Extract duplicate values from both 1st and 2nd column

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F Z
F Z 2015 年 4 月 9 日
編集済み: Stephen23 2015 年 4 月 20 日
I have a large matrix (120000x4096)to process. There are repetitions in the 1st and 2nd columns. If 2 rows have the same number in the 1st and the 2nd column, only one row is stored. However, if 2 rows have the same number in the 1st columns but different numbers in the 2nd one, both rows are stored. Then extract the other columns related to each set of rows. This can seen in the attached file
Any ideas to do it without a loop please?
Thanks for your help!

採用された回答

Stephen23
Stephen23 2015 年 4 月 9 日
編集済み: Stephen23 2015 年 4 月 20 日
Removing duplicate rows, using fully vectorized code, is fast and easy with unique and its 'rows' and 'stable' options:
>> a = [10,10,10;10,10,10;5,5,5;4,4,4]
a =
10 10 10
10 10 10
5 5 5
4 4 4
>> [~,x] = unique(a(:,1:2),'rows','stable');
>> a(x,:)
ans =
10 10 10
5 5 5
4 4 4
Note how indexing is used to select only the first two columns of a as an input to unique, thus the third column is ignored, as the original questions requests that only the first two columns should be compared.
We can also test it on your test data (although putting it in a PDF is a complete pain: please provide text data instead, either within the question itself, or uploaded as a text file):
>> A = [1,2,120; 1,2,120; 1,3,112; 3,5,100; 3,5,100; 3,6,113; 4,2,106; 4,2,106; 4,6,88; 6,11,0; 6,11,0; 6,12,97; 6,12,97]
A =
1 2 120
1 2 120
1 3 112
3 5 100
3 5 100
3 6 113
4 2 106
4 2 106
4 6 88
6 11 0
6 11 0
6 12 97
6 12 97
>> [~,X] = unique(A(:,1:2),'rows','stable');
>> A(X,:)
ans =
1 2 120
1 3 112
3 5 100
3 6 113
4 2 106
4 6 88
6 11 0
6 12 97
Which exactly matches your required output (did I mention that putting the test-case in a PDF is a pain? Please provide plain text instead!)
  3 件のコメント
Image Analyst
Image Analyst 2015 年 4 月 9 日
Note, this is valid only for integers. Maybe you posted a simplified case - I don't know - but if you ever need to find matches for floating point numbers, see the FAQ.
F Z
F Z 2015 年 4 月 15 日
thank you a lot for your help!

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その他の回答 (1 件)

Thomas Koelen
Thomas Koelen 2015 年 4 月 9 日
編集済み: Thomas Koelen 2015 年 4 月 9 日
a=[10 10 10; 10 10 10; 5 5 5; 4 4 4];
a =
10 10 10
10 10 10
5 5 5
4 4 4
Code:
a=[10 10 10; 10 10 10; 5 5 5; 4 4 4; 4 4 4];
sizea=size(a);
for i=1:sizea(1,1)
for i2=1:sizea(1,2)
sizea=size(a);
if a(i,i2)==a(i+1,i2)
a(i+1,:)=[];
if i==sizea(1,1)-1
return
end
end
end
end
a =
10 10 10
5 5 5
4 4 4
Where a is your scan matrix!
  1 件のコメント
Thomas Koelen
Thomas Koelen 2015 年 4 月 9 日
Made the code a bit nicer since it was hard to read.

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