how to randomly shuffle the row elements of a predefined matrix??
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Suvra Vijay
am 14 Feb. 2017
Beantwortet: Sandip
am 15 Okt. 2023
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
am 14 Feb. 2017
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
am 22 Jan. 2020
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)))
3 Kommentare
Tony Richardson
am 11 Jan. 2021
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)))
Jan
am 14 Feb. 2017
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)
2 Kommentare
Jan
am 14 Feb. 2017
While I'm sure, that this Shuffle is working (I've tried it some minutes ago), it requires to be compiled at first.
Nikolay Petrov
am 22 Feb. 2022
Bearbeitet: Nikolay Petrov
am 22 Feb. 2022
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)))
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)), :)
shuffle the elements of each column in the matrix.
1 Kommentar
Tony Richardson
am 22 Feb. 2022
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.
Tony Richardson
am 24 Jan. 2020
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
3 Kommentare
Tony Richardson
am 11 Jan. 2021
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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