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help n=12 for I=1:n p(I) =rand p(I) =28×p(I) p(I)=ceil(p(I)) end I need to put constraint on this code must be theres no redanandnt values ..so if n=28 p will be all the numbers from 1:28 but in random
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n=12
for I=1:n
p(I) =rand
p(I) =28×p(I)
p(I)=ceil(p(I))
end
I need to put constraint on this code must be theres no redanandnt values ..so if n=28 p will be all the numbers from 1:28 but in random
1 件のコメント
Adam
2015 年 2 月 10 日
Please keep code in the body of the question not in the title. The title should be a short description of the problem.
Also please use the {} Code block to include formatted code.
回答 (2 件)
Star Strider
2015 年 2 月 10 日
‘I need to put constraint on this code must be theres no redanandnt values ..so if n=28 p will be all the numbers from 1:28 but in random’
The easiest way is to use the randperm function:
n = 12;
p = randperm(n);
5 件のコメント
Star Strider
2015 年 2 月 10 日
@Adam — Thank you for your Comment!
Our friendly randperm appears frequently in Answers. That’s how I learned about it, since I don’t otherwise need to use it that often.
Star Strider
2015 年 2 月 10 日
Adam
2015 年 2 月 10 日
Something like this would work:
nums = 1:28;
n = 28;
p = zeros(1,n);
count = numel( nums );
for i = 1:n
idx = randi(count);
p(i) = nums( idx );
nums( idx ) = [];
count = count - 1;
end
Not necessarily the most efficient way of doing it, but it will guarantee uniqueness so long as n is at most 28 (or whatever value you set as the maximum of nums.
2 件のコメント
Adam
2015 年 2 月 10 日
replace:
randi( count )
with
round( rand( count ) * n )
and I think that should give you the correct answer.
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