how to randomly shuffle the row elements of a predefined matrix??

i have a matrix , a= [1 2 4 6; 5 8 6 3;4 7 9 1] i want to randomly shuffle the elements of each row. how to do it?? please help

Answer 1

[KSSV]

a= [1 2 4 6; 5 8 6 3;4 7 9 1] ;
[m,n] = size(a) ;
idx = randperm(n) ;
b = a ;
b(1,idx) = a(1,:)  % first row arranged randomly

[Suvra Vijay]

Thank you so much sir, but how to arrange all three rows at a time

[KSSV]

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

[Ren Li]

nice !

[zeu]

I try to make the answer of KSSV work on a 3d array. I only want to randomly permute one dimension of that array. But it doesnt work. Can someone help?

[jaah navi]

Could you please help me to randomly arrange the elements in each column.

[Willingo]

The code above to do this for all rows will not work due to what I expect to be a typo. https://www.mathworks.com/matlabcentral/answers/324891-how-to-randomly-shuffle-the-row-elements-of-a-predefined-matrix#comment_428742

It will just create a permutation of the first row three times and stack them on top of eachother. Regadless, a for loop is not needed in the first place. A modified and more succinct example is created below.

Lastly, for those who want to do this for columns, a lazy way would be to just transpose a before all of this and then at the end. More properly, you set idx = randperm(m) and do b(idx,:) = a

for i=1:m 
    idx = randperm(n) ; 
    %b(i,idx) = a(1,:) ; % This returns error.
    b(i,idx) = a(i,:) ; %corrected
end

Achieve result without a for loop. I replaced the original matrix a with a matrix whose indices' values correspond to their linera index, so it's a bit easier to track what is going on.

a= NaN(3,4);
a(1:end) = 1:numel(a);
[m,n] = size(a) ;
idx = randperm(n) ;

b(:,idx) = a

[Bruno Luong]

@Hamza Shami it shuffle randomly assignmet of columns of a to b

[Hamza Shami]

@Bruno Luong yea got it, thanks

Answer 2

[Tony Richardson]

Perhaps not very efficient, but uses only built-in functions and randomizes all elements of all columns:

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

[m, n] = size(a);

[~, idx] = sort(rand(m,n));

b = a(sub2ind([m, n], idx, ones(m,1)*(1:n)))

[munazah lyle]

How can we get the original sequence back for this.

[Tony Richardson]

You can recover the original a matrix with (here c will equal a):

c = ones(size(b));

c(sub2ind([m, n], idx, ones(m,1)*(1:n))) = b

I expect that you want the reverse indexing to go back from b to a, though. This seems to do that (here d will be equal to a):

[~, rdx] = sort(idx);

d = b(sub2ind([m, n], rdx, ones(m,1)*(1:n)))

[munazah lyle]

Thank you

Answer 3

[Jan]

If you have a C compliler installed, you can try https://www.mathworks.com/matlabcentral/fileexchange/27076-shuffle:

a = [1, 2, 4, 6;  5, 8, 6, 3;  4, 7, 9, 1];
b = Shuffle(a, 2)

[Suvra Vijay]

thank you sir, but I don't have C compiler installed and this 'Shuffle' is not working

[Jan]

While I'm sure, that this Shuffle is working (I've tried it some minutes ago), it requires to be compiled at first.

Answer 4

[Nikolay Petrov]

Not sure why this was more complicated to find that it should have been.

a = [1, 2, 4, 6;  5, 8, 6, 3;  4, 7, 9, 1];
a(:, randperm(size(a, 2)))

ans = 3×4

6 4 1 2 3 6 5 8 1 9 4 7

shuffles the elements of each row in the matrix, while

a = [1, 2, 4, 6;  5, 8, 6, 3;  4, 7, 9, 1];
a(randperm(size(a, 1)), :)

ans = 3×4

5 8 6 3 1 2 4 6 4 7 9 1

shuffle the elements of each column in the matrix.

[Tony Richardson]

These do the same rearrangement in all rows or columns. Perhaps that is what the original poster wanted. I am not sure. I interpret it as asking for each column (or row) to be shuffled independently of the others.

Answer 5

[Tony Richardson]

As a variation of my answer above, I'll note that if you want to generate M permutations of N objects (where the N objects are represented by the integers 1-N) you can use:

[~, x] = sort(rand(N, M));

I can generate 100,000 permutations of 52 objects in 0.3 seconds on my machine.

The probability of drawing 3 aces in a 5 card draw can be estimated (using 100,000 dealt hands):

%%%%%%%%%%%%%%%%%

M = 100000; % Number of trials

N = 52; % Number of cards in deck

[~, x] = sort(rand(N, M)); % columns of x are shuffled decks (100,000 shuffled decks)

y = x(1:5,:); % columns of y are the 5 card hands

% N3a is the number of hands containing three Aces out of M (100,000) deals

% I let Aces be cards 1, 14, 27, and 40 (1-13 is one suit, 14-26 is another, etc)

N3a = sum(sum(or(y == 1, y == 14, y == 27, y == 40)) == 3);

P3a = N3a/M

%%%%%%%%%%%%%%%%%

Successive runs of the script gives values of 0.00181, 0.00185, 0.00189, 0.00171.

The theoretical value is 0.001736

[munazah lyle]

How can we get the original sequence back?

[Tony Richardson]

To go from the shuffled decks (x) back to the unshuffled decks (z):

z = sort(x)

This would just give a matrix whose columns go from 1-52 in order.

Again, I expect what you want is really the reverse index matrix (rdx), to get that and then recover the unshuffled decks from the shuffled ones:

[~, rdx] = sort(idx);

o = idx(sub2ind([m, n], rdx, ones(m,1)*(1:n)))

rdx would be the reverse indexing matrix. The matrix o would be the unshuffled deck (the same as the matrix z above).

[munazah lyle]

@Tony Richardson Thank you

Answer 6

[Sandip]

a_random = a(randperm(size(a, 1)), :);