choose 2 random unique element inside cell

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NA
NA am 23 Feb. 2019
Kommentiert: Jos (10584) am 25 Feb. 2019
how to choose 2 random unique element inside a cell?
index={[1,2,5,9,10,13,17,18,21],[4,5,7,12,13,15,18,20,21],[3,4,6,11,12,14,18,19,20],[8,16,20,22]};
example of result
p={[1,9],[7,21],[3,4],[16,8]}

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per isakson
per isakson am 23 Feb. 2019
Bearbeitet: per isakson am 23 Feb. 2019
and try
>> cellfun( @(row) datasample( row, 2, 'replace',false ), index, 'uni',false )
ans =
1×4 cell array
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
"[...] random unique element inside a cell" does that mean without replacement?
If it means literal "unique elements" then run this is a first step to remove duplicates.
>> index = cellfun( @unique, index, 'uni',false )
ix =
1×4 cell array
{1×9 double} {1×9 double} {1×9 double} {1×4 double}
Or should the propability to pick a specific value be proportional to the number of duplicates of that value?
  8 Kommentare
NA
NA am 24 Feb. 2019
Bearbeitet: per isakson am 24 Feb. 2019
amount is calculated by adding normrnd. so it changes in each iteration.
per isakson
per isakson am 24 Feb. 2019
"amount is calculated by adding normrnd" That means that you need to be careful regarding Accuracy of Floating-Point Data
So far, I have assumed whole numbers

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Weitere Antworten (1)

Jos (10584)
Jos (10584) am 23 Feb. 2019
If I understand you correctly, the final output should have 8 unique numbers, in four groups of two, where the numbers in each group is drawn from a cell in the cell array index. This is not a trivial problem to answer! There might situations where this is simply impossible given a cell array index (for instance index= {[1 2 3], [1 2 4], [2 3 4],... }
You might want to try a brute-force method
index={[1,2,5,9,10,13,17,18,21],[4,5,7,12,13,15,18,20,21],[3,4,6,11,12,14,18,19,20],[8,16,20,22]};
fn = @(v) sort(v(randperm(numel(v),2))) ; % function to draw 2 randomelements from v
k = 1 ;
while k < 1e5 % try a lot of times
C = cellfun(fn, index,'un',0) ;
if numel(unique([C{:}])) == 8,
break
else
C = [] ; k = k+1 ;
end
end
celldisp(C)
  1 Kommentar
Jos (10584)
Jos (10584) am 25 Feb. 2019
This comment has nothing to do with my answer ...

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