使用Matlab查找并索引数组中的重复元素

Question

I have an array A = [10 20 20 30 40 10 50];

Is there any smart way to find and locate the repeating elements of this array ?

ie

10: [1 6]; 
20: [2 3];

I tried to use unique but I failed ...

Answer 1

Here is one solution:

% input array
A = [10 20 20 30 40 10 50];

% find unique elements (vals), and map values of A to indices (valsIdx)
[vals,~,valsIdx] = unique(A);

% group element locations by the above indices mapping
locs = accumarray(valsIdx, 1:numel(valsIdx), [], @(x){sort(x)});

% keep only values that are repeated
idx = cellfun(@numel, locs) > 1;
vals = vals(idx);
locs = locs(idx);

The result:

vals =
    10    20
locs = 
    [2x1 double]
    [2x1 double]

>> celldisp(locs)
locs{1} =
     1
     6
locs{2} =
     2
     3

Answer 2

Here is another:

>> A = [10 20 20 30 40 10 50];
>> S = sort(A);
>> S = arrayfun(@(x) [x find(x==A)], unique(S(diff(S)==0)), 'UniformOutput', false);
>> S{:}
ans =
    10     1     6
ans =
    20     2     3

If you don't have or want to use arrayfun , you can use a plain loop:

A = [10 20 20 20 30 40 10 50];

S = sort(A);
S = unique(S(diff(S)==0));

R = cell(size(S'));
for ii = 1:numel(S)
    R{ii} = [S(ii) find(A==S(ii))]; end

Answer 3

With bsxfun and arrayfun :

comp = tril(bsxfun(@eq, A(:), A(:).')); %'// compare all pairs of values
ind = find(sum(comp)>1); %// find repeated values
values = A(ind);
positions = arrayfun(@(n) find(comp(:,n).'.*(1:numel(A))), ind, 'uni', 0);

This gives:

>> values
values =
    10    20

>> positions{:}
ans =
     1     6

ans =
     2     3

Answer 4

This solution only returns the values not the indices(location) of the values.

%Your Data
A=[10 20 20 30 40 10 50]; 
%sorted Data
A_sorted=sort(A);
%find the duplicates
idx=find(diff(A_sorted)==0);
% the unique is needed when there are more than two duplicates.
result=unique(A_sorted(idx));