我试图通过比较从pointsToSearchFrom
列表中的每个点到pointCloudToSearchList
中的每个点的距离来检索集合中最远的点。您可以从所附图像中看到示例方案。我仍然不是导航这些数据结构的专家,并且该算法超出了我目前对遍历二维数组的了解。
这是我直到知道的代码。任何帮助都会很棒。
public static void Test(List<Point3d> pointsToSearchFrom, List<Point3d> pointCloudToSearch)
{
int rows = pointsToSearchFrom.Count;
int columns = pointCloudToSearch.Count;
double[,] arrayDistance = new double [rows, columns];
for (int i = 0; i < pointsToSearchFrom.Count; i++)
{
for (int j = 0; j < pointCloudToSearch.Count; j++)
{
arrayDistance[i, j] = (pointsToSearchFrom[i] - pointCloudToSearch[j]).magnitude;
}
}
}
答案 0 :(得分:0)
您可以为此使用二维数组,但不必这样做。
您需要的是可以在此处找到的MaxBy方法:
https://github.com/morelinq/MoreLINQ/blob/master/MoreLinq/MaxBy.cs
我们首先构建一对配对:
var pairs = from first in pointsToSearchFrom
from second in pointCloudToSearch
select new { first, second };
请注意,在C#7中,您将使用元组,而不是匿名类型。
现在我们有了所有可能对的序列。使用MaxBy查找使给定数量最大化的对:
var maxPair = pairs.MaxBy(pair => (pair.first - pair.second).magnitude);
现在您已完成:
Console.WriteLine($"{maxPair.first} --> {maxPair.second});
如果您有大量的点集(例如成千上万个),则必须使用特殊的技术,因为如果有成千上万的点,那么将有数百万个对,并且如果有数百万个,点的数量将达到数万亿。但是,如果您只有少数几个,则此技术可以正常工作。
答案 1 :(得分:0)
起初我并没有意识到这是3D。我使用了一个类和列表来存储2点及其之间的距离。这样,我不仅可以得到最长的距离,而且可以得到产生该距离的2个点。
public class Distances
{
public double Distance { get; set; }
public Point3D FromPoint { get; set; }
public Point3D ToPoint { get; set; }
public Distances(Point3D from, Point3D to)
{
FromPoint = from;
ToPoint = to;
Distance = (to - from).Length;
}
}
private void OP2()
{
List<Point3D> searchFrom = new List<Point3D>()
{
new Point3D(20,30,50),
new Point3D(10,50,10),
new Point3D(30,40,20),
new Point3D(60,30,10)
};
List<Point3D> searchCloud = new List<Point3D>()
{
new Point3D(60,70,80),
new Point3D(110,30,30),
new Point3D(80,110,55),
new Point3D(90,20,90)
};
List<Distances> resultDistances = new List<Distances>();
foreach (Point3D p1 in searchFrom)
{
foreach (Point3D p2 in searchCloud)
{
Distances d = new Distances(p1, p2);
resultDistances.Add(d);
}
}
//The following is just for testing purposes, skip to var longestDistance
List<Distances> distancesInOrder = resultDistances.OrderByDescending(i => i.Distance).ToList<Distances>();
foreach (Distances d in distancesInOrder)
{
Debug.Print($"From Point ({d.FromPoint.X}, {d.FromPoint.Y}, {d.FromPoint.Z}) To Point ({d.ToPoint.X}, {d.ToPoint.Y}, {d.ToPoint.Z}) Distance = {d.Distance}");
}
var longestDistance = resultDistances.OrderByDescending(i => i.Distance).FirstOrDefault();
//this gives you the longest distance and the two points
MessageBox.Show($"First Point ({longestDistance.FromPoint.X}, {longestDistance.FromPoint.Y}, {longestDistance.FromPoint.Z}) Cloud Point ({longestDistance.ToPoint.X}, {longestDistance.ToPoint.Y}, {longestDistance.FromPoint.Z}) Distance = {longestDistance.Distance}");
}