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 2018-2-7

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https://ww2.mathworks.cn/matlabcentral/answers/381273-how-to-shuffle-rows-of-a-matrix-say-order-4-with-the-help-of-set-having-4-elements-several-times

评论： Abdul Gaffar 2018-2-7

采纳的回答： Jan

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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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Abdul Gaffar 2018-2-7

Thanks, but still having problem

Birdman 2018-2-7

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采纳的回答

Jan 2018-2-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/381273-how-to-shuffle-rows-of-a-matrix-say-order-4-with-the-help-of-set-having-4-elements-several-times#answer_303919

编辑：Jan 2018-2-7

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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, :)

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Jan 2018-2-7

编辑：Jan 2018-2-7

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@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?

Abdul Gaffar 2018-2-7