randperm in while loop

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MiauMiau
MiauMiau 2015-4-29
编辑： pfb 2015-4-29
Hi, I want to achieve the following with my code:
For A(1) I can chose randomly from 1, 2 or 3 - but if A(1) say is equal to 1, then A(2) should be randomly chosen from either 2 or 3, and if A(2) would be say 3, then A(3) = 2. And with A(4) the game starts again. So I wrote the following code - and run it with RandomizeTrials(3,12) - why is it not running?
function [Array] = RandomizeTrials(numberofSeq, lengthofArray)
Array = zeros(1,lengthofArray);
k = 1;
while k < lengthofArray-1
m = k+numberofSeq-1;
Array(k:m) = randperm(numberofSeq,numberofSeq,false);
k+numberofSeq;
end
end
  1 个评论
Adam
Adam 2015-4-29
What do you mean by "not running"?
When I run it I get a clear error message which tells me exactly what the problem is which I can consult the randperm help page for to confirm. I assume you can do likewise if you are getting the same error.

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回答（2 个）

pfb
pfb 2015-4-29
If I get it right, you want a vector A that is "divided" into bunches of 3 consecutive elements. Each of these "bunches" should be a random permutation of the integers 1, 2, 3.
This should be it
A = zeros(3*N);
for k=1:N
A((k-1)*3+(1:3))=randperm(3);
end
You can also create a Nx3 matrix of random permutations, and then possibly "unwind" it to a single vector.
A = zeros(N,3);
for k =1:N
A(k,:)=randperm(3);
end
A = A'; % transpose so that the permutations are along columns
A = A(:); % the columns of A are concatenated into a single column vector
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MiauMiau
MiauMiau 2015-4-29
Hi, thanks, but what does A((k-1)*3+(1:3)) exactly do? I mean we would end up with A((1:3)) or then A((1:3)+3) etc. - what does eg. A((1:3)+3) or A((1:3)+6) exaclty do? I hadn't encounter this syntax. But I see that if we would use A((1:3)+N)) = randperm(3), we would not neet a loop altogether, right?
pfb
pfb 2015-4-29
编辑：pfb 2015-4-29
Hi
(k-1)*3+(1:3)
is the same as
((k-1)*3+1):(k*3)
i.e. the indices
k*3-2, k*3-1, k*3
if k=1 these are 1,2,3; if k=2 these are 4,5,6, and so on... This particular choice is related to the fact that indices start from 1 in matlab.
The second choice above is more or less what Thorsten suggested below, although he did that in two lines, probably for clarity reason.
I thought that
(1:3)+(k-1)*3
was sufficiently clear:
(1:3)
is the vector [1 2 3]. Then I add (k-1)*3, and I use the result as the index for A.

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Thorsten
Thorsten 2015-4-29
编辑：Thorsten 2015-4-29
In this solution you have a growing array in a loop, which is not the fasted way, but you do not have to fiddle with indices:
N = length_of_array/number_of_sequence;
x = [];
for i=1:N
x = [x randperm(number_of_sequence)];
end
If you insist on preallocation and indices
x = zeros(1, length_of_array);
for i=1:N
i1 = (i-1)*number_of_sequence;
x(i1+1:i1+number_of_sequence) = randperm(number_of_sequence);
end

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