Finding where a set of values lie within a matrix

Question

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=

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.

You wrote, *"j is always to the right of k"*. But your matrix `M` negate this statement! — Sardar Usama, Aug 12 '16 at 10:51
"I'm not sure how to look for a combination of them." But if you **know** that "j is always to the right of k", then you don't need to find the combination. Or did you mean that *if* `j` is on the same rows as `k`, you know it will be (immediately?) to the right of `j`? Your specification is still unclear. — beaker, Aug 12 '16 at 17:45

Nishant · Answer 1 · 2016-08-12T14:01:06.087

1

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.

edited Aug 12 '16 at 14:01

answered Aug 12 '16 at 10:44

Nishant

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1

but this assumes that `j` is in the first column and `k` in the second. I think the problem is more general, so `j` can be in any column except the last one (otherwise there will be no room for `k`). In this case, since the matrix has 3 columns, `j` could be in the second one and `k` in the last one – Aug 12 '16 at 11:38
@SembeiNorimaki Thank you for pointing it out. Fixed it, not sure if it is optimal though. – Nishant Aug 12 '16 at 11:52
just remove the part where j is in column 1 and k in column 3, since I think is not a valid answer, the other two cases are ok. – Aug 12 '16 at 12:22

score 1 · Answer 2 · edited May 23 '17 at 12:08

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

score 1 · Answer 3 · answered Aug 12 '16 at 12:45

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

Finding where a set of values lie within a matrix

3 Answers3