CODEWITHSUNDEEP
matlabmatrixfind
Carly Coop
Carly Coop

Reputation: 47

Finding where a set of values lie within a matrix

I have two values (k and j) which I know are within an nx3 matrix (M). I know that they're and on the same row and that j is always to the right of k, so if k is in M(2,1), then j will be in M(2,2). I tested for this earlier in the function, but now I want to know which row that is for a given k and j. I need the row number of their location to proceed. There are no duplicate combinations of k and j in the matrix.

So if I have the matrix

M= 

1 4 5
1 5 7
k j 5
4 5 6
2 3 1

Then I want to know that they're in row 3. None of the columns are ordered.

What I have tried:

I used the code below

[row,~] = find(M==k);

I'm not sure how to look for a combination of them. I want to avoid using the find function. I hope to potentially use logical indexing.

How do I go about doing this? I hope this question makes sense.

Upvotes: 4

Views: 104

Answers (3)

Nishant
Nishant

Reputation: 2619

Use this:

row = find(((M(:,1) == k ) & ( M(:,2) == j)) | ((M(:,1) == k ) & ( M(:,3) == j)) | ((M(:,2) == k ) & ( M(:,3) == j)) )

Also, logical indexing can only give you a matrix with zeros at all other positions and one at your required position. But to get the index of that position, you will have to use find.

Upvotes: 1

kkuilla
kkuilla

Reputation: 2256

Slightly different version on bsxfun. This one doesn't limit the matrix to three columns.

find(sum(((bsxfun(@eq,M,j) + bsxfun(@eq,M,k)) .* M).' ) == j+k >0)

Case 1:

M = [
     1     4     5
     1     5     7
     6     7     1
     4     5     6
     2     3     1]
k=6;j=7; 

ans =  3

Case 2:

M=[
     1     4     5
     1     5     7
     1     6     7
     4     5     6
     2     3     1
     ];
k=6;j=7;

ans =  3

Upvotes: 1

Divakar
Divakar

Reputation: 221554

You can use bsxfun -

find(all(bsxfun(@eq,A(:,1:2),[k,j]),2) | all(bsxfun(@eq,A(:,2:3),[k,j]),2))

Being a relational operation with bsxfun, according to this post on benchmarked results, this should be pretty efficient.

Sample runs

Case #1 :

A =
     1     4     5
     1     5     7
     6     7     1
     4     5     6
     2     3     1
k =
     6
j =
     7
>> find(all(bsxfun(@eq,A(:,1:2),[k,j]),2) | all(bsxfun(@eq,A(:,2:3),[k,j]),2))
ans =
     3

Case #2 :

A =
     1     4     5
     1     5     7
     1     6     7
     4     5     6
     2     3     1
k =
     6
j =
     7
>> find(all(bsxfun(@eq,A(:,1:2),[k,j]),2) | all(bsxfun(@eq,A(:,2:3),[k,j]),2))
ans =
     3

Upvotes: 1

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