Relation of Elements of matrix

A = [1 2

2 6

2 3

2 7

3 4

5 4

6 3

4 7]

i have matrix A,

if i select B=2, i want output as 1 2 3 6 7, it shows relatiionship of 2

if i select B=3, i want output as 2 3 4 6, it shows relatiionship of 3

[KSSV]

What do you mean by relationship of 2? What is B?

Answer 1

[Walter Roberson]

A = [1   2
       2   6
       2   3
       2   7
       3   4
       5   4
       6   3
       4   7]

A = 8×2

1 2 2 6 2 3 2 7 3 4 5 4 6 3 4 7

B = 2

B = 2

unique([B; A(A(:,1) == B,2); A(A(:,2) == B,1)]).'

ans = 1×5

1 2 3 6 7

However, the basis for including the number itself is not clear. If B does not occur at all in A, then I would argue it would be more appropriate for the result to be empty than for it to include B itself.

[Paul Hoffrichter]

>> If B does not occur at all in A, then I would argue it would be more appropriate for the result to be empty than for it to include B itself.

I agree. Here r is empty and so is z if B does not occur in A.

[r, ~] = ind2sub( size(A), find(A==B) );
z = unique(  A(  sort(r) ,: )  )'

[Walter Roberson]

You do not need to sort r. It does not matter if you access A rows out-of-order, since the values are going to be out of order and you are doing unique() of them.

[Paul Hoffrichter]

Yes, This is better since unique does the sort. Thanks.

[r, ~] = ind2sub( size(A), find(A==B) );
z = unique(  A( r ,: )  )'

[Walter Roberson]

A = [1   2
       2   6
       2   3
       2   7
       3   4
       5   4
       6   3
       4   7];
B = 2;

unique(A(any(A == B,2),:)).'

ans = 1×5

1 2 3 6 7

[Paul Hoffrichter]

After adding a note that removing the sort was faster, I then started thinking that the two line solution might actually be faster than the more one-line solution. (Usually, I think of one-line solutions as not only more elegant, but also, usually also faster.)

Yet, the two line solution was 4-5 times faster than the one line solution. Not that it matters much since the time for either one is negligible.

B = 12345;
n = 1e8;
A = randi(n/2000,n,2);

tic
z1 = unique([B; A(A(:,1) == B,2); A(A(:,2) == B,1)]).';
toc

tic
[r, ~] = ind2sub( size(A), find(A==B) );
z2 = unique(  A( r ,: )  )';
toc

Elapsed time is 0.665452 seconds.
Elapsed time is 0.154779 seconds.

[Walter Roberson]

B = 12345;
n = 5e7;
A = randi(n/2000,n,2);

tic
z1 = unique([B; A(A(:,1) == B,2); A(A(:,2) == B,1)]).';
toc

Elapsed time is 0.759719 seconds.

tic
[r, ~] = ind2sub( size(A), find(A==B) );
z2 = unique(  A( r ,: )  )';
toc

Elapsed time is 0.251981 seconds.

tic
z3 = unique(A(any(A == B,2),:)).';
toc

Elapsed time is 0.306312 seconds.

Answer 2

[Paul Hoffrichter]

B = 3;
[r, ~] = ind2sub( size(A), find(A==B) );
z = A(sort(r),:);
zu = unique(z(:))'

[Paul Hoffrichter]

B = 2;
[r, ~] = ind2sub( size(A), find(A==B) );
z = unique(  A(  sort(r) ,: )  )'

[Paul Hoffrichter]

Don't need the sort, so this is faster:

[r, ~] = ind2sub( size(A), find(A==B) );
z = unique(  A( r ,: )  )'