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

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Suvra Vijay
Suvra Vijay 2017-2-14
回答： Sandip 2023-10-15
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

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KSSV
KSSV 2017-2-14
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
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Tony Richardson
Tony Richardson 2020-1-22
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)))
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Tony Richardson
Tony Richardson 2021-1-11
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)))

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Jan
Jan 2017-2-14
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)
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Suvra Vijay
Suvra Vijay 2017-2-14
thank you sir, but I don't have C compiler installed and this 'Shuffle' is not working
Jan
Jan 2017-2-14
While I'm sure, that this Shuffle is working (I've tried it some minutes ago), it requires to be compiled at first.

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Nikolay Petrov
Nikolay Petrov 2022-2-22
编辑：Nikolay Petrov 2022-2-22
Ran in:
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.
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Tony Richardson
Tony Richardson 2022-2-22
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.

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Tony Richardson
Tony Richardson 2020-1-24
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
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Tony Richardson
Tony Richardson 2021-1-11
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).

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Sandip
Sandip 2023-10-15
a_random = a(randperm(size(a, 1)), :);

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