How can I rearrange a list of numbers so that every N numbers is nonrepeating? - matlab

Question

So I have a list of 190 numbers ranging from 1:19 (each number is repeated 10 times) that I need to sample 10 at a time. Within each sample of 10, I don't want the numbers to repeat, I tried incorporating a while loop, but computation time was way too long. So far I'm at the point where I can generate the numbers and see if there are repetitions within each subset. Any ideas?
N=[];
for i=1:10
N=[N randperm(19)];
end
B=[];
for j=1:10
if length(unique(N(j*10-9:j*10)))<10
B=[B 1];
end
end
sum(B)
Below is an updated version of the code. this might be a little more clear in showing what I want. (19 targets taken 10 at a time without repetition until all 19 targets have been repeated 10 times)
nTargs = 19;
pairs = nchoosek(1:nTargs, 10);
nPairs = size(pairs, 1);
order = randperm(nPairs);
values=randsample(order,19);
targs=pairs(values,:);
Alltargs=false;
while ~Alltargs
targs=pairs(randsample(order,19),:);
B=[];
for i=1:19
G=length(find(targs==i))==10;
B=[B G];
end
if sum(B)==19
Alltargs=true;
end
end

Answer 1

Here are some very simple steps to do this, basically you just shuffle the vector once, and then you grab the last 10 unique values:
v = repmat(1:19,1,10);
v = v(randperm(numel(v)));
[a idx]=unique(v);
result = unique(v);
v(idx)=[];
The algorithm should be fairly efficient, if you want to do the next 10, just run the last part again and combine the results into a totalResult

Answer 2

You want to sample the numbers 1:19 randomly in blocks of 10 without repetitions. The Matlab function 'randsample' has an optional 'replacement' argument which you can set to 'false' if you do not want repetitions. For example:
N = [];
replacement = false;
for i = 1:19
N = [N randsample(19,10,replacement)];
end
This generates a 19 x 10 matrix of random integers in the range [1,..,19] without repetitions within each column.
Edit: Here is a solution that addresses the requirement that each of the integers [1,..,19] occurs exactly 10 times, in addition to no repetition within each column / sample:
nRange = 19; nRep = 10;
valueRep = true; % true while there are repetitions
nLoops = 0; % count the number of iterations
while valueRep
l = zeros(1,nRep);
v = [];
for m = 1:nRep
v = [v, randperm(nRange,nRange)];
end
m1 = reshape(v,nRep,nRange);
for n = 1:nRep
l(n) = length(unique(m1(:,n)));
end
if all(l == nRep)
valueRep = false;
end
nLoops = nLoops + 1;
end
result = m1;
For the parameters in the question it takes about 300 iterations to find a result.

Answer 3

I think you should approach this constructively.
It's easy to initially find a 19 groups that fulfill your conditions just by rearranging the series 1:19: series1 = repmat(1:19,1,10); and rearranged= reshape(series1,10,19)
then shuffle the values
I would select two random columns copy them and switch the values at two random positions
then make a test if it fulfills your condition - like: test = #(x) numel(unique(x))==10 - if yes replace your columns
just keep shuffling till your time runs out or you are happy
of course you might come up with more efficient shuffling or testing

Answer 4

I was given another solution through the MATLAB forum that works pretty well (Credit to Niklas Nylen over on the MATLAB forum). Computation time is pretty low too. It basically shuffles the numbers until there are no repetitions within every 10 values. Thanks all for your help.
y = repmat(1:19,1,10);
% Run enough iterations to get the output random enough, I selected 100000
for ii = 1:100000
% Select random index
index = randi(length(y)-1);
% Check if it is allowed to switch places
if y(index)~=y(min(index+10, length(y))) && y(index+1)~=y(max(1,index-9))
% Make the switch
yTmp = y(index);
y(index)=y(index+1);
y(index+1)=yTmp;
end
end