Question

所以基本上我有2个字符串列表:(列表点，列表连接）。列表＆＃34;点＆＃34;是一组xyz坐标和列表＆＃34;连接＆＃34;显示如何在列表中连接点以创建三角形。我想要做的是填充包含三角形的列表。请注意＆＃34;连接＆＃34;中每一行末尾的值-1可以排除（可以说新的三角形即将到来）。

我创建的类ProcessTriangles应该得到那些三角形但是当我编译代码时没有任何反应。

有人可以帮忙吗？

我的代码：

using System;
using System.Collections.Generic;

struct Point
{
    public readonly double X;
    public readonly double Y;
    public readonly double Z;

    public Point(double x, double y, double z)
    {
        X = x;
        Y = y;
        Z = z;
    }
    public override string ToString()
    {
        return String.Format("{0} {1} {2},", X, Y, Z);
    }
    public static Point FromString(string p)
    {
        string[] items = p.Replace(",", "").Split(' ');
        double[] coords = new double[3];
        for (int i = 0; i < 3; i++) coords[i] = double.Parse(items[i]);
        return new Point(coords[0], coords[1], coords[2]);
    }
}

struct Triangle
{
    public readonly Point P1;
    public readonly Point P2;
    public readonly Point P3;

    public Triangle(Point p1, Point p2, Point p3)
    {
        P1 = p1;
        P2 = p2;
        P3 = p3;
    }
    public override string ToString()
    {
        return String.Format("{0} {1} {2},", P1, P2, P3);
    }
}

class Program
{
    static void Main()
    {
        #region listDefinition
        var points = new List<string>
      {
         "0 0 128.588459085565,",
         "25 0 134.979628462965,",
         "0 0 134.979628462965,",
         "25 0 128.588459085565,",
         "25 0 140.100717207301,",
         "0 0 140.100717207301,",
         "25 0 143.87494372892,",
         "0 0 143.87494372892,",
         "25 0 146.245863353464,",
         "0 0 146.245863353464,",
         "25 0 147.181161545899,",
         "0 0 147.181161545899,", 
         "25 0 146.668001779529,",
         "0 0 146.668001779529,", 
         "25 0 144.711312619297,",
         "0 0 144.711312619297,", 
         "25 0 141.340784276868,",
         "0 0 141.340784276868,",
         "25 0 136.604499944182,",
         "0 0 136.604499944182,", 
         "25 0 130.573315654463,",
         "0 0 130.573315654463,",
         "25 0 123.341482952817,",
         "0 0 123.341482952817,",
         "25 0 115.031289382498,", 
         "0 0 115.031289382498,",
         "25 0 105.793445439687,",
         "0 0 105.793445439687," 
       };

        var connections = new List<string>
       {
         "0, 1, 2,-1",
         "1, 4, 2,-1",
         "2, 4, 5,-1",
         "4, 6, 5,-1",
         "5, 6, 7,-1",
         "6, 8, 7,-1",
         "7, 8, 9,-1",
         "8, 10, 9,-1",
         "9, 10, 11,-1",
         "10, 12, 11,-1",
         "11, 12, 13,-1",
         "12, 14, 13,-1",
         "13, 14, 15,-1",
         "14, 16, 15,-1",
         "15, 16, 17,-1",
         "16, 18, 17,-1",
         "17, 18, 19,-1",
         "18, 20, 19,-1",
         "19, 20, 21,-1",
         "20, 22, 21,-1",
         "21, 22, 23,-1",
         "22, 24, 23,-1",
         "23, 24, 25,-1",
         "24, 26, 25,-1",
         "25, 26, 27,-1",
         "0, 3, 1,-1"         
       };
        #endregion

        ProcessTriangles(points, connections);
        PrintGroup(points, connections);
    }

    public static List<string> ProcessTriangles(List<string> points, List<string> connections)
    {
        List<string> trianglesOut = new List<string>();

        for (int i = 0; i < connections.Count; i++)
        {
            string[] items = connections[i].Split(',');
            int[] indices = new int[6];
            for (int j = 0; j < 3; j++) indices[j] = int.Parse(items[j]);

            Point v0 = Point.FromString(points[indices[0]]);
            Point v1 = Point.FromString(points[indices[1]]);
            Point v2 = Point.FromString(points[indices[2]]);

            Triangle tri = new Triangle(v0, v1, v2);
            trianglesOut.Add(tri.ToString());

            --i;                
        }

        return trianglesOut;
    }

    static void PrintGroup(List<string> points, List<string> connections)
    {
        Console.WriteLine("-- GROUP --");
        Console.WriteLine("\nLIST1\n");
        foreach (string point in points) Console.WriteLine(point);
        Console.WriteLine("\nLIST2\n");
        foreach (string connection in connections) Console.WriteLine(connection);
        Console.WriteLine();
    }
}

Answer 1

在ProcessTriangles()方法中，在for循环中，您--i;只需删除它。

for (int i = 0; i < connections.Count; i++)
{
    string[] items = connections[i].Split(',');
    int[] indices = new int[6];
    for (int j = 0; j < 3; j++) indices[j] = int.Parse(items[j]);

    Point v0 = Point.FromString(points[indices[0]]);
    Point v1 = Point.FromString(points[indices[1]]);
    Point v2 = Point.FromString(points[indices[2]]);

    Triangle tri = new Triangle(v0, v1, v2);
    trianglesOut.Add(tri.ToString());

   //deleted --i;
}

Answer 2

您的ProcessTriangles循环永远不会以app/mock-heroes.ts

结束

--i

Answer 3

以下LINQ语句将根据示例数据创建三角形：

        var triangles = connections
            .SelectMany((cs, i) => cs.Split(',')
                .Select(c => new
                {
                    triangleIndex = i,
                    pointIndex = int.Parse(c.Trim())
                }))
                .Where(tp => tp.pointIndex != -1)
            .Join(points
                .Select((ps, i) => new
                {
                    index = i,
                    coords = ps.TrimEnd(new[] { ',' }).Split(' ')
                        .Select(p => double.Parse(p.Trim()))
                        .ToArray()
                }), tp => tp.pointIndex, p => p.index, (tp, p) => new
                {
                    tp.triangleIndex,
                    point = new Point(p.coords[0], p.coords[1], p.coords[2])
                })
            .GroupBy(j => j.triangleIndex, i => i.point)
            .Select(g => g.ToArray())
            .Select(g => new Triangle(g[0], g[1], g[2]))
            .ToList();

输出：

0 0 128.588459085565,25 0 134.979628462965,0 0 134.979628462965 ,,

25 0 134.979628462965,25 0 140.100717207301,0 0 134.979628462965 ,,

0 0 134.979628462965,25 0 140.100717207301,0 0 140.100717207301 ,,


25 0 140.100717207301,25 0 143.87494372892,0 0 140.100717207301 ,,

等...

通过重新排列其他两个列表来创建新列表

3 个答案: