Finding Duplicate Values per Column

Greetings, suppose Column A has these values - 7 18 27 42 65 49 54 65 78 82 87 98
Is there a way to compare the values (row by row) and search for duplicates? (I'm using Matlab R2010b)I don't want the duplicated values to be removed.
Thanks.

 Akzeptierte Antwort

Jan
Jan am 22 Okt. 2011

10 Stimmen

A = [7 18 27 42 65 49 54 65 78 82 87 98];
[n, bin] = histc(A, unique(A));
multiple = find(n > 1);
index = find(ismember(bin, multiple));
Now the values A(index) appear mutliple times.

Weitere Antworten (4)

the cyclist
the cyclist am 22 Okt. 2011

6 Stimmen

Here's a slightly different way:
X = [1 2 3 4 5 5 5 1];
uniqueX = unique(X);
countOfX = hist(X,uniqueX);
indexToRepeatedValue = (countOfX~=1);
repeatedValues = uniqueX(indexToRepeatedValue)
numberOfAppearancesOfRepeatedValues = countOfX(indexToRepeatedValue)

4 Kommentare

Harold
Harold am 13 Apr. 2012
Is there a way that I can use this approach for a cell array. Suppose I have a 1x4 cell array and each cell contains a (x,y) co-ordinates. Can I somehow modify your solution to search for duplicate cells and get rid of them?
Yowh
Yowh am 19 Nov. 2012
Harold can you give an example of your cell array variable ?
Jan
Jan am 19 Nov. 2012
@Yowh: It is unlikely, that Harold listens to comments after 8 months.
Anurag Pujari
Anurag Pujari am 25 Mär. 2016
Bearbeitet: Anurag Pujari am 25 Mär. 2016
Accurate. What an excellent piece of code.

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Hannes Greim
Hannes Greim am 19 Jan. 2013

0 Stimmen

you can use "tabulate" for cell arrays.

1 Kommentar

Clemens
Clemens am 19 Mär. 2014
Bearbeitet: Clemens am 19 Mär. 2014
Working only for versions after 2012 with the intro of tables.

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Wesley Allen
Wesley Allen am 9 Feb. 2018
Bearbeitet: Wesley Allen am 9 Feb. 2018

0 Stimmen

Duplicate Finding with Tolerance
If you want to find duplicates with tolerances (e.g., for non-integers), I use the following:
A = [1.313;2.4;2.400000001;1.31299999999;2.25;2.25;2.25000000001;3.7];
TOL = 1e-5;
uniqueA = uniquetol(A,TOL);
duplicateBool = abs(repmat(A,size(uniqueA.'))-repmat(uniqueA.',size(A))) < max(abs(uniqueA))*TOL;
duplicateCount = sum(duplicateBool).';
Just like with the cyclist's answer, if you want to isolate only the values that have more than one instance:
iDuplicate = (duplicateCount ~= 1);
repeatedValues = uniqueA(iDuplicate);
numberOfAppearancesOfRepeatedValues = duplicateCount(iDuplicate);
repeatedBool = duplicateBool(:,iDuplicate);
Using the Results
The unique values are in uniqueA:
>> uniqueA
uniqueA =
1.3130
2.2500
2.4000
3.7000
The quantity of each unique value is in duplicateCount:
>> duplicateCount
duplicateCount =
2
3
2
1
To get the indices of A corresponding to the n-th unique value, uniqueA(n)
>> n = 2;
>> uniqueA(n)
ans =
2.2500
>> duplicateIndex = find(duplicateBool(:,n))
duplicateIndex =
5
6
7
Fernando Meo
Fernando Meo am 13 Aug. 2018

0 Stimmen

Here is another answer (a one liner)
If AA is a 2D matrix and you wish to find the rows which have a duplicate values in its columns,
RowsWhichHaveDuplicates = find(arrayfun(@(i (~isequal(length(unique(AA(i,:))),size(AA,2))), [1:size(AA,1)]));
Example
AA = [6 7 11 6; 7 11 4 8; 11 15 1 10; 15 4 14 12;
18 13 18 8; 12 13 18 1; 3 14 6 18];
>> RowsWhichHaveDuplicates = RowsWhichHaveDuplicates = find(arrayfun(@(i) (~isequal(length(unique(AA(i,:))),size(AA,2))), [1:size(AA,1)]))
RowsWhichHaveDuplicates =
1 5
If your values are real, then a tolerance can be set by using the matlab "round" function to the decimal places you wish to use.
AA = round(rand(10)*10,1); % First decimal place
AA =
6.0000 2.0000 0.4000 6.7000 9.4000 0.6000 8.3000 3.1000 1.0000 3.0000
9.1000 7.5000 0.6000 6.0000 0.7000 3.1000 0.3000 4.2000 9.0000 3.7000
2.5000 8.9000 5.0000 3.4000 7.2000 6.6000 8.4000 9.3000 9.0000 7.6000
8.6000 1.0000 4.1000 4.0000 8.3000 4.6000 2.6000 0.6000 0.8000 3.1000
7.6000 5.2000 2.2000 3.9000 7.3000 0.2000 6.6000 8.2000 5.2000 9.6000
2.2000 6.0000 4.3000 7.0000 5.1000 6.9000 6.7000 6.4000 2.8000 2.1000
4.2000 9.8000 9.5000 1.4000 5.2000 4.1000 2.6000 8.2000 8.8000 7.3000
1.3000 6.7000 2.0000 3.8000 7.6000 5.7000 3.3000 3.3000 6.7000 2.5000
9.2000 8.5000 7.1000 2.2000 6.3000 9.9000 2.5000 9.5000 1.2000 8.9000
2.9000 1.7000 7.8000 4.1000 0.7000 8.6000 7.1000 9.1000 3.7000 7.1000
RowsWhichHaveDuplicates = find(arrayfun(@(i) (~isequal(length(unique(AA(i,:))),size(AA,2))), [1:size(AA,1)]))
RowsWhichHaveDuplicates =
5 8 10
Hope this helps

Tags

Gefragt:

RDG
am 22 Okt. 2011

Beantwortet:

am 13 Aug. 2018

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