Asked by Clémentine OU
on 30 Aug 2017

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:

- get one row from L in a loop;
- extract the first column and the last column of this row;
- compare the extracted number with the related location in every row of K;
- If the numbers are matched, get the index from K.Can anyone help me how to do that, please?Thanks in advance.

Answer by José-Luis
on 30 Aug 2017

Accepted Answer

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.

Clémentine OU
on 31 Aug 2017

@ 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
on 31 Aug 2017

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)

Please accept the answer that best solves your problem.

As an exercise, you could try to adapt it so you can specify the column indexes instead of hardcoding them.

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Answer by Jan
on 30 Aug 2017

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);

Clémentine OU
on 31 Aug 2017

@Image Analyst and @Jan Simon: Thank you for your suggestion. Actually I just simply my problem. But I should post my original problem. I have two matrices 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 want to check that some elements in 1st, 2nd and 5th columns of every row in L are in which locations in K. Take an example of 1st row of L. I want to locate '1 4 1'(in 1st, 2nd and 5th columns of 1st row)in L is in which line of P. Obviously the answer is 1st,3rd,11th. As for elements in 2nd row of L,'2 4 1', the answers in K are 7th,12th and 13th rows.Just like it. I want to the exact locations of every row of L in P.

@KL: I am a newbie. I don't understand what you mean 'OP' and 'homework'. Some terminologies?

Jan
on 31 Aug 2017

"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
on 1 Sep 2017

@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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## 7 Comments

## José-Luis (view profile)

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## Jan (view profile)

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## Jan (view profile)

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