# Find different arrays in a matrix

조회 수: 2(최근 30일)
EldaEbrithil 20 Aug 2020
편집: EldaEbrithil 20 Aug 2020
Hi all
i have a matrix like that A=[1 2 3; 4 5 6; 7 8 9] i want the index of the rows that contain B=[1 3;4 5;7 9;4 6;2 3] so in this case the rows are:1, 2, 3, 2, 1
Thank you for the help
Regards
##### 댓글 수: 1표시숨기기 없음
EldaEbrithil 20 Aug 2020
I have tried
[r,c]=ismember(A,B,'rows')
but it gives me error due to the fact that B and A must have the same dimension

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### 채택된 답변

David Hill 20 Aug 2020
A=[1 2 3; 4 5 6; 7 8 9];
B=[1 3;4 5;7 9;4 6;2 3];
a=zeros(size(B,1),1);
for k=1:size(B,1)
a(k)=find(sum(ismember(A,B(k,:)),2));
end
##### 댓글 수: 1표시숨기기 없음
EldaEbrithil 20 Aug 2020
Hi David thank you for answer me
i also found a solition of this type:
A = [1 14 24;2 15 25;1 16 72;1 14 24] % A sample matrix for demonstration...
c=[14 24;15 25;14 24]
[rig,col]=size(c);
for i=1:rig
I(:,i) = double(ismember(A(:,2:3),c(i,:),'rows'));
end
[rig,colo]=ind2sub(size(I),find(I>0))
the only inconvenient is that, due to the fact that A contain repeated lines, I is a unitary matrix that contains multiple (in this case two) ones in the same column( in this case in the first and last columns)
1 0 1
0 1 0
0 0 0
1 0 1
i don't want that, i want a single one for each column: i could i remove the multiple one in the columns? Whether it is the first one in the first or last row makes no difference, the important thing is that there is only one per column

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R2019a

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