Is there a better/faster way of randomly shuffling a matrix in MATLAB?

Question

In MATLAB, I am using the shake.m function (http://www.mathworks.com/matlabcentral/fileexchange/10067-shake) to randomly shuffle each column. For example:

a = [1 2 3; 4 5 6; 7 8 9]
a =

     1     2     3
     4     5     6
     7     8     9

b = shake(a)
b =

     7     8     6
     1     5     9
     4     2     3

This function does exactly what I want, however my columns are very long (>10,000,000) and so this takes a long time to run. Does anyone know of a faster way of achieving this? I have tried shaking each column vector separately but this isn't faster. Thanks!

Answer 1

You can use randperm like this, but I don't know if it will be any faster than shake:

[m,n]=size(a)
for c = 1:n
    a(randperm(m),c) = a(:,c);
end

Or you can try switch the randperm around to see which is faster (should produce the same result):

[m,n]=size(a)
for c = 1:n
    a(:,c) = a(randperm(m),c);
end

Otherwise how many rows do you have? If you have far fewer rows than columns, it's possible that we can assume each permutation will be repeated, so what about something like this:

[m,n]=size(a)
cols = randperm(n);
k = 5;  %//This is a parameter you'll need to tweak...
set_size = floor(n/k);
for set = 1:set_size:n
    set_cols = cols(set:(set+set_size-1))
    a(:,set_cols) = a(randperm(m), set_cols);
end

which would massively reduce the number of calls to randperm. Breaking it up into k equal sized sets might not be optimal though, you might want to add some randomness to that as well. The basic idea here though is that there will only be factorial(m) different orderings, and if m is much smaller than n (e.g. m=5, n=100000 like your data), then these orderings will be repeated naturally. So instead of letting that occur by itself, rather manage the process and reduce the calls to randperm which would be producing the same result anyway.

Answer 2

Here's a simple vectorized approach. Note that it creates an auxiliary matrix (ind) the same size as a, so depending on your memory it may be usable or not.

[~, ind] = sort(rand(size(a))); %// create a random sorting for each column
b = a(bsxfun(@plus, ind, 0:size(a,1):numel(a)-1)); %// convert to linear index

Answer 3

Obtain shuffled indices using randperm

idx = randperm(size(a,1));

Use the indices to shuffle the vector:

m = size(a,1);
for i=1:m
 b(:,i) = a(randperm(m,:);
end

Look at this answer: Matlab: How to random shuffle columns of matrix

Answer 4

Here's a no-loop approach as it processes all indices at once and I believe this is as random as one could get given the requirements of shuffling among each column only.

Code

%// Get sizes
[m,n] = size(a);

%// Create an array of randomly placed sequential indices from 1 to numel(a)
rand_idx = randperm(m*n);

%// segregate those indices into rows and cols for the size of input data, a
col = ceil(rand_idx/m);
row = rem(rand_idx,m);
row(row==0)=m;

%// Sort both these row and col indices based on col, such that we have col
%// as 1,1,1,1 ...2,2,2,....3,3,3,3 and so on, which would represent per col
%// indices for the input data. Use these indices to linearly index into a
[scol,ind1] = sort(col);
a(1:m*n) = a((scol-1)*m + row(ind1))

Final output is obtained in a itself.