How to find nearest two points each other ?
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Ender Rencuzogullari
il 29 Nov 2015
Hi Everyone,
I have
X = (Yt.*sin(W))-(Xt.*cos(W));
Y = (Yt.*cos(W))+(Xt.*sin(W)); which give the coordinates X and Y.
X_inv = R.*sin(B_involute);
Y_inv = R.*cos(B_involute); which give the coordinates X_inv and Y_inv.
I need to find the nearest two points between X,Y and X_inv,Y_inv.
lots of thanks from now.
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Image Analyst
il 29 Nov 2015
What is W? What is B_involute? Can you post a diagram or plot of this and point out X, Y, X_inv, Y_inv what two points on there you'd like to have it return - which are the "two points"? And it's ambiguous. What's nearest to what? Two points nearest to each other? Two points nearest to either X,Y or X_inv,Yinv? For example if you have 2,6,7,50, then 6 and 7 are nearest to each other but 6 and 7 are closer to 2 than they are to 50, so would you return 6 and 7, or return 2???
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Star Strider
il 29 Nov 2015
If you have the Statistics Toolbox, use the pdist2 function.
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Star Strider
il 29 Nov 2015
Modificato: Star Strider
il 29 Nov 2015
My pleasure!
You would simply use your ‘X’, ‘Y’, ‘X_inv’ and ‘Y_inv’ in my code, eliminating the lines I used to create them to test it.
You would use only:
for k1 = 1:length(X)
for k2 = 1:length(X_inv)
DE(k1,k2) = hypot(X(k1)-X_inv(k2), Y(k1)-Y_inv(k2)); % Euclidean Distance
end
end
[DEmin,ix] = min(DE(:));
[K1,K2] = ind2sub(size(DE),ix);
fprintf(1,'\nNearest Points:\n\tDistance = %.3f\n', DEmin)
fprintf(1,'\t\tX(%d), Y(%d) \t\t\t= %.2f, %.2f\n', K1, K1, X(K1), Y(K1))
fprintf(1,'\t\tX_inv(%d), Y_inv(%d) \t= %.2f, %.2f\n', K2, K2, X_inv(K2), Y_inv(K2))
Più risposte (2)
Image Analyst
il 29 Nov 2015
Use a nested for loop and the sqrt() function, then sort() and find() to find the 8 closest distances at the two points where your curves intersect.
Or maybe you could use roots(curve1-curve2).
Semion
il 14 Apr 2020
Hi. Could you explain, how does method "dsearchn" select an index of multi closest points with the same distance to target point? BW, the method "dnsearch" with and without triangulation produce different results.
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