Exchange position of elements in an array

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luca
luca 2019-10-7
回答： David Hill 2019-10-7
Hi given two arrays
SP = [1 2 3 4 8 11 13 14 15 16 18 19 20];
Y= [24 18 18 18 18 18 18 18 18 18 18 18 18];
I create an array
S= [19 1 4 14 20 11 20 13 18 11 8 16 14 13 3 13 8 1 15 13 16 14 1 11 3 4 20 19 8 20 14 18 3 11 13 13 11 20 11 20 16 2 20 15 1 11 4 13 18 20 3 15 11 8 16 8 14 14 1 8 14 8 1 16 3 20 8 18 16 4 15 2 8 2 3 16 14 3 15 2 1 18 13 4 18 4 20 18 14 4 2 2 2 4 8 11 20 13 3 11 16 19 2 18 3 11 19 1 13 3 19 11 15 3 14 19 16 2 8 13 3 11 20 19 18 18 19 15 14 16 8 1 3 18 18 4 14 15 15 18 1 3 4 1 15 2 11 1 16 15 8 13 16 16 1 19 16 13 19 8 19 11 20 1 15 2 13 14 20 1 2 8 14 4 15 15 4 1 1 3 18 1 4 4 1 8 11 2 1 18 14 18 13 8 14 19 3 18 11 4 11 20 16 1 15 15 19 2 19 3 2 20 13 19 4 13 15 4 16 1 8 19 1 3 19 16 2 4 14 2 14 1 16 19 15 20 2 20 13 18];
Where 1 is repeated 24 times, 2 is repeated 18 times, 3 is repeated 18 times and so on. So the array SP tells us the diversity of elements and Y how many times they are repeated in S.
Now what I want to do is to exchange randomly the position of the elements inside S.
In particular I want to follow two rules:
1) Just some elements can exchange with eachother. In particular considering the matrix BB
BB= [1 2 3 4 0 11 14 15;
0 0 0 8 13 16 0 0;
0 0 0 0 0 18 0 0;
0 0 0 0 0 19 0 0;
0 0 0 0 0 20 0 0];
We can exchange each other just the element on the same column. It means that a 11 can just exchange with 16,18,19,20. While 4 can just exchange with 8. Obviously if there is just one number on the columns it will remain in its position.
2) I want that the elements exchange their position randomly with the one on the same columns of BB. But I don't want to substitute a value with another random value on the same colums! This means that at the end, the number of repetitons indicated in Y will be respected in the new array
May someone can help me with this code?

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采纳的回答

David Hill
David Hill 2019-10-7
for i=1:size(BB,2)
x=S(ismember(S,BB(:,i)));
S(ismember(S,BB(:,i)))=x(randperm(length(x)));
end

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