How can I arrange or matrix columns, depending on the total sum that has the columns?

Hi, I need to arrange a matrix A to look like matrix B.

Matrix A generate new positions at every iteration, so the positions are not fixed.

If there is a column in A (in this case, column 3) that has sum=0, and after it has column that sum=1 (column 4), I want column 4 to move to column 3. Then column 5 move to column 4 and the last column is the one that keeps the sum=0 column.

This is a sequencing problem, so my desire is to have a smooth sequence.

What is the best way to do this?

Thanks!

if true
  A = [0  1  0  0  0;
       1  0  0  0  1;
       0  0  0  1  0;
       0  0  0  0  0;
       0  0  0  0  0];
B = [0  1  0  0  0;
     1  0  0  1  0;
     0  0  1  0  0;
     0  0  0  0  0;
     0  0  0  0  0];
end

Answer 1

[Roger Stafford]

Do

 s = sum(A,1);
 p = find(s(1:end-1)==0&s(2:end)==1);
 B = A(:,[setdiff(1:size(A,2),p),p]);

(Corrected)

[Patty]

Thanks!! I tried, but it seems that only works for this example matrix.

I already wrote a small code that works fine. I share here for any body that might need someday.

Seq is a 3-D matrix. micro_period= number of columns

if true
newSeq=Seq;

for kk=1:Line

    idx_zero=[];
    idx_one=[];
    for ii=1:micro_period
        col_sum=sum(Seq(:,ii,kk));
        if col_sum==0
            idx_zero=[idx_zero;ii];
        else
            idx_one=[idx_one;ii];
        end

    end

    for jj=1:length(idx_one)
        newSeq(:,jj,kk)=Seq(:,idx_one(jj),kk);
    end

    for zz=1:length(idx_zero)
        newSeq(:,length(idx_one)+zz,kk)=Seq(:,idx_zero(zz),kk);
    end
end
Seq=newSeq;
end

Maybe its a little bit rustic haha, but it's working! :)

[Roger Stafford]

Patty, in your description you stated "If there is a column in A (in this case, column 3) that has sum=0, and after it has column that sum=1 (column 4), I want ...". However, the code you wrote never checks for a succeeding sum of one. To accomplish what your code does, you can just remove the test for a sum of 1 in the code I gave you:

   s = sum(A,1);
   p = find(s(1:end-1)==0);
   B = A(:,[setdiff(1:size(A,2),p),p]);

It will always agree with the result you obtained.

Answer 2

[Image Analyst]

Patty: Here's an alternate way using sort() that I think does what you want:

% Generate random A.
A = randi(9, [20,6])
% Sum the columns.
sumA = sum(A, 1)
% Sort A in ascending order.
% The sortOrder is what we need here.
[sortedSums, sortOrder] = sort(sumA, 'Ascend')
% Make B sorted in order of increasing column sums of A.
B = A(:, sortOrder)
% Get sums of B to verify it worked
sumB = sum(B, 1)

A =
     2     7     1     9     6     7
     8     4     2     2     4     7
     3     5     2     3     3     7
     6     6     2     4     5     1
     9     1     3     1     4     7
     4     3     3     7     4     5
     7     7     2     4     6     2
     7     7     3     9     7     1
     4     2     9     4     4     8
     6     2     7     6     4     2
     1     1     6     2     2     2
     9     1     2     4     1     6
     2     4     2     2     3     9
     3     6     1     7     3     5
     8     7     9     8     6     7
     5     5     7     4     9     2
     7     1     6     7     9     9
     4     6     3     3     5     5
     3     2     2     5     3     7
     1     2     6     8     7     1
sumA =
    99    79    78    99    95   100
sortedSums =
    78    79    95    99    99   100
sortOrder =
     3     2     5     1     4     6
B =
     1     7     6     2     9     7
     2     4     4     8     2     7
     2     5     3     3     3     7
     2     6     5     6     4     1
     3     1     4     9     1     7
     3     3     4     4     7     5
     2     7     6     7     4     2
     3     7     7     7     9     1
     9     2     4     4     4     8
     7     2     4     6     6     2
     6     1     2     1     2     2
     2     1     1     9     4     6
     2     4     3     2     2     9
     1     6     3     3     7     5
     9     7     6     8     8     7
     7     5     9     5     4     2
     6     1     9     7     7     9
     3     6     5     4     3     5
     2     2     3     3     5     7
     6     2     7     1     8     1
sumB =
    78    79    95    99    99   100