Select a point on the graph

Hi given the following code,

figure(1);
scatter(x(:,1),x(:,2));
hold on;
scatter(member_value(:,1),member_value(:,2),'r');
legend({'Data','Pareto Frontier'})

I obtain a graph like this

And I want to select the red point that is closest to the origin.

May someone help me with the code?

[Adam]

What do you mean by 'select' it? You can click on it and select it if you wish, but that depends what you want to do having 'selected' it.

If you mean programmatically find it then isn't it just a simple case of pythagoras, having subtracted your origin from all points? (Or some built-in distance function that does the maths for you anyway)

[luca]

Yes the idea is to use something like pythagoras, that able me to find the point with the minimum distance from the origin. But do you know how can I implement it?

Answer 1

[Adam Danz]

To find the coordinate closest to the origin (0,0),

d = hypot(member_value(:,1),member_value(:,2)); 
[~, minIdx] = min(d); 

plot(member_value(minIdx,1),member_value(minIdx,2),'ks','MarkerSize',12);

hypot() method avoids potential under/overflow: https://www.mathworks.com/help/matlab/ref/hypot.html

[luca]

Due to the fact that the red point are the one farther from (0,0)

in case I want to find the fartermost point, how can I modify the code?

[Adam Danz]

Simply exchange min() with max().

[~, maxIdx] = max(d); 

[Adam]

Change min to max...

[luca]

thanks Adam

[Adam Danz]

Thanks Adams. :D

@luca, the hypot() function merely computes the hypotenuse ('h') of a right triangle with sides 'a' and 'b'. All of your points form a right triangle if you imagine a line going straight down to the x axis and across to the y axis. The line from (0,0) to each of your points in the hypotenuse.

[luca]

I don't get what you mean Adam

[Adam Danz]

Let's say you have 4 coordinates (black dots, below).

Each of those coordinates forms a right-triangle where 1 side of the triangle is the 'x-coordinate and the other side of the triangle is the y-coordinate. The hypotenuse of the triangle is the line that extends from (0,0) to the dot.

hypot() inputs are the x and y coordinates which are the sides of each triangle. It outputs the length of the hypotenuse which is the distance of the dot from (0,0). The hypotenuse are the diagonal lines in the image below.

Hypotenuse demo code for the image above:

clf()
x = rand(4,2).*8+1;
scatter(x(:,1),x(:,2));
hold on;
axis equal
xlim([0,9])
ylim([0,9])
grid on

leg = plot(x(:,[1,1])', [x(:,2),zeros(size(x,1),1)]', '-','LineWidth',2);
hyp = plot([x(:,1),zeros(size(x,1),1)]', [x(:,2),zeros(size(x,1),1)]', '-','LineWidth',2);
set(hyp,{'Color'},{leg.Color}')


d = hypot(x(:,1),x(:,2)); 
labels = strsplit(sprintf('h=%.1f\n',d'));
t = text(x(:,1)-.2,x(:,2),labels(1:end-1),'HorizontalAlignment','right');

[luca]

Thanks Adam ! now its clear how it works

Answer 2

[Turlough Hughes]

You can do the following:

[~,ind]=min(sqrt(member_value(:,1).^2+member_value(:,2).^2)); %find index for point closest to origin
hold on; plot(member_value(ind,1),member_value(ind,2),'.k');

Note, that if x was arranged as a row vector this will not work, but this is not the case for you.