How to find the closest point to a line

Hello,

I am having a bit of a hard time finding which point (i.e centroid) is closest to a line formed by the start and end coordinates. The lines are always in a straight line.

points = array2table([1,48,-55,48.1,-55.1; 2, 50,-55,50.1,-55.1], 'VariableNames', {'name', 'start_lat','start_long','end_lat','end_long'})

centroids = cell2table({'A', 47, -56; 'B', 51,-54}, 'VariableNames', {'name', 'centroid_lat','centroid_long'})

desired_output = cell2table({1,48,-55,48.1,-55.1, 'A'; 2, 50,-55,50.1,-55.1, 'B'}, 'VariableNames', {'name', 'start_lat','start_long','end_lat','end_long', 'closet_centroid'})

Thank you,

[Davide Masiello]

Obviously, the centroid of a line will be the closest point to that line because it lies on it.

What you want to do looks like just appending the centroid identifier ('A' and 'B') on an additional column of the first table containing the lines start and end coordinates.

If that is not the only thing you want to achieve, could you please clarify your objective?

[Blue]

I have edited the question to change the centroids positions. I simply want to associate each centroid to the closest line formed by the start and end positions of each point

[Davide Masiello]

How do you determine the distance between a line and a point?

Is the distance between the point and the line centroid ok?

Answer 1

[Walter Roberson]

See https://www.mathworks.com/matlabcentral/answers/95608-is-there-a-function-in-matlab-that-calculates-the-shortest-distance-from-a-point-to-a-line#answer_281258 -- and it looks like @Rik created https://www.mathworks.com/matlabcentral/fileexchange/64396-point-to-line-distance based on that.

[Blue]

Based on the posts suggested by Walter I would do this to solve my question:

points = array2table([1,48,-55,48.1,-55.1; 2, 50,-55,50.1,-55.1], 'VariableNames', {'name', 'start_lat','start_long','end_lat','end_long'});

centroids = cell2table({'A', 47, -56; 'B', 51,-54}, 'VariableNames', {'name', 'centroid_lat','centroid_long'});

for i = 1:height(points)

    v1 = [points.start_lat(i), points.start_long(i)];
    v2 = [points.end_lat(i), points.end_long(i)];

    if length(v1) == 2, v1(3) = 0; end    %extend 1x2 to 1x3 if needed
    if length(v2) == 2, v2(3) = 0; end    %extend 1x2 to 1x3 if needed

    for  j = 1:height(centroids)

        pt = [centroids.centroid_lat(j), centroids.centroid_long(j)];

        if size(pt, 2) == 2, pt(1, 3) = 0; end  %extend nx2 to nx3 if needed

        v1_ = repmat(v1, size(pt, 1), 1);
        v2_ = repmat(v2, size(pt, 1), 1);

        a = v1_ - v2_;

        b = pt - v2_;

        distance(j) = sqrt(sum(cross(a,b,2).^2,2)) ./ sqrt(sum(a.^2,2));
    end

    [~, idx] = min(distance);
    points.associated_centroid(i) = centroids.name(idx)

end

[Rik]

Since you know you only put in 2 points, why not create the vector including the 0? That way you also get to skip the test with length (which I wouldn't use anymore.

I also removed the repmat call, as you only provide a single point.

points = array2table([1,48,-55,48.1,-55.1; 2, 50,-55,50.1,-55.1],...
    'VariableNames', {'name', 'start_lat','start_long','end_lat','end_long'});
centroids = cell2table({'A', 47, -56; 'B', 51,-54},...
    'VariableNames', {'name', 'centroid_lat','centroid_long'});

for i = 1:height(points)
    v1 = [points.start_lat(i), points.start_long(i), 0];
    v2 = [points.end_lat(i)  , points.end_long(i)  , 0];

    for  j = 1:height(centroids)
        pt = [centroids.centroid_lat(j), centroids.centroid_long(j), 0];

        a = v1 - v2;
        b = pt - v2;

        distance(j) = sqrt(sum(cross(a,b,2).^2,2)) ./ sqrt(sum(a.^2,2));
    end

    [~, idx] = min(distance);
    points.associated_centroid(i) = centroids.name(idx);
end
points

points = 2×6 table
    name    start_lat    start_long    end_lat    end_long    associated_centroid
    ____    _________    __________    _______    ________    ___________________

     1         48           -55         48.1       -55.1             {'A'}       
     2         50           -55         50.1       -55.1             {'B'}       

[012786534]

I agree !

Answer 2

[Image Analyst]

Does my attached function help any?