Ordering the closest points to a certain point

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Zach
Zach 2011 年 6 月 27 日
I have a randomly generated set of k points on a graph. I'm trying to order each points distance in reference to the other. Each point is labelled 1 to n with no particular order. I want each of the n points to create a list that has the x and y coordinates of the other points all in increasing order of distance. Here's what I tried so far but it's not working for me.
for n=1:k
x(1,n) = 50.*rand(1,1)+25;
y(1,n) = 50.*rand(1,1)+25;
end
for n=1:k;
for i=1:k;
if n~=i
d(n,i)= sqrt(((x(1,i)-x(1,n))^2)+(y(1,i)-y(1,n))^2);
else
d(n,i)=NaN
end
end
sort(d(n,:));
end
for n=1:k;
for i=1:k;
for j=1:k;
if (sqrt(((x(1,i)-x(1,n))^2)+(y(1,i)-y(1,n))^2)) == d(n,j)
x(n,j)=x(1,i);
y(n,j)=y(1,i);
end
end
end
end
So the idea is that x(n,1) would be the x coordinate of the closest point to the nth point.

採用された回答

Walter Roberson
Walter Roberson 2011 年 6 月 27 日
Sounds like a nearest-neighbour computation.
x = 50.*rand(1,k)+25;
y = 50.*rand(1,k)+25;
dists = cell2mat(arrayfun(@(K) bsxfun(@(J,K) hypot(x(J)-x(K),y(J)-y(K)), K, 1:k), 1:k, 'Uniform',0).');
[sorteddists, rowidx] = sort(dists,2);
This does need some enhancement, though: before doing the sort, you should remove the diagonal, as points are always distance 0 from themselves. You could sort and then remove the first column, but there is a danger in doing so: if any of the points are in exactly the same place, then because the sort is "stable", there will be a row in which the point that is in the same place sorts before the point to itself. Replacing the diagonal with -inf before doing the sort would probably work.
  3 件のコメント
Zach
Zach 2011 年 6 月 27 日
I see that you sorted the distances and created and index for the distances. If you could get a way to create a list so that the x and y coordinates are arranged by this distance index. So that the I can get the x and y coordinate of the closest element to another element even though that might not be the closest x and y coordinate to that element if that makes any sense.
Walter Roberson
Walter Roberson 2011 年 6 月 28 日
Closest to [x(K) y(K)] is
J = rowidx(K,1);
[x(J) y(J)]

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