# Compare some parts from one row in two matrices

조회 수: 3 (최근 30일)
Clémentine OU 2017년 8월 30일
편집: Clémentine OU 2017년 9월 1일
Hello, I have two matrices, one is
K=[1 4 2 0 1;1 4 0 0 1;2 4 0 0 1;2 4 3 0 1;5 4 2 0 2];
the other is
L=[1 4 0 0 1; 2 4 0 0 1; 4 2 0 0 2].
I want to do somethings:
1. get one row from L in a loop;
2. extract the first column and the last column of this row;
3. compare the extracted number with the related location in every row of K;
4. If the numbers are matched, get the index from K.Can anyone help me how to do that, please?Thanks in advance.
##### 댓글 수: 7이전 댓글 5개 표시이전 댓글 5개 숨기기
Jan 2017년 8월 30일
Corresponding to what?
Clémentine OU 2017년 8월 30일
I have extracted the first column and the last column of every row in L. I want to compare whether they are matched with the first column and the last column of every row in K. Is it more clear?

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### 채택된 답변

José-Luis 2017년 8월 30일
K_sub = K(:,[1,end]);
L_sub = L(:,[1,end])
[ii,kk] = size(L_sub);
L_sub=reshape(L_sub',[1,kk,ii]);
result = squeeze(sum(bsxfun(@eq,K_sub,L_sub),2) == 2)
Where each column will be a logical index to where your condition is true.
##### 댓글 수: 2없음 표시없음 숨기기
Clémentine OU 2017년 8월 31일
편집: Clémentine OU 2017년 8월 31일
@ José-Luis :Thanks for your reply. I've tried your method which can solve my problem well. You use 'reshape', 'bsxfun' and 'squeeze' commands which are enlarged my knowledge. Could you help me to solve bigger dimension problem, please? Now I want to locate that the elements(in 1st, 2nd and 5th columns) of every row in L is in which line of P.
K =
1 4 2 0 1
4 3 0 0 1
1 4 0 0 1
3 4 0 0 2
1 4 3 0 2
4 1 0 0 2
2 4 0 0 1
3 4 0 0 1
4 1 0 0 1
4 2 0 0 1
1 4 3 0 1
2 4 3 0 1
2 4 1 0 1
3 4 2 0 1
3 4 1 0 1
L =
1 4 0 0 1
2 4 0 0 1
3 4 0 0 1
4 1 0 0 1
4 2 0 0 1
4 3 0 0 1
1 4 0 0 2
3 4 0 0 2
4 1 0 0 2
4 3 0 0 2
I tried the similar code as follows:
K_sub = K(:,[1,2,end])
L_sub = L(:,[1,2,end])
[ii,kk,mm] = size(L_sub)
L_sub=reshape(L_sub',[1,kk,ii,mm])
result = squeeze(sum(bsxfun(@eq,K_sub,L_sub),3) == 3)
why my result shows nothing?
result =
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
José-Luis 2017년 8월 31일
편집: José-Luis 2017년 8월 31일
You don't need mm. You still need to perform the sum along the second dimension.
K_sub = K(:,[1,2,end])
L_sub = L(:,[1,2,end])
[ii,kk] = size(L_sub)
L_sub=reshape(L_sub',[1,kk,ii])
result = squeeze(sum(bsxfun(@eq,K_sub,L_sub),2) == 3)
As an exercise, you could try to adapt it so you can specify the column indexes instead of hardcoding them.

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### 추가 답변 (1개)

Jan 2017년 8월 30일
K = [1 4 2 0 1;1 4 0 0 1;2 4 0 0 1;2 4 3 0 1;5 4 2 0 2];
L = [1 4 0 0 1; 2 4 0 0 1; 4 2 0 0 2];
[match, locK] = ismember(L(:, [1,end]), K(:, [1,end]), 'rows')
Result = locK(match);
##### 댓글 수: 9이전 댓글 7개 표시이전 댓글 7개 숨기기
Jan 2017년 8월 31일
"OP" is the "original poster", the person who has asked the question.
"Homework" is a question given to pupils or students to be solved at home. If a user posts a homework question, the forum tend to explain, who it can be solved by your own. Posting a working solution could cause troubles, because you submitting this solution would be cheating and you had missed the chance to learn Matlab.
It is strange, that you do not want to post the wanted output for the shown input, although we have asked repeatedly for this. This is not useful. What's wrong with provided the information, which would allow for solving the problem?
Following your explanations here, this is a job for ismember('rows') and I still think that ismember('rows') solves the problem directly. At least you did not explain, what the differences between the result and your wanted results is.
Clémentine OU 2017년 9월 1일
편집: Clémentine OU 2017년 9월 1일
@Jan Simon: Oh,thank you for your explanation. I'm a researcher. What I posted is just an issue in a part of my code. I tried to modify your code to show my meaning. Please see as follows:
Q1n=K(:, [1,2,end])
L1n=L(:, [1,2,end])
[nbL1n,~]=size(L1n);
[nbQ1n,~]=size(Q1n);
for i=1:nbL1n
l=L1n(i,:)
for j=1:nbQ1n
q=Q1n(j,:)
[match, locK] = ismember(l,q,'rows')
end
end
Then I could find the elements at 1st,2nd and 5th columns in every row of L whether they are included in P. But the problems coded in this way are that I can't get the exact indices from P. Because I want to extract all lines from P when the condition is true. Then I don't really understand why you add this syntax 'Result = locK(match)'. I can't see how it works. As for using 'ismember('rows')', you are right that it's a more direct way to solve my problem. But LocK can only get the lowest index in Q1n. I want to get all indices when match=1. I'm not sure that 'ismember('rows')' can make it true.

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