

[Star Strider]

I used the R2017a version of the circshift function.

This will work in all versions:

Mr(:,k1) = circshift(M(:,k1), [1-idx 0]);   % Rotate To Put First Minimum In First Row

[Utsav Vishal]

yaa...Thank You, sir. I will check with my data. Best Regards.

[Star Strider]

My pleasure.

Answer 2

[KSSV]

doc min

A = rand(5,2) ;
[val,idx] = min(A,[],1) ;
B = [val ; A]

[Utsav Vishal]

A matrix(5,2), is changing matrix size. B(6,2). The sequence is changing. The command is just duplicating the minimum value to 1st position. The output matrix B should be also 5 X 2 matrix. eg: 6 9 8 4 5 6 3 4 12 8 As minimum value(3) is in the 7th position. such as: 3 4 12 8 6 9 8 4 5 6. Thank You.

[KSSV]

A = rand(5,2) ;
[nx,ny] = size(A) ;
[val,idx] = min(A,[],1) ;
p = sub2ind(size(A),idx,1:ny) ;
B = A ;
B(idx) = [] ;
B =reshape(B,nx-1,ny) ;
B = [val; B] ;

[Utsav Vishal]

The code is just taking the minimum value to the 1st position, and deleting it from its original position. also, the 1st element of second colomn becomes the last element of 1st column. I have to deal with each column separately. eg[5 4 3 1 2, 8 6 4 5 7] to [1 2 5 4 3, 4 5 7 8 6]. Thank you.

Answer 3

[Roger Stafford]

[~,I] = min(A,[],1);
for k = 1:size(A,2);
  A(:,k) = circshift(A(:,k),1-I(k),1);
end

[Roger Stafford]

@Utsav: You stated that the circular shift was to be along each column, but in your example you have shown a row vector being shifted. If it is the latter you actually want, the above code must be modified accordingly.

[Utsav Vishal]

 for k = 1:size(A,2);
  A(:,k) = circshift(A(:,k),1-I(k),1);
end
Error using circshift
Too many input arguments.

[Utsav Vishal]

No sir, circular shift may apply on each column. I may have got wrong in the example.

[Roger Stafford]

Given the error you received, try this instead:

   [~,I] = min(A,[],1);
   for k = 1:size(A,2);
     A(:,k) = circshift(A(:,k),1-I(k));
   end

[Utsav Vishal]

Thank You, sir. I will check with my data. Best Regards.