Retrieving a random element from each row in a matrix

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

[Alex Mcaulley]

What is your question? Do you have any issue?

[Lui]

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

Answer 1

[Star Strider]

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.

[Adam Danz]

I'd do the same (except R is a 10x10 matrix in the description but 20x10 matrix in the OP's example which is confusing).

[Star Strider]

My code adapts automatically to the size of ‘R’.

[madhan ravi]

@Adam: Whatever matrix it might be, this approach yields the desired result.

[Adam Danz]

Of course.... I was about to post the exact same response (down to the character) ;)

Just pointing out that the example data in R is a [20 x 10] matrix where the OP is expecting a final vector of length 10.

"I need to create a 10 by 1 matrix at the end. "

[madhan ravi]

I need to create a 10 by 1 matrix at the end.

I maybe blind but I don't see that statement anywhere in the question.

[Adam Danz]

Well, now it says 20 by 1 but I copy pasted that statement just seconds ago.

[madhan ravi]

XD looks like our beloved OP played a game inbetween.

[Lui]

Sorry guys. That was my mistake. I edited the question to suit what I am working with right now. It was a 10 by 10 initially. It is my mistake. Nut star riders answer worked well.

I am not sure if I wanted to do it n times if it would still work. I appreciate

[Star Strider]

As always, my pleasure!

[Lui]

May I add another question.

I have added a for loop to iterate the process for n times and it still works great. I can get the 20 by 1 matrix just fine. However, is it possible for me to generate 20 by20 matrix instead. This means that for every first n iterations, the loop returns the first colum then another n iterations it returns yet another column until I have a 20 by 20 matrix generated by the same process.

R=randi(1000,20,10);
for n=1:100
idx = sub2ind(size(R), (1:size(R,1))', randi(size(R,2),size(R,1),1));
B = R(idx)
end

How do I change this code to yield a 20 by 20 matrix instead of just one. And for each of the 20 columns, the above line of code should be applicable.

thanks

[Star Strider]

To get your (20x20) matrix, you code needs to change a bit:

R=randi(1000,20,10);
B = zeros(20);                                                              % Preallocate
for n=1:20
    idx = sub2ind(size(R), (1:size(R,1))', randi(size(R,2),size(R,1),1));
    B(:,n) = R(idx);
end

That worked when I tried it.

[Lui]

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]

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.