如何逆时针订购积分 [英] How to order points anti clockwise

问题描述

让这些得分.

pt={{-4.65371,0.1},{-4.68489,0.103169},{-4.78341,0.104834},{-4.83897,0.100757},
{-4.92102,0.0949725},{-4.93456,0.100181},{-4.89166,0.122666},{-4.78298,0.129514}, 
{-4.72723,0.121442},{-4.68355,0.11023},{-4.65371,0.1},{-4.66924,0.10173}, 
{-4.93059,0.0966989},{-4.93259,0.105094},{-4.91074,0.116966},{-4.90635,0.094878}, 
{-4.66846,0.105327},{-4.92647,0.0956182},{-4.93433,0.102498},{-4.9333,0.0982262},
{-4.66257,0.10102}};

现在它们处于特定顺序(对我来说是一种疾病！)，如果我们查看ListLinePLot

Now they are in certain order (for me is a disorder!) which can be seen if we look at the ListLinePLot

picUnorder=ListLinePlot[pt,Frame-> True,Mesh-> All,MeshStyle-> PointSize[Large]];
SeepicUnorder=ListLinePlot[pt,Frame-> True,Mesh-> All,MeshStyle-> 
PointSize[Large]]/.Line[rest_]:>{Arrowheads[Table[0.02,{i,0,1,.02}]],Arrow[rest]};
GraphicsGrid[{{picUnorder,SeepicUnorder}}]

但是我们需要像下面的图片一样订购它们.

But we need to order them like the picture below.

有人对逆时针方向对此类2D点进行排序的算法提出任何建议，以便我们可以通过在重新排列的点上使用ListLinePlot来重新排列点列表以创建类似上图所示的几何图形? ?

Does anybody has some suggestion for a algorithm to sort such 2D points in counter clockwise direction so that we can rearrange the list of points to create a geometry like the last pic just by using ListLinePlot on the rearranged points????

使用建议，我们将获得以下内容.

Using the suggestion we get something like the following.

center=Mean[pt];
pts=SortBy[pt,Function[p,{x,y}=p-center;ArcTan[x,y]]];
Show[ListPlot[pt],ListLinePlot[pts,Mesh-> All,MeshStyle->
PointSize[Large]],Frame-> True]

BR

推荐答案

我刚刚阅读了

I've just read in a comment to nikie's answer that what you really want is the algorithm for an airfoil. So, I am posting another (unrelated) answer to this problem:

似乎比一般问题更容易，因为它几乎是凸的".我认为以下算法可降低FindShortestTour本质上在急性顶点处所具有的风险:

Seems easier than the general problem, because it is "almost convex". I think the following algorithm reduce the risks that FindShortestTour inherently has at the acute vertex:

1. 找到ConvexHull(占上表面和攻击面)
2. 从集合中删除凸包中的点
3. 用剩余的点执行FindShortestTour
4. 在最近的端点处连接两条曲线
5. Voilà

1. Find the ConvexHull (that accounts for the upper and attack surfaces)
2. Remove from the set the points in the convex hull
3. Perform a FindShortestTour with the remaining points
4. Join both curves at the nearest endpoints
5. Voilà

赞:

pt1 = Union@pt;
<< ComputationalGeometry`
convexhull = ConvexHull[pt1, AllPoints -> True];
pt2 = pt1[[convexhull]];
pt3 = Complement[pt1, pt2];
pt4 = pt3[[(FindShortestTour@pt3)[[2]]]];
If[Norm[Last@pt4 - First@pt2] > Norm[Last@pt4 - Last@pt2], pt4 = Reverse@pt4];
pt5 = Join[pt4, pt2, {pt4[[1]]}];
Graphics[{Arrowheads[.02], Arrow@Partition[pt5, 2, 1], 
          Red, PointSize[Medium], Point@pt1}]

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