Split non-bijective function into two bijective regions
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I have a model which generates a large number of samples (i.e. a set of x values and a set of y values). The output roughly follows the shape of the curve pictured, so that there are two y values for each x value. The data is not sorted by x or y.
I would like to split this dataset into two bijective datasets, representing points above and below the marked X. (This will then allow me to interpolate between points on the upper portion of the curve, where otherwise interpolation would bounce up and down between the two parts of the curve.)
My initial approach was to find the co-ordinates of the marked X, xX and yX, and assign the two datasets according to y > yX and y < yX. However, there are a handful of points on the bottom portion of the curve that can sometimes have y > yX, so this doesn't work.
Can anyone suggest a potential solution to this problem?
4 个评论
Dyuman Joshi
2023-10-3
What is the criteria for defining the point X on the graph to be used for separation?
Bruno Luong
2023-10-3
Please share your data
Thanks for the comments.
Dyuman: The marked point X has an x co-ordinate which is the maximum value of x (i.e. max(x))
Bruno: This is not my actual data but is a simple version that has the essence of the problem. My approach would put the points (1,2) and (2.5,1.8) into the top half of the curve
x = [4;2;3;6;6.5;4.5;1;1.5;3.5;2.5;5;6.2;2.8;6.3];
y = [5.5;6.5;1;4;1.5;1;2;7;6;1.8;5;1;6.2;3];
scatter(x,y)
采纳的回答
x = [4;2;3;6;6.5;4.5;1;1.5;3.5;2.5;5;6.2;2.8;6.3];
y = [5.5;6.5;1;4;1.5;1;2;7;6;1.8;5;1;6.2;3];
% TSP to order the data, See here
% https://www.mathworks.com/help/optim/ug/travelling-salesman-problem.html
nStops = length(x);
idxs = nchoosek(1:nStops,2);
% Bruno comment: you might normalize x and y before comuting dist if they have different units
dist = hypot(x(idxs(:,1)) - x(idxs(:,2)), ...
y(idxs(:,1)) - y(idxs(:,2)));
lendist = length(dist);
G = graph(idxs(:,1),idxs(:,2));
Aeq = spalloc(nStops,length(idxs),nStops*(nStops-1)); % Allocate a sparse matrix
for ii = 1:nStops
whichIdxs = (idxs == ii); % Find the trips that include stop ii
whichIdxs = sparse(sum(whichIdxs,2)); % Include trips where ii is at either end
Aeq(ii,:) = whichIdxs'; % Include in the constraint matrix
end
beq = 2*ones(nStops,1);
intcon = 1:lendist;
lb = zeros(lendist,1);
ub = ones(lendist,1);
opts = optimoptions('intlinprog','Display','off');
[x_tsp,costopt,exitflag,output] = intlinprog(dist,intcon,[],[],Aeq,beq,lb,ub,opts);
x_tsp = logical(round(x_tsp));
Gsol = graph(idxs(x_tsp,1),idxs(x_tsp,2),[],numnodes(G));
%%
[~,minloc]=min(x);
[~,maxloc]=max(x);
p1 = Gsol.shortestpath(minloc,maxloc);
A=Gsol.adjacency;
A(p1(1),p1(2))=0;
A(p1(2),p1(1))=0;
Gtmp = graph(A);
p2 = Gtmp.shortestpath(minloc,maxloc);
p2(1) = [];
x1 = x(p1); y1 = y(p1);
x2 = x(p2); y2 = y(p2);
figure()
plot(x1,y1,'Linewidth',2)
hold on
plot(x2,y2,'Linewidth',2)
4 个评论
Dan
2023-10-3
Torsten
2023-10-3
For a simple solution, choose the point furthest to the right :-)
Bruno Luong
2023-10-3
The challenge is reorder the data, selecting the turning point is easy.
Torsten
2023-10-3
You are right (as always) - I didn't read the question carefully.
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