How to find the closest value pair in a matrix?

I have two numbers 0.85 and 1.25. I would like to find the closest pair of columns that contains both these values. I.e. A = [0.84 1.26; 0.90 1.24; 0.83 1.30]; and intersted pair is [0.85 1.25]. The closest pair from [0.85 1.25] is [0.84 1.26]. The difference is [-0.01 0.01], while for other pairs the difference is [0.04 -0.01] and [0.02 0.04]. So is there any way, I can make trial and error and find the closest pair of columns that has closest values? I tried following code, but it is not sufficient.

A = [0.84 1.26; 0.90 1.24; 0.83 1.30];
Pairs = [0.85 1.25]; 
for i = 1:size(Pairs,1)
    loc = ismember(A,Pairs(i,:));
    index(i,:)=find(sum(loc,2)==numel(Pairs(i,:)))';
end

Any help will be greatly appriciated. Thank you so much in advance!

[Torsten]

Is it given that A has only two columns and that the numbers in A and Pairs are in ascending order ?

[Megha]

Yes, we can consider that A has only two columns but Pairs may not be in ascending order always..@Torsten

Answer 1

[John D'Errico]

Assuming that you have a much larger problem, and that you have multiple points, to test, knnsearch is the correct tool.

A = [0.84 1.26; 0.90 1.24; 0.83 1.30];

B = [0.85 1.25;.80 1.31];
IDX = knnsearch(A,B)

IDX = 2×1

1 3

So point number 1 in A is the closest to B(1,:), and point 3 in A is the closest to B(2,:).

Note that knnsearch will be much faster than exhaustively testing the points to see which is closest.

A = rand(1e6,2);
B = rand(1e5,2);
tic, IDX = knnsearch(A,B);toc

Elapsed time is 1.404760 seconds.

So only a little over a second to identify the nearest point in A for all of the 100000 points in B.

Answer 2

[Voss]

A = [0.84 1.26; 0.90 1.24; 0.83 1.30];
Pairs = [0.85 1.25];

[~,idx] = min((Pairs(:,1).'-A(:,1)).^2 + (Pairs(:,2).'-A(:,2)).^2, [], 1);
closest = A(idx,:)

closest = 1×2

0.8400 1.2600

Note that the above method also works when you have multiple rows in Pairs:

% 3 points to find the closest point in A to:
Pairs = [0.85 1.25; 0.9 1.1; 0.95 1.4];

[~,idx] = min((Pairs(:,1).'-A(:,1)).^2 + (Pairs(:,2).'-A(:,2)).^2, [], 1);
closest = A(idx,:)

closest = 3×2

0.8400 1.2600 0.9000 1.2400 0.8300 1.3000

[Steven Lord]

You can simplify and generalize the computation of the data you pass into your min call using vecnorm.

A = [0.84 1.26; 0.90 1.24; 0.83 1.30];
Pairs = [0.85 1.25];

distance1 = (Pairs(:,1).'-A(:,1)).^2 + (Pairs(:,2).'-A(:,2)).^2

distance1 = 3×1

0.0002 0.0026 0.0029

distance2 = vecnorm(Pairs-A, 2, 2).^2 % 2-norm squared along dimension 2

distance2 = 3×1

0.0002 0.0026 0.0029

This also lets you experiment with other norms.

distance3 = vecnorm(Pairs-A, 1, 2).^2 % 1-norm squared

distance3 = 3×1

0.0004 0.0036 0.0049

distance4 = vecnorm(Pairs-A, Inf, 2).^2 % Inf-norm squared

distance4 = 3×1

0.0001 0.0025 0.0025

Answer 3

[Torsten]

A = [0.84 1.26; 0.90 1.24; 0.83 1.30];
Pairs = [0.85 1.25];
Ah     = sort(A, 2, 'ascend')

Ah = 3×2

0.8400 1.2600 0.9000 1.2400 0.8300 1.3000

Pairsh = sort(Pairs,'ascend')

Pairsh = 1×2

0.8500 1.2500

[~,idx] = min(sum(abs(Ah-Pairsh),2))

idx = 1

A(idx,:)

ans = 1×2

0.8400 1.2600

Answer 4

[Bruno Luong]

For A that has at least 3 rows and the closest in the sense of l2 norm (auclidian distance), without the need of toolbox.

But if you have the toolbox @John D'Errico solution is faster.

A = [0.84 1.26; 0.90 1.24; 0.83 1.30]

A = 3×2

0.8400 1.2600 0.9000 1.2400 0.8300 1.3000

Pairs = [0.85 1.25]

Pairs = 1×2

0.8500 1.2500

T = delaunayTriangulation(A)

T = 
  delaunayTriangulation with properties:

              Points: [3×2 double]
    ConnectivityList: [2 3 1]
         Constraints: []

T.Points(T.nearestNeighbor(Pairs),:)

ans = 1×2

0.8400 1.2600

It can also handle a very large number of points

A = rand(1e6,2);
B = rand(1e5,2);

tic
T = delaunayTriangulation(A);
T.nearestNeighbor(B);
toc

Elapsed time is 4.397585 seconds.

[Megha]

Thank you @Torsten, @John D'Errico, @Voss, @Steven Lord for giving your valuable time. I agree that other methods works too but knnsearch was the quick and easy one for me to adapt! I appriciate your efforts.