Finding the indices of duplicate values in one array
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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(8 个)
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
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.
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.
5 个评论
Arthur Souza
2018-5-12
I have the 2013a version. and... IT WORKED! Thank you so much Image Analyst! I think my problem is solved now! Have a nice weekend!
Tyann Hardyn
2022-1-21
You save my life (indirectly) again, Mr Image Analyst. Thank you so much. You helped someone else, then your help will be a good answer for the others, like me, lol.
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.
9 个评论
CompViscount
2022-9-20
编辑:CompViscount
2022-9-20
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]
[~, uniqueIdx] = unique(A);
dupeIdx = ismember( A, A( setdiff( 1:numel(A), uniqueIdx ) ) );
dupes = A(dupeIdx)
dupeLoc = find(dupeIdx)
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
2 个评论
MRINAL BHAUMIK
2021-6-28
A=[ 1 1 2 3 5 6 7 6]
B = A'./A
B = B-diag(diag(B))
[pos1 pos2]=find(B==1)
o/p
pos1 =
2
1
8
6
0 个评论
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
0 个评论
Eduardo Gonzalez Rodriguez
2023-7-13
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) = [];
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Anamika
2023-7-17
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);
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.
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