Finding Indices of Duplicate Values
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Suppose, I have a variable,
a{1}=[
2 2 1 3
5 2 1 1
5 2 1 4
5 2 1 2
1 1 2 1
2 2 2 1
1 1 3 4
1 1 3 3
4 1 3 5
1 1 4 3
1 1 4 4
1 1 4 1
2 2 4 1
2 2 4 2
4 1 4 2
4 1 4 6
2 2 5 2
2 2 5 6
4 1 5 1
4 1 5 2
4 1 5 6
4 1 5 5]
How can I find the indices of duplicate values for column 3:4?
I am using Matlab R2013a. Thank you in advance.
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추가 답변 (2개)
B Mohandes
2016년 11월 10일
5 개 추천
by coincidence i found another way to do it, but thanks to those who answered already. the outputs of the command "unique" can be tricky to deal with, hence, here is a straight forward way.
>> find(hist(a,unique(a))>1)
the command (hist) counts the frequency (number of repetitions) of a certain value in a vector. if you use: hist(a), matlab will divide the whole range of values to 10 periods, and count the repetitions of values lying within these ranges. however, if you use: hist(a,b), then the repetitions are counted against the reference (b). so when you count the occurrences of each element in (a) against the unique elements of (a), and you find the results that are >2, then you're finding the elements that occurred more than once.
regards
댓글 수: 4
Chunli
2018년 11월 28일
But you can work like this:
[C,ia,ib]=unique(a,'rows','stable');
find(hist(ib,unique(ib))>1)
Yavor Kamer
2019년 9월 10일
i think you can use 1:max(ib) in your second unique
Matt Fetterman
2020년 3월 2일
I think a has to be sorted for this to work properly
If I understand correctly, you want the row indices where the pair a{1)}(3,4) are duplicated.
a{1}=[
2 2 1 3
5 2 1 1
5 2 1 4
5 2 1 2
1 1 2 1
2 2 2 1 % me!
1 1 3 4
1 1 3 3
4 1 3 5
1 1 4 3
1 1 4 4
1 1 4 1
2 2 4 1 % and me!
2 2 4 2
4 1 4 2 % and me!
4 1 4 6
2 2 5 2
2 2 5 6
4 1 5 1
4 1 5 2 % and me!
4 1 5 6 % and me!
4 1 5 5]
[~,ixu] = unique( a{1}(:,3:4), 'rows'); % gives indices of the first of the unique rows
ixd = setdiff( 1:size(a{1},1), ixu); % (duplicate rows) = (all rows) - (unique rows)
clear ixu
ixd
% show rows that are duplicates in columns 3:4
a{1}(ixd, :)
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