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

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inzamam shoukat 2018-12-5

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评论： inzamam shoukat 2018-12-8

采纳的回答： Stephen23

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

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madhan ravi 2018-12-5

what do you mean 2 and 4 are repeated twice? what‘s the rule?

inzamam shoukat 2018-12-5

编辑：inzamam shoukat 2018-12-5

Actually i have to scan only column 2 to chk repeated terms....and then then arrange the matrix as given i can do it by sortroes cammand but it can give me only 1 answer i need factorial times arrangments...... column 1 enteries are just serial number there is nothing to do with it imagine it was not even there

repeated terms=2

Arrangments are

1 4

2 4

3 2

&

2 4

1 4

3 2

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Answer 1

Stephen23 2018-12-5

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编辑：Stephen23 2018-12-5

在 MATLAB Online 中打开

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 =
 4
 9
 2
 9
 4
 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 =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 9
ans =
 4
 9
 2
 9
 4
 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....