Select a point on the graph
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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?
2 个评论
Adam
2019-10-9
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
2019-10-9
采纳的回答
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
8 个评论
luca
2019-10-9
Adam Danz
2019-10-9
Simply exchange min() with max().
[~, maxIdx] = max(d);
Adam
2019-10-9
Change min to max...
luca
2019-10-9
Adam Danz
2019-10-9
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
2019-10-9
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
2019-10-9
更多回答（1 个）
Turlough Hughes
2019-10-9
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.
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