Finding the indices of duplicate values in one array

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Luis Alves
Luis Alves 2017-4-21
编辑： Walter Roberson 2025-4-3，20:25
Given one array A=[ 1 1 2 3 5 6 7].
I need help to known the indices where there are duplicate values.
Thanks

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回答（9 个）

Stephan Koehler
Stephan Koehler 2019-7-16
A = [1 2 3 2 5 3]
[v, w] = unique( A, 'stable' );
duplicate_indices = setdiff( 1:numel(A), w )
this should work too, and is elegant
  2 个评论
Jun W
Jun W 2019-11-11
How about finding how many times are those elements repeated?
Image Analyst
Image Analyst 2019-11-11
Use histcounts and look for bins with more than 2 counts.
A = [1 2 3 2 5 3]
[counts, edges] = histcounts(A)
A =
1 2 3 2 5 3
counts =
1 2 2 0 1
edges =
Columns 1 through 5
0.5 1.5 2.5 3.5 4.5
Column 6
5.5
You can see that the bins for 2 and 3 both have 2 counts so there are multiples of 2 and 3 in A.
Note: This will find any repeats, and they don't have to be consecutive. If you want to look for consecutive repeats, call the diff() function and look for zeros.

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Image Analyst
Image Analyst 2018-5-11
编辑：Image Analyst 2018-5-12
Here's one way:
A = [-2 0 1 1 2 3 5 6 6 6 7 11 40]
% Elements 3, 4, 8, 9, and 10 are repeats.
% Assume A is integers and get edges
edges = min(A) : max(A)
[counts, values] = histcounts(A, edges)
repeatedElements = values(counts >= 2)
% Assume they're integers
% Print them out and collect indexes of repeated elements into an array.
indexes = [];
for k = 1 : length(repeatedElements)
indexes = [indexes, find(A == repeatedElements(k))];
end
indexes % Report to the command window.
You get [3,4,8,9,10] as you should.
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Steven Lord
Steven Lord 2025-3-10，16:06
Ran in:
The last bin includes both the left and right edges, while the earlier bins include only the left edges. This is stated in the description of the edges input on the histcounts documentation page: "Bin edges, specified as a vector. The first vector element specifies the leading edge of the first bin. The last element specifies the trailing edge of the last bin. The trailing edge is only included for the last bin."
So add on a number that's greater than the maximum element in your data. Inf is a good choice.
A = [-2 0 1 1 2 3 5 6 6 6 7 11 40 40]
A = 1×14
-2 0 1 1 2 3 5 6 6 6 7 11 40 40
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edges = [min(A):max(A) Inf];
[counts1, edges1] = histcounts(A, edges);
repeatedElements = edges1(counts1 >= 2)
repeatedElements = 1×3
1 6 40
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Or you could use non-equally spaced bins containing the unique elements from your data.
[counts2, edges2] = histcounts(A, [unique(A) Inf])
counts2 = 1×10
1 1 2 1 1 1 3 1 1 2
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<mw-icon class=""></mw-icon>
edges2 = 1×11
-2 0 1 2 3 5 6 7 11 40 Inf
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repeatedElements = edges2(counts2 >= 2)
repeatedElements = 1×3
1 6 40
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If your data spans a wide range this can reduce the number of bins histcounts uses.
whos counts1 edges1 counts2 edges2
Name Size Bytes Class Attributes counts1 1x43 344 double counts2 1x10 80 double edges1 1x44 352 double edges2 1x11 88 double
The unique approach uses 10 bins, the non-unique approach uses 43. This is a fairly small difference for your sample A, but the impact is much larger if you have a distant outlier.
B = [A 5000]; % 5000 is far from the rest of the elements in A
edges = [min(B):max(B) Inf];
[counts1, edges1] = histcounts(B, edges);
[counts2, edges2] = histcounts(B, [unique(B) Inf]);
whos counts1 edges1 counts2 edges2
Name Size Bytes Class Attributes counts1 1x5003 40024 double counts2 1x11 88 double edges1 1x5004 40032 double edges2 1x12 96 double
Walter Roberson
Walter Roberson 2025-3-10，16:57
@Steven Lord
Using edges = [min(B):max(B) Inf]; assumes that the input data is integer.

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Adam
Adam 2017-4-21
编辑：Adam 2017-4-21
[~, uniqueIdx] = unique( A );
duplicateLocations = ismember( A, find( A( setdiff( 1:numel(A), uniqueIdx ) ) ) );
then
find( duplicateLocations )
will give you the indices if you want them rather than a logical vector.
There are probably neater methods though.
If you want only the duplicates after the first then simply
setdiff( 1:numel(A), uniqueIdx )
should do the job.
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CompViscount
CompViscount 2022-9-20
编辑：CompViscount 2022-9-20
Ran in:
Commenting here as it's led me to overall the best answer here, it just has a mistake. The "find" in the 2nd line changes the values into indices before passing to ismember, which just makes the output nonsense. I removed that. Using the same numbers as image analyst above:
A=[ 1 1 2 3 5 6 6 7]
A = 1×8
1 1 2 3 5 6 6 7
[~, uniqueIdx] = unique(A);
dupeIdx = ismember( A, A( setdiff( 1:numel(A), uniqueIdx ) ) );
dupes = A(dupeIdx)
dupes = 1×4
1 1 6 6
dupeLoc = find(dupeIdx)
dupeLoc = 1×4
1 2 6 7

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Jan
Jan 2018-5-12
编辑：Jan 2021-7-2
function Ind = IndexOfMultiples(A)
T = true(size(A));
off = false;
A = A(:);
for iA = 1:numel(A)
if T(iA) % if not switched already
d = (A(iA) == A);
if sum(d) > 1 % More than 1 occurrence found
T(d) = off; % switch all occurrences
end
end
end
Ind = find(~T);
end
If the input has more than 45 elements, this is faster:
function T = isMultiple(A)
% T = isMultiple(A)
% INPUT: A: Numerical or CHAR array of any dimensions.
% OUTPUT: T: TRUE if element occurs multiple times anywhere in the array.
%
% Tested: Matlab 2009a, 2015b(32/64), 2016b, 2018b, Win7/10
% Author: Jan, Heidelberg, (C) 2021
% License: CC BY-SA 3.0, see: creativecommons.org/licenses/by-sa/3.0/
T = false(size(A));
[S, idx] = sort(A(:).');
m = [false, diff(S) == 0];
if any(m) % Any equal elements found:
m(strfind(m, [false, true])) = true;
T(idx) = m; % Resort to original order
end
end
MRINAL BHAUMIK
MRINAL BHAUMIK 2021-6-28
编辑：Walter Roberson 2025-4-3，20:25
Ran in:
A=[ 1 1 2 3 5 6 7 6]
A = 1×8
1 1 2 3 5 6 7 6
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
B = A'./A
B = 8×8
1.0000 1.0000 0.5000 0.3333 0.2000 0.1667 0.1429 0.1667 1.0000 1.0000 0.5000 0.3333 0.2000 0.1667 0.1429 0.1667 2.0000 2.0000 1.0000 0.6667 0.4000 0.3333 0.2857 0.3333 3.0000 3.0000 1.5000 1.0000 0.6000 0.5000 0.4286 0.5000 5.0000 5.0000 2.5000 1.6667 1.0000 0.8333 0.7143 0.8333 6.0000 6.0000 3.0000 2.0000 1.2000 1.0000 0.8571 1.0000 7.0000 7.0000 3.5000 2.3333 1.4000 1.1667 1.0000 1.1667 6.0000 6.0000 3.0000 2.0000 1.2000 1.0000 0.8571 1.0000
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
B = B-diag(diag(B))
B = 8×8
0 1.0000 0.5000 0.3333 0.2000 0.1667 0.1429 0.1667 1.0000 0 0.5000 0.3333 0.2000 0.1667 0.1429 0.1667 2.0000 2.0000 0 0.6667 0.4000 0.3333 0.2857 0.3333 3.0000 3.0000 1.5000 0 0.6000 0.5000 0.4286 0.5000 5.0000 5.0000 2.5000 1.6667 0 0.8333 0.7143 0.8333 6.0000 6.0000 3.0000 2.0000 1.2000 0 0.8571 1.0000 7.0000 7.0000 3.5000 2.3333 1.4000 1.1667 0 1.1667 6.0000 6.0000 3.0000 2.0000 1.2000 1.0000 0.8571 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[pos1 pos2]=find(B==1)
pos1 = 4×1
2 1 8 6
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
pos2 = 4×1
1 2 6 8
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Anamika
Anamika 2023-7-17
Ran in:
In MATLAB, you can find the indices of duplicate values in an array using the `find` function along with the `unique` function. Here's how you can do it:
A = [1 1 2 3 5 6 7];
% Finding the unique elements in the array
unique_elements = unique(A);
% Initializing an empty array to store the indices of duplicate values
duplicate_indices = [];
% Iterating through each unique element
for i = 1:numel(unique_elements)
% Finding the indices of occurrences of the current unique element
indices = find(A == unique_elements(i));
% If there are more than one occurrence, add the indices to the duplicate_indices array
if numel(indices) > 1
duplicate_indices = [duplicate_indices indices];
end
end
% Displaying the indices of duplicate values
disp(duplicate_indices);
1 2
Running this code will give you the indices of the duplicate values in the array A. In this case, the output will be: 1 2
This means that the duplicate values are located at indices 1 and 2 in the array A.
Eduardo Gonzalez Rodriguez
Here is my solution to find repeated values and their counts
function [dup, counts] = duplicates(A)
[dup,~,n] = unique(A, 'rows', 'stable');
counts = accumarray(n, 1, [], @sum);
dup(counts==1) = [];
counts(counts==1) = [];
Piotr
Piotr 2023-5-11
Hello,
here is my attempt to solve it. I faced similar problem but in my case I wanted to have the result in two column representation. Each row contains indices of repeated values.
A = [ 1 1 2 3 5 6 7 6];
nk = nchoosek(1:length(A),2);
nk(diff(A(nk),[],2)~=0,:) = [];
disp(nk)
Cheers, Piotr
Tim
Tim 2025-4-3，18:27
编辑：Tim 2025-4-3，18:28
Ran in:
Posting @CompViscount's comment as an answer because it gives a logical array identifying which elements have more than one entry in the array. This is the answer I need.
https://www.mathworks.com/matlabcentral/answers/336500-finding-the-indices-of-duplicate-values-in-one-array#comment_2372095
A=[ 1 1 2 3 5 6 6 7]
A = 1×8
1 1 2 3 5 6 6 7
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~, uniqueIdx] = unique(A);
dupeIdx = ismember( A, A( setdiff( 1:numel(A), uniqueIdx ) ) )
dupeIdx = 1x8 logical array
1 1 0 0 0 1 1 0
A = [-2 0 1 1 2 3 6 5 6 6 6 7 11 40 40]
A = 1×15
-2 0 1 1 2 3 6 5 6 6 6 7 11 40 40
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[~, uniqueIdx] = unique(A);
dupeIdx = ismember( A, A( setdiff( 1:numel(A), uniqueIdx ) ) )
dupeIdx = 1x15 logical array
0 0 1 1 0 0 1 0 1 1 1 0 0 1 1

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