在 List<List<double[]>> 中合并一些 List<double[]>

Question

我正在努力尝试以特定方式合并列表列表。我想要做的如下所述。

我的输入列表是：

列表 A = List<List<double[]>>

列表 B = List<int[]>

而期望的输出是：

列表 C = List<List<double[]>>

我想要合并列表 A 中的一些 List<double[]>（例如，使用 linq 中的联合功能）并将结果列表存储在名为“ListC”的 List<List<double[]>> 中。

要在 ListA 中合并的 List 的索引存储在 ListB 中。如果不应该合并元素（它们的索引没有出现在 ListB 中），那么我们直接将它们添加到 ListC。

示例：

输入：

ListA = new List<List<double[]>>() { a0, a1, a2, a3, a4, a5}; 

其中 ai = List<double[]>

ListB = new List<int[]>() { new int[]{0,1,2}, new int[]{3,5} }; 

表示列表 0-1-2 必须合并为一个。与列表 3-5 相同

输出：ListC = List<List<double[]>>包含了List A所有合并的List<double[]>以及唯一的List<double[]>。即：

--> c0= union of a0, a1, a2 "三个要合并为1的列表"

--> c1= union of a3, a5 "两个列表合并为2"

--> c2= a4 "是唯一列表"

注意：重复不是问题。

什么是实现这一目标的好方法？

代码：

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


class _unionLists
{

    public static void Main(string[] args)
    {

        #region input lists
        //--------------------LIST A---------------------------------------------------
        //List a0
        var point1 = new double[] { 3, 3, 0 };
        var point2 = new double[] { 9, 6, 0 };
        var point3 = new double[] { 12, 8, 0 };
        var point4 = new double[] { 18, 8, 0 };
        List<double[]> a0 = new List<double[]>() { point1, point2, point3, point4 };

        //List a1
        var point5 = new double[] { 3, 3, 0 };
        var point6 = new double[] { 9, 7, 0 };
        var point7 = new double[] { 15, 7, 0 };
        var point8 = new double[] { 21, 15, 0 };
        List<double[]> a1 = new List<double[]>() { point5, point6, point7, point8 };

        //List a2
        var point9 = new double[] { 20, 13, 0 };
        var point10 = new double[] { 22, 16, 0 };
        List<double[]> a2 = new List<double[]>() { point9, point10 };

        //List a3
        var point11 = new double[] { 15, 19, 0 };
        var point12 = new double[] { 27, 19, 0 };
        List<double[]> a3 = new List<double[]>() { point11, point12 };

        //List a4
        var point13 = new double[] { 18, 20, 0 };
        var point14 = new double[] { 21, 19, 0 };
        List<double[]> a4 = new List<double[]>() { point13, point14 };

        //List a5
        var point15 = new double[] { 27, 19.5, 0 };
        var point16 = new double[] { 30, 5, 0 };
        List<double[]> a5 = new List<double[]>() { point15, point16 };          

        var ListA = new List<List<double[]>>() { a0, a1, a2, a3, a4, a5};

        //--------------------LIST B---------------------------------------------------
        var ListB = new List<int[]>() { new int[]{0,1,2} , new int[]{3,5} };

        #endregion

        //There are some of the List<double[]> within List A that I want to merge (e.g. using  the Union capability in linq) and store the resultant list in a List<List<double[]>> called "ListC".
        //The indices of the Lists to be merged within ListA are stored in the ListB. If the elements are not suposed to be merged (their index doesn't appear in ListB) then we add them directly to ListC.
        //
        //Example: 
        //  Inputs: ListA = new List<List<double[]>>() { a0, a1, a2, a3, a4, a5}; where ai=List<double[]>;
        //          ListB = new List<int[]>() { new int[]{0,1,2} , new int[]{3,5} }; indicates that the lists 0-1-2 have to be merged into one. Same with lists 3-5
        //  Output: ListC= List<List<double[]>> contains all the merged List<double[]> as well as all the unique List<double[]> of List A. That is:
        //          c0= union of a0, a1, a2 "three lists to be merged into 1"
        //          c1= union of a3, a5 "two lists to be merged into 2"
        //          c2= a4 "is a unique list"

        var ListC = new List<List<double[]>>();


    }
}

Answer 1

试试这个：

var elementsACount = ListA.Count();
var groupedElementsIdxs = ListB.SelectMany(e => e);
var ungroupedElementsIdxs = Enumerable.Range(0, elementsACount).Except(groupedElementsIdxs);

var result = new List<List<double[]>>();
// Merge and add the grouped elements.
foreach (var el in ListB)
{
    result.Add(el.Select(e => ListA[e]).SelectMany(e => e).ToList());
}
// Merge and add the ungrouped elements.
result.Add(ungroupedElementsIdxs.Select(e => ListA[e]).SelectMany(e => e).ToList());

Answer 2

var result = new List<List<double[]>>();

foreach (var array in listB)
{
    var ld = new List<double[]>();
    foreach (var index in array)
    {
        ld.Concat(listA[index]);
    }

    result.Add(ld);
}

var indices = listB.SelectMany(arr => arr);

result.AddRange(listA.Where((list, index) => !indices.Contains(index)));

Answer 3

你可以写

var C = A.SelectMany(x => x).Union(B);