Random rows in matrix so no element from the first column occur twice in a row
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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
Adam Danz on 22 Mar 2023
Edited: Adam Danz on 23 Mar 2023
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)
accepted = false;
nrows = height(M);
nPossibilities = factorial(nrows); % number of possible permutations
count = 0;
% 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;
accepted = true;
% Final check: throw warning if no permutations worked.
warning('No permutations result in non-consecutive values.')
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.
Adam Danz on 23 Mar 2023
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.
More Answers (1)
Cameron on 22 Mar 2023
A = randi(100,5,8);
[~,randrow] = sort(rand(size(A,1),1));
RandA = A(randrow,:)
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