choose 2 random unique element inside cell
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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]}
采纳的回答
per isakson
2019-2-23
编辑：per isakson
2019-2-23
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 个评论
NA
2019-2-23
编辑：per isakson
2019-2-24
per isakson
2019-2-24
编辑：per isakson
2019-2-24
Problem: You have a working script that runs the simulation once and you want to run it automatically many times
The first approach you shall consider is a for-loop.
- Convert the script to a function
- Create a second function that calls the first many times
- Run the second function
Example
>> out = cssm( index, 5 )
out =
5×1 cell array
{1×4 cell}
{1×4 cell}
{1×4 cell}
{1×4 cell}
{1×4 cell}
>> out{3}
ans =
1×4 cell array
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
where
function out = cssm( index, n )
out = cell( n, 1 );
for jj = 1 : n
out{jj} = run_once( index );
end
end
function out = run_once( index )
out = cellfun( @(row) datasample( row, 2, 'replace',false ) ...
, index, 'uni',false );
end
BTW:
%%
index = repmat( index, 5,1 );
out = cellfun( @(row) datasample( row, 2, 'replace',false ), index, 'uni',false )
outputs
out =
5×4 cell array
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
{1×2 double} {1×2 double} {1×2 double} {1×2 double}
per isakson
2019-2-24
What happened to the comments on the interpretation of "unique"?
per isakson
2019-2-24
编辑：per isakson
2019-2-24
"you mean I do something like this" No!
What happend to "I want to run simulation 5 times"?
for iteration=1:5
...
end
isn't meaningful. It does the same thing five times and overwrites result. How is amount supposed to change?
It's hard to help, since you don't try to understand my answer and move the goalposts in the comments.
And I repeat, what happened to the comments on the interpretation of "unique"?
NA
2019-2-24
NA
2019-2-24
编辑：per isakson
2019-2-24
per isakson
2019-2-24
"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
更多回答（1 个）
Jos (10584)
2019-2-23
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 个评论
Jos (10584)
2019-2-25
This comment has nothing to do with my answer ...
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