How to sort a matrix in matlab

Assume matrix A as follows:

A = [...
1  50  0  10
2  0  0  0
3  0  30  0
4  0  15  0
5  45  0  65
6  55  0  0
7  0  10  0
8  0  0  90
9  0  0  5
10  10  0  0
];

I want to sort the matrix A column 2:4 and produce matrix B. In matrix B pair of successive columns are represent to sorted array of first and corresponded column in matrix A.

B = [...
10  10  7  10  1  10
5  45  4  15  5  65
1  50  3  30  8  90
6  55  1  0  2  0
2  0  2  0  3  0
3  0  5  0  4  0
4  0  6  0  6  0
7  0  8  0  7  0
8  0  9  0  9  0
9  0  10  0  10  0
];

Answer 1

[John Chilleri]

Hello,

The sort function can do this for you,

% Given your A - first column:
A =    [50  0  10
  0  0  0
  0  30  0
  0  15  0
  45  0  65
  55  0  0
  0  10  0
  0  0  90
  0  0  5
  10  0  0];

% Set 0s to infinity to place them after in sort (will switch back to 0s):
A(A==0)=Inf;
[C,I] = sort(A);
B = [I(:,1) C(:,1) I(:,2) C(:,2) I(:,3) C(:,3)];
B(B==Inf)=0;

which results with,

>> B

B =

    10    10     7    10     9     5
     5    45     4    15     1    10
     1    50     3    30     5    65
     6    55     1     0     8    90
     2     0     2     0     2     0
     3     0     5     0     3     0
     4     0     6     0     4     0
     7     0     8     0     6     0
     8     0     9     0     7     0
     9     0    10     0    10     0

Hope this helps!

[John Chilleri]

To generalize the code to any sized matrix:

A(A==0)=Inf;
[C,I] = sort(A);
B = [];
for i = 1:size(A,2)
   B = [B I(:,i) C(:,i)];
end
B(B==Inf)=0;

[Alex Rob]

Great job John!

Can I have now matrix B with the following format:

B = [1  10  10
1  5  45
1  1  50
1  6  55
2  7  10
2  4  15
2  3  30
3  9  5
3  1  10
3  5  65
3  8  90
];

% in the new format all rows include 0 are removed and first column got similar index 1:3 (because we had 3 columns in matrix A)

[John Chilleri]

Yes,

If you use:

count = 1;
for i = 2:2:size(B,2)
   for j = 1:size(B,1)
      if (B(j,i) ~= 0)
         C(count,1:3) = [i/2 B(j,i-1) B(j,i)];
         count = count + 1;
      end
   end
end

It will produce the desired C,

>> C

C =

     1    10    10
     1     5    45
     1     1    50
     1     6    55
     2     7    10
     2     4    15
     2     3    30
     3     9     5
     3     1    10
     3     5    65
     3     8    90

Hope this helps!

Answer 2

[Andrei Bobrov]

A = [1  50  0  10
2  0  0  0
3  0  30  0
4  0  15  0
5  45  0  65
6  55  0  0
7  0  10  0
8  0  0  90
9  0  0  5
10  10  0  0
];

[m,n] = size(A);
B = zeros([m,2*(n-1)]); 
B(:,2:2:end) = A(:,2:end);
B(B == 0) = nan;
[B(:,2:2:end),ii] = sort(B(:,2:2:end));
B(:,1:2:end) = ii;
B(isnan(B)) = 0;

Bout = permute(reshape(B,m,2,[]),[1,3,2]);
t = Bout(:,:,2) ~= 0;
[~,jj] = find(t);
out = [jj,reshape(Bout(repmat(t,1,1,2)),[],2)];