hi , i have two graph one is fixed and the other one is changing (64 graphe )i want to find the most closest graphe to the fixed one .please help??

hi , i have two graph one is fixed and the other one is changing i want to find the most close graphe to the fixed one .please help??

[Bob Thompson]

What determines 'close'?

[nihed sboui]

my goal is to find the curve(from 64 curves) that matches my fixed curve

Answer 1

[Walter Roberson]

One approach is to use https://www.mathworks.com/help/stats/knnsearch.html knnsearch . That would get you a list of the closest point in the fixed to each point in the varying, and you could then find the minimum distance from there.

Another approach, assuming Euclidean distance:

dsq = bsxfun(@minus, xvarying(:), xfixed(:).').^2 + bsxfun(@minus, yvarying(:), yfixed(:).').^2;
mindsq = min(dsq(:));
[varying_idx, fixed_idx] = find(dsq == mindsq);

Then it is the case that for each element in varying_idx, x(varying_idx(K)), y(varying_idx(K)) is the closest point to x(fixed_idx(K)), y(fixed_idx(K))

The reason why there can be multiple elements in varying_idx is that it is possible that there are two different points that are exactly as far as each other to their closest corresponding points in the fixed curve.

[nihed sboui]

for i=1:64

idx=knnsearch(fixed,QE_C{i})

end

figure

plot(lambdainter,idx)

=>Undefined function 'knnsearch' for input arguments of type 'double'.

[nihed sboui]

acctually my goal is to find the curve(from 64 curves) that matches my fixed curve .

the idea is to find the min distance between the fixed curve and the 64 curves .

and finally to plot the most closest one to my fixed curve .thanks for help

[Walter Roberson]

knnsearch requires the statistics toolbox and a release from the last few years.

Your problem description is not clear. If I have one point on a curve that is quite close to the fixed curve but the other points are far away, then is that to be considered the closest? If I have a curve with widely spaced points that are a distance from the fixed curve, but interpolation between the points implies that the two cross, then in one sense the minimum distance between them is 0 which makes them as close as they can be, even if they are perpendicular and not much alike.

Do all of the curves have the same number of points? Are they in the same relative order with the same x, such that it could make sense to calculate sum((yvary-yfixed).^2) as the distance?

[nihed sboui]

@walter yes all the curves have the same number of points and they are relative order with the same x .

i have to find the one that matches the most the fixed curve .i think i have to find the curve that contain the maximum of number of points that are quite close of my fixed curve's points .

[Walter Roberson]

 All_residue = sum((all_varying_y - fixed_y).^2), 2);
 [~, best_curve_idx)] = min(All_residue)

This assumes that the fixed_y is a row vector and that all_varying_y is a 2d array with any one curve being along the row.

This algorithm does not count the number of points that are "quite close" the way you asked. Instead it calculates sum of squared distance. For example if you had a curve in which the first three y were quite far away but the rest were nearly identical to the fixed curve, then the number that are nearly the same would be high, but the three points that are wildly different would result in quite an overall difference.