How to shuffle rows of a matrix, say order 4, with the help of set having 4 elements, several times?
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Abdul Gaffar
el 7 de Feb. de 2018
Comentada: Abdul Gaffar
el 7 de Feb. de 2018
Let
R=[1,2,3,4;5,6,7,8;9,10, 11,12;13,14,15,16];
k=[3,4,1,2];
for i=1:4
R1(i,:)=R(k(i),:);
end
for i=1:4
R2(i,:)=R1(k(i),:);
end
for i=1:4
R3(i,:)=R2(k(i),:);
end
R3
My output will be R3. I want to do this using loop
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Jan
el 7 de Feb. de 2018
Editada: Jan
el 7 de Feb. de 2018
The for i loop over k is not needed:
R = [1,2,3,4; 5,6,7,8; 9,10,11,12; 13,14,15,16];
k = [3,4,1,2];
for i = 1:3
R = R(k, :);
end
Now R is modified inplace.
If you R is huge, it is cheaper to shuffle an index vector instead:
k = [3,4,1,2];
v = 1:4;
for i = 1:3
v = v(k);
end
Rfinal = R(v, :)
3 comentarios
Jan
el 7 de Feb. de 2018
Editada: Jan
el 7 de Feb. de 2018
@ABDUL: And this is exactly what my code does. Did you try it?
R = randi([1,10], 4, 4)
k = [3,4,1,2]
R = R(k, :)
R = R(k, :)
R = R(k, :)
Now the rows of R are shuffled according to k 3 times. Because the 1st and 3rd, as well as 2nd an 4th rows are swapped, the sequence of R is repeating. But for longer k there will be more states.
The same in a loop:
for j = 1:3
R = R(k, :)
end
The line
R = R(k, :)
is just a simpler version of
for i = 1:4
R(i, :) = R(k(i), :);
end
As far as I understand your question, you do not need R1 and R2, so I've overwritten R.
Applying the permutation to the index vector is more efficient, because you do not access the complete matrix R in each iteration and the result is the same.
What is the different to what you want? Do you want to change the index vector k in each iteration?
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