Sorting and randomising a matrix according to a particular constraint.

Hi There,

I have a 32 x 3 matrix as follows:

A =

[ 1 1 1;

1 1 2;

1 1 3;

1 1 4;

1 1 5;

1 1 6;

1 1 7;

1 1 8;

1 2 1;

1 2 2;

1 2 3;

1 2 4;

1 2 5;

1 2 6;

1 2 7;

1 2 8;

2 1 1;

2 1 2;

2 1 3;

2 1 4;

2 1 5;

2 1 6;

2 1 7;

2 1 8;

2 2 1;

2 2 2;

2 2 3;

2 2 4;

2 2 5;

2 2 6;

2 2 7;

2 2 8]

What I need to do is randomise the order of the rows, while keeping the row values together, however, this needs to be constrained such that for A(:, 4), every 8 rows contain only the numbers 1 - 8. So for example, you could have something like:

A(1:8, :) =

[ 1 2 4;

1 1 5;

2 1 6;

1 1 8;

2 2 1;

2 1 2;

2 1 7;

1 1 3]

The occurrence of 1 and 2 in the first two columns needs to be random, 1 - 8 needs to be randomised for every 8 values of the third column. Initially I tried to use the function randswap within a loop with an added constraint, but this has only led to an infinite loop. Also important that the rows stay together as the 1s and 2s in the first two columns need to appear alongside the last column an equal number of times. Pretty stumped. Any help much appreciated.

Answer 1

[Lauren Shone]

Someone has kindly provided this very excellent solution that eradicates a lot of work and gives the perfect solution if anyone is interested:

   %# calculate index for the 1's and 2's
   r = rand(8,4);
   [~,idx12] = sort(r,2);

   %# calculate index for the 1's through 8's
   [~,idx8] = sort(r,1);

   %# use idx8 to shuffle idx12
   idx8into12 = bsxfun(@plus,idx8,[0 8 16 24]);

   %# now we can construct the output matrix
   B = [1 1;1 2;2 1;2 2];
   out = [B(idx12(idx8into12),:),idx8(:)];

Answer 2

[Andrei Bobrov]

variant

m = size(A,1);
n = 1:8;
out = [1 1 1];
i2 = 1;
while ~all(ismember(n,out(:,3))) && i2 < 100
    i1 = randperm(m);
    out = A(i1(n),:);
    i2 = i2 + 1;
end

more variant

eg

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

[i1,i1] = sort(A(:,3));
k = mat2cell(A(i1,:),histc(A(:,3),unique(A(:,3))),size(A,2));
out = cell2mat(arrayfun(...
                 @(i3)k{i3}(randi(size(k{i3},1)),:),randperm(numel(k))','un',0))

more variant for your case

A1 = sortrows(A,3);
out = A1(4*(randperm(8)-1)+randi(4,1,8),:)

last variant

d = rand(8,4);
[~,i1] = sort(d(:,1),1);
[~,i2] = sort(d,2);
A1 = sortrows(A,3);
out = A1(4*(i1-1)+i2(:,1),:)

[Lauren Shone]

Hi, Thanks for you answer. Not quite sure it works, for example, out =

1 1 5

2 1 7

1 1 8

1 2 2

1 1 4

2 1 6

2 1 2

1 2 3

where numbers are repeating in the last column.

[Andrei Bobrov]

corrected :)

[Lauren Shone]

awesome! Thanks, you answer is actually bit more useful and flexible (for my limited capabilities), many thanks :)

Answer 3

[Lauren Shone]

I thought maybe I could add a loop like this:

ng = true; while ng ng = false;

    X = randswap(trialmat, 1);
    for i = 1:32
        if X(i, 3) ~= O_All(i)
            ng = true;
        end
    end
end

where O_All is a 32x1 with repeating values of 1-8 every 8 rows in a random order, however this leads to an infinite loop.