Simulating a random sample from a matrix

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Frank
Frank el 16 de Abr. de 2014
Respondida: Star Strider el 16 de Abr. de 2014
I am trying to randomly sample a 88x2 matrix. The sample should select an element at random from each of the 88 rows and be simulated 1,000 times.
  4 comentarios
Star Strider
Star Strider el 16 de Abr. de 2014
The impression I got from your original question was that you want to select a random 2-element row from your (88x2) matrix. So do you want to first select a random row and then a random column element from that row?
Frank
Frank el 16 de Abr. de 2014
Not a random row. If I break the matrix apart, I have 2 arrays (A or B). I'd like to go to the first row and randomly select A or B. Then I'd like to go to the next row and select A or B, and so on until I reach all 88 rows. Then I'd like to loop to simulate this 1000 times.

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Respuestas (3)

Andrew Newell
Andrew Newell el 16 de Abr. de 2014
Editada: Andrew Newell el 16 de Abr. de 2014
Suppose M is your matrix. The following code will produce a matrix of random selections with one row for each selection.
idx = randi(88,88,1000);
Mrand = M(idx);
  2 comentarios
Frank
Frank el 16 de Abr. de 2014
Not a random row. If I break the matrix apart, I have 2 arrays (A or B). I'd like to go to the first row and randomly select A or B. Then I'd like to go to the next row and select A or B, and so on until I reach all 88 rows. Then I'd like to loop to simulate this 1000 times. I thing the output should be a matrix 88x1000.
Andrew Newell
Andrew Newell el 16 de Abr. de 2014
Sorry, I didn't read the question carefully enough. I have edited the above code to answer the correct problem.

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W. Owen Brimijoin
W. Owen Brimijoin el 16 de Abr. de 2014
So for each row you want to randomly pick either the first or the second column, correct? This would result in 88 total values that are coin flips between one of the two elements in each row. There are ways of doing this all at once, but an easily understandable way is to use sub2ind to make yourself an index of random column choices, and do this 1000 times in a loop.
Try this:
Mrand = zeros(88,1000); %create an empty array for the results
for n = 1:1000,%step through the following 1000 times:
%make an index of one random col per row:
idx = sub2ind([88 2],[1:88]',randi(2,88,1));
%use this index to pick values in M:
Mrand(:,n) = M(idx);
end

Star Strider
Star Strider el 16 de Abr. de 2014
With the clarifications (and a night’s sleep for me), I suggest:
M = rand(88,2); % Data Matrix
R = []; % Initialise ‘R’ vector
for k1 = 1:1000 % Run outer loop 1000 times
for k2 = 1:size(M,1) % Run inner loop the row size of ‘M’
ix = randi(size(M,2)); % Choose either colunm 1 or 2
R = [R; M(k2,ix)]; % Accumulate result in ‘R’
end
end
It took 21.2 seconds to generate the 88000-element ‘R’ vector on my machine.

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