如何通过组合每个列表中的一个元素将两个不同对象的列表组合成一个列表？

Question

嗨，我正在尝试将两个列表与不同的对象结合起来。

[{"SaveValues":[{"id":1,"allposition":{"x":-0.015270888805389405,
"y":9.267399787902832},"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":1}],

"NoteValues":[{"movenumber":1,"notemsg":"Added one"}]},
{"SaveValues":[{"id":1,"allposition":{"x":-0.015270888805389405,
"y":9.267399787902832},"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":2},{"id":1,
"allposition":{"x":-0.02840942144393921,"y":6.721944808959961},
"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":2}],
"NoteValues":[{"movenumber":2,"notemsg":"Added two"}]},
{"SaveValues":[{"id":1,"allposition":{"x":-0.015270888805389405,
"y":9.267399787902832},"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":3},{"id":1,
"allposition":{"x":-0.02840942144393921,"y":6.721944808959961},
"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},"allscale":{"x":1.0,"y":1.0},
"linepos0":{"x":0.0,"y":0.0,"z":0.0},"linepos1":{"x":0.0,"y":0.0,"z":0.0},
"movetype":3},{"id":1,"allposition":{"x":-0.10085266828536987,
"y":4.49822473526001},"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},
"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":3}],"NoteValues":
[{"movenumber":3,"notemsg":"Added three"}]},{"SaveValues":
[{"id":1,"allposition":{"x":-0.015270888805389405,"y":9.267399787902832}
,"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},

"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":4},{"id":1,
"allposition":{"x":-0.02840942144393921,"y":6.721944808959961},
"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},"allscale":{"x":1.0,"y":1.0},"linepos0":{"x":0.0,"y":0.0,"z":0.0},
"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":4},{"id":1,"allposition"
:{"x":-0.10085266828536987,"y":4.49822473526001},
"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},
"allscale":{"x":1.0,"y":1.0},
"linepos0":{"x":0.0,"y":0.0,"z":0.0},"linepos1":{"x":0.0,"y":0.0,"z":0.0},
"movetype":4},{"id":1,"allposition":{"x":0.17862117290496827,"y":1.5408382415771485},
"allrotation":{"x":0.0,"y":0.0,"z":0.0,"w":1.0},"allscale":{"x":1.0,"y":1.0},
"linepos0":{"x":0.0,"y":0.0,"z":0.0},"linepos1":{"x":0.0,"y":0.0,"z":0.0},"movetype":4}],
"NoteValues":[{"movenumber":4,"notemsg":"Added four"}]}]

Class 下面给出。

[System.Serializable]
public class PlayerHandler 
{
public int id; 
public Vector2 allposition;
public Quaternion allrotation;
public Vector2 allscale;

public Vector3 linepos0;
public Vector3 linepos1;
public int movetype;


public PlayerHandler(int ids,Vector2 allpos,Quaternion allrot,Vector2 allscal,Vector3 Line0,Vector3 Line1,int Moves)

{
    this.id = ids;
    this.allposition = allpos;
    this.allrotation = allrot;
    this.allscale = allscal;
    this.linepos0 = Line0;
    this.linepos1 = Line1;
    this.movetype = Moves;
}



}

[System.Serializable]
public class PlayerMovement
{
public int movenumber;
public string notemsg;

public PlayerMovement(int Movenum,string Note)
{
    this.movenumber = Movenum;
    this.notemsg = Note;

}

}

现在我有两个列表及其对应的值。

List<PlayerHandler> SaveValuesDeserialize = new List<PlayerHandler>();
List<PlayerMovement> NoteValuesDeserialzeList = new List<PlayerMovement>();

我想将上述两个列表组合成第三个列表，其中第 0 个元素来自 NoteValuesDeserialzeList[0]，第 1 个元素来自 SaveValuesDeserialize[0]，第 2 个元素来自 NoteValuesDeserialzeList[1]，第 3 个元素来自 SaveValuesDeserialize[1]，直到所有值相加。

各种疑问出现。 新增列表List<?>AllcombinedList =new List<?>();的 object 会是什么？ .

加起来新列表的计数是多少（两个列表的组合计数）？

Answer 1

因为 PlayerHandler 和 PlayerMovement 的唯一祖先是 object，所以只能创建一个对象列表：

List<object> AllcombinedList = new List<object>();

int count = Math.Min(SaveValuesDeserialize.Count, NoteValuesDeserialzeList.Count);
// Or raise an exception if SaveValuesDeserialize.Count != NoteValuesDeserialzeList.Count
// Or ask the user what to do with the rest if it can happen
// Else use count = SaveValuesDeserialize.Count for example if always same
for ( int index = 0; index < count; index++ )
{
  AllcombinedList.Add(NoteValuesDeserialzeList[index]);
  AllcombinedList.Add(SaveValuesDeserialize[index]);
}

也许您可以使用元组列表来实现您的目标：

var AllcombinedList = new List<Tuple<PlayerHandler, PlayerMovement>>();

int count = Math.Min(SaveValuesDeserialize.Count, NoteValuesDeserialzeList.Count);
// Or raise an exception if SaveValuesDeserialize.Count != NoteValuesDeserialzeList.Count
// Or ask the user what to do with the rest if it can happen
// Else use count = SaveValuesDeserialize.Count for example if always same
for ( int index = 0; index < count; index++ )
{
  var item = new Tuple<PlayerMovement, PlayerHandler>(NoteValuesDeserialize[index],
                                                      SaveValuesDeserialzeList[index]);
  AllcombinedList.Add(item);
}

或匿名元组：

var AllcombinedList = new List<(PlayerHandler, PlayerMovement)>();

int count = Math.Min(SaveValuesDeserialize.Count, NoteValuesDeserialzeList.Count);
// Or raise an exception if SaveValuesDeserialize.Count != NoteValuesDeserialzeList.Count
// Or ask the user what to do with the rest if it can happen
// Else use count = SaveValuesDeserialize.Count for example if always same
for ( int index = 0; index < count; index++ )
{
  AllcombinedList.Add((NoteValuesDeserialize[index], SaveValuesDeserialzeList[index]));
}

或者您可以创建一个 class 提供打字和命名：

var AllcombinedList = new List<PlayerItem>();

int count = Math.Min(SaveValuesDeserialize.Count, NoteValuesDeserialzeList.Count);
// Or raise an exception if SaveValuesDeserialize.Count != NoteValuesDeserialzeList.Count
// Or ask the user what to do with the rest if it can happen
// Else use count = SaveValuesDeserialize.Count for example if always same
for ( int index = 0; index < count; index++ )
  AllcombinedList.Add(new PlayerItem
  {
    Movement = NoteValuesDeserialzeList[index],
    Handler = SaveValuesDeserialize[index]
  });

public class PlayerItem
{
  public PlayerMovement Movement { get; set; }
  public PlayerHandler Handler { get; set; }
}

如果您想管理其他项目，例如可以这样做：

int count = Math.Max(SaveValuesDeserialize.Count, NoteValuesDeserialzeList.Count);
for ( int index = 0; index < count; index++ )
{
  var movement = index < NoteValuesDeserialzeList.Count 
               ? NoteValuesDeserialzeList[index] 
               : null;
  var handler = index < SaveValuesDeserialize.Count 
              ? SaveValuesDeserialize[index] 
              : null;
  AllcombinedList.Add(new PlayerItem
  {
    Movement = movement,
    Handler = handler
  });
}

您必须自己决定如何处理其他项目（当其中一个计数多于另一个时）：什么都不做，将另一个添加为 null 或创建一个默认实例。

Answer 2

泛型List<T>在其定义中仅接受一种类型T 因此，您不能在其中包含不同类型的元素，所有元素都必须是相同类型。

您可以使用class或struct来保存和公开这两种数据类型，然后将此 class 或结构用作通用List<T>定义中的类型T 或者，您可以使用tuples 。

如果这种方法对您来说似乎过于复杂，那么您可以将PlayerHandler和PlayerMovement的祖先Object用作List<T>的类型T 。