Random rows in matrix so no element from the first column occur twice in a row

Hi,

I have a matrix of x rows and 8 columns and need to random the rows so no element from the first column occur twice in a row.

It is a x times 8 double.

I know how to random the first coulumn but end up with a x times 1 array instad.

Thank you in advance

Answer 1

[Adam Danz]

There is no general solution since the values in column 1 may not sufficiently vary. For example, what if all of the values in column 1 were identical? Or what if the values in column 1 were [1;1;1;2]?

Assuming the data in column 1 sufficiently varies, you can use randperm to randomly permute the rows of the matrix. But that doesn't guarantee a lack of consecutive identical values. You could pemute the rows within a while-loop that continues to apply random permutations until the condition is satisfied or until all possibilities are exhausted, though this approach challenges the notion of randomness.

Here's a demo. If there are no permutations that satisfy the constraints, a warning is thrown (you can change that however you'd like).

rng default % for demo purposes
M = randi(3,8,3)

M = 8×3

3 3 2 3 3 3 1 1 3 3 3 3 2 3 2 1 2 1 1 3 3 2 1 3

accepted = false; 
nrows = height(M);
nPossibilities = factorial(nrows);  % number of possible permutations
count = 0; 
while ~accepted
    % If there are still consecutive identical values or 
    % if we ran out of possible permutations
    if any(diff(M(:,1))==0) && count <= nPossibilities
        pidx = randperm(nrows);
        M = M(pidx,:);
        count = count+1;
    else
        accepted = true;   
        % Final check: throw warning if no permutations worked.
        if any(diff(M(:,1))==0)
            warning('No permutations result in non-consecutive values.')
        end
    end
end

disp(M)

     3     3     3
     1     2     1
     3     3     3
     2     1     3
     1     3     3
     3     3     2
     2     3     2
     1     1     3

Replace the demo matrix with M = randi(2,8,3) to see the warning.

Note, if the matrix has many rows and the values in column 1 do not sufficiently vary, this could run a very long time. For example, a matrix with 10 rows has 3,628,800 possible permutations. If you expect to have this issue, you'll need a smarter solution that initially determines whether a permutation is possible.

[Måns]

Hi, Thank you for your reply. Yes the list can become fairly long, at the moment I was working on a 49 rows long matrix but it can be up towards ~800 rows long. The values will generally be in the range of 70-230.

I tried it out and it took about 10 seconds but it seems to work. Thank you very much!

[Adam Danz]

If the values sufficiently vary, a permutation that satisfies the constraints should be found quickly. For example, if all values are unique, the first permutation will be accepted.

Answer 2

[Cameron]

A = randi(100,5,8);
A

A = 5×8

53 42 74 79 12 36 34 40 67 49 33 93 83 60 86 45 48 93 26 6 31 41 66 64 18 12 64 87 4 30 59 36 76 96 71 46 28 97 40 42

[~,randrow] = sort(rand(size(A,1),1));
RandA = A(randrow,:)

RandA = 5×8

18 12 64 87 4 30 59 36 76 96 71 46 28 97 40 42 53 42 74 79 12 36 34 40 48 93 26 6 31 41 66 64 67 49 33 93 83 60 86 45