Retrieving a random element from each row in a matrix

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Lui 2019-5-23

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评论： Star Strider 2019-5-24

采纳的回答： Star Strider

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hi everyone

I have a 20 by 10 matrix. I am running the following code to retrieve a random element from each row of the matrix. In this case, I need to create a 20 by 1 matrix at the end. I have written the following code but it does give me a shuffle of the first column alone.

sz=size(R,1);
B=zeros(sz,1);
i=1:20
% for n=1:100
for j=1
    for i=1:20
        for row=1:20
            M(i,j)=randi(length(R),20,1);
            B(i,j)=R(M(i,j));
        end
    end
end
% end

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Alex Mcaulley 2019-5-23

What is your question? Do you have any issue?

Lui 2019-5-23

The code does not yield the desired results and would appreciate if someone helped point out where it is wrong or an alternative methods

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Star Strider 2019-5-23

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It is likely easier to use the sub2ind funciton to create linear indices into ‘R’ from a defined list of random column subscripts:

R=randi(1000,20,10);
idx = sub2ind(size(R), (1:size(R,1))', randi(size(R,2),size(R,1),1));
B = R(idx)

The second argument defines the row indices and the third defines the random column indices.

See if that does what you want.

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Lui 2019-5-24

I am grateful.

Just for curiosity sake, if I were to use randi with data which is not normally distributed, the results would be normally distributed. What is the alternative to ensure that the sample data maintains follows its initial distribution?

Star Strider 2019-5-24

As always, my pleasure.

The output of the randi function is uniformly distributed, not normally distributed. If you were to use the normal distribution instead (the randn function being one such), and you did not change the mean and standard deviation to create the ‘R’ matrix, the data would remain normally distributed.

If you do not otherwise change the parameters of the distribution you are using to create the matrix, the matrix should retain the properties of that distribution, regardless of how you sample it.

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