Creating random order of trials (numbers in a matrix) with restrictions

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natalie222
natalie222 am 29 Apr. 2015
Kommentiert: natalie222 am 2 Mai 2015
Hi! I am new to MATLAB and I am struggling with creating a matrix with numbers which would indicate the different trials of an experiment. The experiment consists of 5 blocks with 50 trials in each block (10 different trials, each repeated 5 times per block). I would like to always start with a different first trial (so not start for instance with trial index 3 in two blocks) and not to repeat the same trial more than once in any given run (so for instance 5 6 2 2 9 ... is ok, but not 5 6 2 2 2 9). I only have the core of the structure - just the randomization, but not the two constraints.
Many thanks!
A=zeros(5,50);
for i=1:5,
p=randi(10,1,50);
if p(1)==A(1:5,1)
???????? How to make the array re-do itself so that the first number will definitely be different?
else
end
A(i,:)=p;
end
  3 Kommentare
natalie222
natalie222 am 29 Apr. 2015
Indeed, dimensions are correct and like you say - no number in the five rows of column 1 can be repeated. Secondly, no number can be present more than twice consequently in any of the five rows - they can (and indeed will be repeated), but should not be repeated more than twice one after another (e.g. 3 3 3). Is this clear?
Joseph Cheng
Joseph Cheng am 29 Apr. 2015
yes... the math didn't add up on the repetition portion but in your clarification you were talking about consecutive trials. I'll come back if i think of anything that doesn't require lots of loops.

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Antworten (2)

Joseph Cheng
Joseph Cheng am 29 Apr. 2015
Well initial thoughts are to generate the initial matrix using
for ind = 1:5
A(ind,:)=mod(randperm(49,49),10)+1;
end
to generate columns 2 to 50. The chances of 3 repeats are low but not infeasible. Then attach column 1 using the randperm(10,5) to get non repeating numbers.
Then go through and detect of groups of consecutive numbers. This can be done through something like this
consecutive=diff(diff(A(ind,:))==0)
and then look for the pattern of [1 0 -1] which would mean there would be 3 consecutive same numbers. more consecutive numbers would be more zeros between the 1 and -1. Then knowing the index locations substitute these numbers with another random set of numbers and recheck/repeat till no consecutive numbers are repeated >2 times in a row.

pfb
pfb am 29 Apr. 2015
So you need a 5x50 matrix of numbers between 1 and 10, such that:
- all the numbers in the first column are different - no number can appear more than two consecutive times in a row
I think I have it. Since it is not a lot of numbers, I did this in a perhaps trivial way, using 2 loops and some parsing
A = zeros(5,50);
pf=0; % previous element in the 1st column. It is 0 at the beginning
for r =1:5
% first trial
t=randi(10);
% this takes care of the constraint in 1st column
while(t==pf)
t=randi(10);
end
A(r,1)=t;
pf=t;
cons=1;
for c=2:50
t=randi(10);
% this should avoid that more than 2 consecutive elements appear in
% each row
cons=cons+(t==A(r,c-1));
if(cons==3)
t=randi(10);
while(t==A(r,c-1))
t=randi(10);
end
cons=1;
end
A(r,c)=t;
end
end

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