How can I generate a random vector between 1-9 with 10 numbers and at least one number should be >2.

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X=rand(10,1)*(9-1)+1 How should I change this code to make at least one number >2.

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John D'Errico
John D'Errico 2014-12-21
编辑:John D'Errico 2014-12-21
Generate 9 (NINE) random numbers with no constraint from 1:9. Then generate a 10th number that is assuredly greater than 2. Randomly permute the set.
R = [ceil(rand(1,9)*9), ceil(rand*7 + 2)];
R = R(randperm(10));
Easy peasy. If you prefer to use randi, it is just as easy, and perhaps easier to read.
R = [randi([1 9],[1 9]), randi([3,9],1)];
R = R(randperm(10));
The general idea is a common one to be found in probability. If you need one number to have some property, then generate it as such.
Edit: to generate real numbers instead of integers...
R = [rand([1 9])*8 + 1, rand(1)*7 + 2];
R = R(randperm(10));
The basic idea remains the same. Note that the rand function generates a value in the strict open interval (0,1), so the result of our 10th number will never be exactly 2. Thus, that value must always be strictly greater than 2.
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John D'Errico
John D'Errico 2014-12-21
编辑:John D'Errico 2014-12-21
Sorry. I presumed (incorrectly) that your numbers be integers given a too fast reading of your question. Even so, it is trivial to change to generate real numbers as I show in the edit.

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更多回答(2 个)

Matt J
Matt J 2014-12-21
编辑:Matt J 2014-12-21
One way,
X=rand(10,1)*(8-1)+1 +ceil(rand(10,1))

Azzi Abdelmalek
Azzi Abdelmalek 2014-12-21
X=rand(10,1)*8+1;
idx=X>2;
if nnz(idx)==0
X(randi(numel(X)))=7*rand+2;
end

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