Randomise vector numbers to maximum variance

I have a vector that contains numbers 1, 0 and 10. In partcular, I have 100 times the 1, 400 times the 0 and 20 times the 10.

Is there any function or any way to shuffle the numbers to the maximum variance regarding their repetition? Namely, I want to shuffle them in a way to minimise their sequential repeatibility, so a to achive the low number of times that I get 1,1 or 0,0 or 10,10 in my vector?

Thank you so much in advance!

[Torsten]

0,1,0,1,...,0,1,0,1,(100 times)

0,10,0,10,...,0,10, (20 times)

0,0,0,0,....0 (280 times)

Or is it important how often the 0 is repeated ?

You should define an exact expression of what you want to minimize or maximize.

[abaza]

Thank you fo the comment!

I mean that I have this vector: [1,1,1,1...1,0,0,0,0,...0,10,10,10....10]

And I want to shuffle the numbers so as to minimize the repetition of 1 and 0 and 10.

It is obvious that I would get repetiton of 0 since the 1 and 10 are less than 1.

[Torsten]

Yes, it's obvious that 1 and 10 don't need to repeat and that 0 needs to repeat 279 times. That's optimal and the solution I already posted. Or do you think there is a better one ?

With maximum variance you mean

sum(abs(x(i+1)-x(i)),i=1,...,519)

?

[abaza]

I would like to minimise the repetition.... I cannot avoid repetition of some ZEROS since they are more than 1 and 10, but I would like a more automated way (e.g. when you have more numbers...)

Answer 1

[Bruno Luong]

A = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);

[abaza]

Thank you Bruno for your reply!

Could you please make it more general?? It works for this case, but if I had different cases of numbers and sizes?

[Bruno Luong]

Essentially you just need put each chunk of same value regularly spaced, so only largest chunk repeat.

I let you make the code.

[abaza]

I see, this is totally empirical...

[Bruno Luong]

Not more empirical then what you call "maximum variance", since the true variance estimated from such sequence doesn't change at all with order.

A = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);
for k=1:10
    Ak = A(randperm(length(A)));
    var(Ak)
end

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

ans = 3.7128

You must have different definition of variance than standard statisics variance anyone knows. But as long as you don't define it, any sequence will be empirical.

[abaza]

Thank you! Probably variability or maximum randomisation should be more correct!

[Torsten]

Probably variability or bst randomisation should be more correct!

And what's the meaning of those ? You operate with vague associations, not with precise definitions.

[abaza]

Minimum number of pairs...

Thanks! Or

if D= my_vector

MINIMUM number of:

unique(find(diff(D)==0))

[Torsten]

D = repmat([10 repmat([repmat(0, 1, 2), 1 repmat(0, 1, 2)], 1, 5)], 1, 20);
nnz(diff(D)==0)

ans = 280

D = [repmat([0,1],1,100),repmat([0,10],1,20),zeros(1,280)];
nnz(diff(D)==0)

ans = 279

[Bruno Luong]

This is a method for generic case

v=[0 1 10];
n=[400 100 20];
A = shuffle(v, n)

A = 1×520

0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

function A = shuffle(v, n)
[m,i] = maxk(n,2);
mm = min(m);
if mm > 0
    n(i) = n(i)-mm;
    A = [repmat(v(i), 1, mm), shuffle(v, n)];
else
    [mm,j] = max(m);
    A = repmat(v(i(j)), 1, mm);
end
end

[abaza]

That is amazing!!! Thank you Bruno!