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choose 2 random unique element inside cell

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NA
NA on 23 Feb 2019
Commented: Jos (10584) on 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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Accepted Answer

per isakson
per isakson on 23 Feb 2019
Edited: per isakson on 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?

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NA
NA on 24 Feb 2019
I want to do something like this
if iteration==1
changed_mes_index=cellfun( @(m) datasample( m, changed_num, 'replace',false ), index, 'uni',false );
end
this code
index = repmat( index, 5,1 );
out = cellfun( @(row) datasample( row, 2, 'replace',false ), index, 'uni',false )
does not give what I want
NA
NA on 24 Feb 2019
amount is calculated by adding normrnd. so it changes in each iteration.
per isakson
per isakson on 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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More Answers (1)

Jos (10584)
Jos (10584) on 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 Comment

Jos (10584)
Jos (10584) on 25 Feb 2019
This comment has nothing to do with my answer ...

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