首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >算法-按行分组地理点

算法-按行分组地理点
EN

Stack Overflow用户
提问于 2018-06-28 15:22:07
回答 1查看 60关注 0票数 1

我有一个果园的x,y坐标集合(以米为单位)。我试图自动对行进行分组,并从上到下对组(行)中的树进行编号(基于对顶部和底部的定义)。不幸的是,我还没有想出解决办法。有关图片和指向数据集的链接,请参见下面。

树的数字的一个例子是:

5-1

5-2

其中5是行号,1和2是行内的树数。

一排树之间的距离约为6米,行间的距离约为12米。这样,就可以用欧几里得距离来定义相邻树在7米以内的行。按y坐标组织数据不起作用,因为行不是直线。

为了使事情变得更复杂,行需要从左到右或从右到左依次排列。

我可以使用现有的算法吗?如果没有,我能做些什么来使这件事成功呢?我们会感谢你的指导!

数据:https://drive.google.com/file/d/1csLM4IpP3tMF0fqQkql6gIANHngX9A3c/view?usp=sharing

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2018-06-29 14:25:57

谢谢你的帮助。关于我的解决方案,见下文。这是非常混乱的,但我相信这个想法会实现的:

代码语言:javascript
复制
public class Group
{
    public int group;
    public int row;
    public double highestRelDistance;

    public Group(int _group)
    {
        group = _group;
    }

}

public class Tree
{
    public string name;
    public Group group = new Group(0);
    public int orderInGroup;
    public double x;
    public double y;
    public string type;
    public double relDistance;
}



public static void FitTreesToLine(List<Tree> treesList, out double m, out double c)
{
    double[] xdata = treesList.Select(x => x.x).ToArray();
    double[] ydata = treesList.Select(x => x.y).ToArray();

    Tuple<double, double> p = Fit.Line(xdata, ydata);
    double a = p.Item1; // == 10; intercept
    double b = p.Item2; // == 0.5; slope

    m = b;
    c = a;
}

public static double FindDistanceBetweenPointAndLine(double m, double c, double point_x, double point_y )
{

    double line_start_x = point_x * 0.5;
    double line_start_y = m * line_start_x + c;

    double line_end_x = point_x * 1.5;
    double line_end_y = m * line_end_x + c;

    double distance = Math.Abs((line_end_x - line_start_x) * (line_start_y - point_y) - (line_start_x - point_x) * (line_end_y - line_start_y)) /
            Math.Sqrt(Math.Pow(line_end_x - line_start_x, 2) + Math.Pow(line_end_y - line_start_y, 2));

    return (distance);

}

 public static void DoCalculations(List<Tree> treeList)
 {

    //Calculate groups
    Group curGroup = new Group(1);
    groupList.Add(curGroup);

    int searchFailures = 0;

    treeGrouping:

    List<Tree> noGroupList = treeList.Where(x => x.group.group == 0).ToList();
    List<Tree> closeTreeList = new List<Tree>();

    if (noGroupList.Count() >= 3 && searchFailures < 1000)
    {

        var refTree = noGroupList[0];
        closeTreeList.Add(refTree);
        for (int i = 1; i < noGroupList.Count(); i++)
        {

            double distance = Math.Sqrt(Math.Pow(refTree.x - noGroupList[i].x, 2) + Math.Pow(refTree.y - noGroupList[i].y, 2));

            if (distance <= 7)
            {
                closeTreeList.Add(noGroupList[i]);

                if (closeTreeList.Count() == 2)
                {
                    //Fit linear curve
                    double m = 0;
                    double c = 0;
                    FitTreesToLine(closeTreeList, out m, out c);

                    //Find all points that is close to the line in original tree list
                    for (int j = 0; j < noGroupList.Count(); j++)
                    {
                        double distanceFromLine = FindDistanceBetweenPointAndLine(m, c, noGroupList[j].x, noGroupList[j].y);

                        if (distanceFromLine <= 8)
                        {
                            noGroupList[j].group = curGroup;
                        }
                    }

                    //Iterate current group
                    curGroup = new Group(curGroup.group + 1);
                    groupList.Add(curGroup);

                    goto treeGrouping;

                }
            }
        }

        refTree.group.group = 9999999;

        //curGroup = new Group(curGroup.group + 1);
        //groupList.Add(curGroup);

        searchFailures++;
        goto treeGrouping;

    }

    //Order trees within their groups
    foreach (var group in groupList)
    {
        var groupTreeList = treeList.Where(x => x.group == group).OrderBy(x => x.y).ToList();
        for (int i = 0;i < groupTreeList.Count();i++)
        {
            groupTreeList[i].orderInGroup = i + 1;
        }
    }

    //Get max group rel distance
    foreach (var group in groupList)
    {

        var items = treeList.Where(x => x.group == group);

        if (items.Count() > 0)
        {
            group.highestRelDistance = items.OrderBy(x=>x.orderInGroup).Last().x;
        }

    }

    //Order tree groups into rows
    groupList = groupList.OrderBy(x => x.highestRelDistance).ToList();
    for (int i = 0;i < groupList.Count();i++)
    {
        var items = treeList.Where(x => x.group == groupList[i]).ToList();

        foreach (var item in items)
        {
            item.group.row = i;
        }

    }

    //Generate tree names
    foreach (var tree in treeList)
    {
        tree.name = "(" + tree.group.row.ToString().PadLeft(2,'0') + "-" + tree.orderInGroup.ToString().PadLeft(3, '0') + ")";
    }

    //Order list
    treeList = treeList.OrderBy(x => x.group.row).ThenBy(x => x.orderInGroup).ToList();

}

票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/51086516

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档