MATLAB: Hello…….i have a matrix…….what i wonaa do is count the repeated entries in column 2 first then make arrangement =(number of repeated termes)! ..for example in given matrix i have two repeated terms so first arrangmnt..1 2 3 the second 2 1 3

makerMATLAB

a=[1 4;2 4;3 2]
required answer is
repeated terms=2
Arrangments are
1 4
2 4
3 2
&
2 4
1 4
3 2

Best Answer

Interesting problem. I used a slightly more complex example, with two values duplicated (4 twice, 9 thrice), for which we would expect twelve possible permutations:
>> a = [1,4;2,9;3,2;4,9;5,4;6,9]
a =
   1   4
   2   9
   3   2
   4   9
   5   4
   6   9
And the code:
[N,B] = histc(a(:,2),unique(a(:,2))); % count instances of values in 2nd column
X = find(N>1);                        % find duplicate values
foo = @(x)perms(find(B==x));          % row permutations for each duplicate value
C = arrayfun(foo,X,'uni',0);          % "


S = cellfun('size',C,1);              % count row permutations
L = arrayfun(@(r)1:r,S,'uni',0);      % row indices for each permutation matrix
[L{:}] = ndgrid(L{:});                % combine row indices
R = cellfun(@(v)v(:),L,'uni',0);      % "
baz = @(m,r)m(r,:);                   % use row indices to select all permutations
M = cellfun(baz,C,R,'uni',0);         % "
V = 1:size(a,1);                      % original row indices of input matrix
Z = cell(1,prod(S));                  % preallocate output cell array
for ii = 1:prod(S)                    % for each possible permutation...
    for jj = 1:numel(X)               % for each duplicate value...
        V(B==X(jj)) = M{jj}(ii,:);    % get row permutation indices
    end                               %
    Z{ii} = a(V,:);                   % use indices to select from input matrix
end
And checking the output:
>> numel(Z) % should be 3!*2!
ans = 12
>> Z{:}
ans =
   1   4
   2   9
   3   2
   4   9
   5   4
   6   9
ans =
   5   4
   2   9
   3   2
   4   9
   1   4
   6   9
ans =
   1   4
   4   9
   3   2
   2   9
   5   4
   6   9
ans =
   5   4
   4   9
   3   2
   2   9
   1   4
   6   9
ans =
   1   4
   2   9
   3   2
   6   9
   5   4
   4   9
ans =
   5   4
   2   9
   3   2
   6   9
   1   4
   4   9
ans =
   1   4
   4   9
   3   2
   6   9
   5   4
   2   9
ans =
   5   4
   4   9
   3   2
   6   9
   1   4
   2   9
ans =
   1   4
   6   9
   3   2
   2   9
   5   4
   4   9
ans =
   5   4
   6   9
   3   2
   2   9
   1   4
   4   9
ans =
   1   4
   6   9
   3   2
   4   9
   5   4
   2   9
ans =
   5   4
   6   9
   3   2
   4   9
   1   4
   2   9
Note that the number of permutations is (# of 1st duplicate value)! * (# of 2nd duplicate value)! * (# of 3rd duplicate value)! * ... , which will clearly explode very quickly into something totally intractable....
Related Question