基于最小坐标标准的排序点列表

Question

我希望你能帮助我解决这个问题。作为较大代码的一部分，我必须对点列表（x、y、z 坐标）进行一些“特殊排序”。

此列表（根据定义）将：

i) 总是有一列的所有值都为零。

ii) 第一个点和最后一个点总是相同的。

列表的排序取决于哪一列为零。我自己可以毫无问题地识别此列（请参阅代码），但是当它带有排序位时我很挣扎。首先我必须找到一个满足某些特定标准的点，然后在此基础上重新组织列表。我已经在代码中将排序标准解释为 cmets（参见第 2 节和第 3 节）。

如果您可以就 2) 和 3) 提供帮助，那就太好了（我认为第一部分还可以）。

非常感谢

代码：

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{

[STAThread]

public static void Main(string[] args)
{
    //Populating the list
    var ListPoints = new List<double[]>();

    double[] P0 = { 10, 10, 0 };
    double[] P1 = { 10, 0, 0 };
    double[] P2 = { 0, 0, 0 };
    double[] P3 = { 0, 10, 0 };

    ListPoints.Add(P0);
    ListPoints.Add(P1);
    ListPoints.Add(P2);
    ListPoints.Add(P3);
    ListPoints.Add(P0);

    ///This list (by definition) will: 
    /// i) Always have a column with all the values equal zero
    /// ii) The first and the last point will always be the same. 
    ///We need to detect the column with all values = zero, because the list sorting will depend on this. 



    /// 1) Detect which columns has all values equal to zero using the columnZero variable
    var counterX = new List<int>();
    var counterY = new List<int>();
    var counterZ = new List<int>();


    for (int i = 0; i < ListPoints.Count - 1; i++)
    {
        //First column with all values equal zero
        if (ListPoints[i][0] == 0 && ListPoints[i][0] == ListPoints[i + 1][0]) { counterX.Add(1); }
        //Second column with all values equal zero
        if (ListPoints[i][1] == 0 && ListPoints[i][1] == ListPoints[i + 1][1]) { counterY.Add(1); }
        //Third column with all values equal zero
        if (ListPoints[i][2] == 0 && ListPoints[i][2] == ListPoints[i + 1][2]) { counterZ.Add(1); }
    }

    if (counterX.Count == ListPoints.Count - 1)
    { Console.WriteLine("all points of the 1st column are zero");}

    if (counterY.Count == ListPoints.Count - 1)
    { Console.WriteLine("all points of the 2nd column are zero");}

    if (counterZ.Count == ListPoints.Count - 1)
    { Console.WriteLine("all points of the 3rd column are zero");}

    /// 2) Now a point "Q" must be found in the list according to this:
    /// 2.1/ If the first column has all values = zero:
    ///      Find the row index of the smallest value in the second column.
    ///         If there are several rows in the second column with the same minimum value go and find between those which one has the smallest value in the third column.
    ///         If there is only one value in the second column keep that one.
    /// 2.2/ If the second column has all values = zero:
    ///      Find the row index of the smallest value in the first column.
    ///         If there are several rows in the first column with the same minimum value go and find between those which one has the smallest value in the third column.
    ///         If there is only one value in the first column keep that one.
    /// 2.3/ If the third column has all values = zero:
    ///      Find the row index of the smallest value in the first column.
    ///         If there are several rows in the first column with the same minimum value go and find between those which one has the smallest value in the second column.
    ///         If there is only one value in the first column keep that one.
    ///

    /// 3) Once this value has been found we have to put the list starting by this point "Q", then copy the previous values at the end of the list  and finally add again "Q". 
    ///    Example:The column with all values = 0  is column 3 and the generic point "Q" is the point "P2" in my list. The input is P0-P1-P2-P3-P0 but I want to have it P2-P3-P0-P1-P2.
}
}

Answer 1

根据您的标准，排序算法的前两点会退化为一个更简单的算法。

即：无论哪一列全是 0，你总是按第一列、第二列、第三列升序排序。这会将您的目标项目留在列表的开头，然后您只需从原始列表中获取正确的顺序即可。

private List<double[]> SpecialSort(List<double[]> list)
{
  // Make an editable duplicate of your original list and remove the duplicated array at the start (or end) of the list.
  List<double[]> orig = new List<double[]>(list);
  orig.RemoveAt(0);

  // Copy and sort the list by 1st column, 2nd column, 3rd column.
  List<double[]> copy = orig.OrderBy(p => p[0]).ThenBy(p => p[1]).ThenBy(p => p[2]).ToList();

  // The first item in the copy-sorted list is your point "Q".
  // Find its index in the original list.
  int index = orig.IndexOf(copy[0]);
  List<double[]> sorted = new List<double[]>();

  // For the list count + 1 (adding point Q twice) add the original list
  // objects in the correct order, starting at "point'Q'".
  for (int i = 0; i <= orig.Count; i++)
  {
    sorted.Add(orig[(index + i) % orig.Count]);
  }

  return sorted;
}

然后调用 as

ListPoints = this.SpecialSort(ListPoints);