How to effectively reorder data
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I have unordered coordinates and when i order it by doing th following,
contorno = readtable('HJ_00-01.xlsx');
contx = table2array(contorno(:,1));
conty = table2array(contorno(:,2));
cx = mean(contx);
cy = mean(conty);
a = atan2(conty - cy, contx - cx);
[~, order] = sort(a);
reorderedx = contx(order);
reorderedy = conty(order);
When I plot, this is what i get
plot(reorderedx, reorderedy)
How do I get a continuous "smoth" line plot without the spikes that are present in the figure.
What is the best way to reorder my data?
采纳的回答
Bruno Luong
2018-11-23
contorno = readtable('HJ_00-01.xlsx');
contx = table2array(contorno(:,1));
conty = table2array(contorno(:,2));
userConfig = struct('xy',[contx(:) conty(:)]);
resultStruct = tsp_nn(userConfig);
更多回答（2 个）
Cris LaPierre
2018-11-23
2 个投票
This is a non-trivial problem, as I'm sure you've discovered. I'm wondering if you can find something useful in this traveling salesman example problem.
Image Analyst
2018-11-23
0 个投票
One way is to sort things angulalry. Compute the center, then compute the angles of each point from the center, then use both outputs of sort(). Easy, but let us know if you can't do it.
4 个评论
Bruno Luong
2018-11-23
@IA: But he just did that, don't you see his code?
Image Analyst
2018-11-23
编辑：Image Analyst
2018-11-23
Honestly I didn't look at the code much. The plot does not look like the points are all going in the same angular direction. There is a lot of backtracking going on.
Bruno Luong
2018-11-23
编辑：Bruno Luong
2018-11-23
Of course it does, depending where the center is (I can see very well where it's located approximatively, up north there), which is in turn depend on the density of the point on the boundary.
In anycase this kind of method (angle sorting) bounds to fail for such complicated boundary.
Image Analyst
2018-11-23
Yes, when I ran the code, and plotted the centroid, it shows the centroid is not where you'd have thought at first. For what it's worth, I also tried the boundary() function. I agree the TSP solution seems to be best. I'm voting for that.
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