Loops 如何使一只海龟能够在Netlogo中保存其他海龟的ID？

以下是我想做的：

我想要一个由三只海龟组成的小组

所有海龟都需要将自己的ID和队友SID存储在名为队友SID的变量中。（假设你想创建一个组，你会把你的名字和你朋友的名字放在一个列表中，这就是为什么这个变量需要存储ownID，我不擅长解释东西。）

我需要他们能够显示所有队友的ID，而不是仅仅显示

（agentset，3只海龟）

在他们收集了所有3名成员后，他们将获得

团队ID

这里的问题是我不知道如何让这些海龟将自己的ID和队友SID存储在名为队友SID的变量中。在一个小组中，他们应该有三个不同的成员。这些队友不是来自同一组的。以及如何给予他们

团队ID

他们获得所有成员后

这是我的代码：

global
[ teamID]

turtles-own
[
  myID
  teammatesID
]

to setup
  clear-all
  set-default-shape turtles "arrow"
  create-turtles 10

  ask turtles [ set myID who]
  reset-ticks
  setup-group
end

to setup-group

   set teamID []
   let agent one-of other turtles
   let  nearestNeighbor one-of other turtles in-radius 1 
   set teamID = 0
   ask agent with [ teammatesID < 3]
   [ ask nearestNeighbor with [teammatesID < 3]
     [ ;set teammatesID = myID ; here is the problem, I need to make sure they did not save the same turtle in one group.
       show teammatesID]]

       ask agent with [teammatesID > 3]
       set teamID fput teamID teamID
        end

全局
[团队ID]
乌龟自己的
[
粘虫
队友席德
]
设置
清除所有
设置默认形状海龟“箭头”
创造海龟10
问海龟[设置myID谁]
重置滴答声
设置组
终止
设置组
设置团队ID[]
让探员找一只海龟
让最靠近的一只海龟在半径1内
设置teamID=0
向代理询问[SID<3]
[与[Sid<3]一起询问最近的邻居]
[；set teambersid=myID；问题是，我需要确保他们没有在同一组中拯救同一只海龟。
显示SID]]
用[SID>3]询问代理
设置teamID fput teamID teamID
终止

我很感激你在这里的额外时间。谢谢。
如果团队是随机的，我认为您不需要为此循环，因为
ask
将以随机顺序调用海龟

我认为您仍然应该使用agentsets，因为它使某些事情变得更容易。例如，一旦海龟知道了它们的
（agentset，3只海龟）
，您就可以轻松地查询该agentset以获得
myID
s或任何您想要的变量

我不完全清楚
teamID
的目的，因此我可能在这里偏离了基准，但如果
teamID
s对于每组三只海龟都是唯一的，我认为您不希望它成为一个全局变量。您可能希望它也是turtles自己的
变量

下面是一个包含上述思想的示例。使用此设置：

turtles-own [ myteamset teamID myID teammatesID ]

to setup 
  ca
  crt 10 [
    set myID who
    set myteamset nobody 
    set teammatesID [ ]
  ]
  setup-groups
  print remove-duplicates [teammatesID] of turtles
  print sort [teamID] of turtles
  reset-ticks
end

以及
设置组
程序（更多详细信息请参见备注）：

如果这是你想要的，请告诉我，希望有帮助

编辑-根据您的评论：

要获得连续的团队编号，您可以使用一个简单的计数器，该计数器分配给一个团队，然后为下一个团队递增-修改后的
设置组
如下所示：

to setup-groups
  ; Create a temporary variable to use as a counter
  let teamCounter 1
  ask turtles [
    if myteamset = nobody [
      let possible-teammates other turtles with [ myteamset = nobody ]
      ifelse count possible-teammates > 1 [
        set myteamset ( turtle-set self n-of 2 possible-teammates )
        let ids sort [myID] of myteamset
        let teammmembers myteamset

        ; Assign the current teamCounter as team number, then 
        ; increment it by one for the next team
        let tempteam teamCounter
        set teamCounter teamCounter + 1
        ask myteamset [
          set teammatesID ids
          set myteamset teammmembers
          set teamID tempteam
        ]
      ] [
        show "Not enough turtles to make a new team!"
      ]
    ]
  ]
end

第二个问题本身可能值得提出一个新问题，这取决于您希望获得的深度，但这里有一个非常简单的方法，您只需获得所有独特的团队ID，并拥有一个具有该ID的海龟，让他们的团队一起移动：

to move-in-groups
  ; Get the unique team IDs
  let teamNumbers remove-duplicates [teamID] of turtles 

  ; Get one member of each team to ask all members
  ; of its team (itself included) to move 
  foreach teamNumbers [
    tn ->
    ask one-of turtles with [ teamID = tn ] [
      let newHeading heading + random 60 - 30
      if myteamset != nobody [
        ask myteamset [
          set heading newHeading
          fd 1
        ]
      ]
    ]
  ]
end

编辑2:NetLogo 5友好版

to move-in-groups
  ; Get the unique team IDs
  let teamNumbers remove-duplicates [teamID] of turtles

  ; Get one member of each team to ask all members
  ; of its team (itself included) to move
  foreach teamNumbers [
    ask one-of turtles with [ teamID = ? ] [
      let newHeading heading + random 60 - 30
      if myteamset != nobody [
        ask myteamset [
          set heading newHeading
          fd 1
        ]
      ]
    ]
  ]
end

如果团队是随机的，我认为你不需要一个循环，因为
ask
无论如何都会以随机顺序调用海龟

我认为您仍然应该使用agentsets，因为它使某些事情变得更容易。例如，一旦海龟知道了它们的
（agentset，3只海龟）
，您就可以轻松地查询该agentset以获得
myID
s或任何您想要的变量

我不完全清楚
teamID
的目的，因此我可能在这里偏离了基准，但如果
teamID
s对于每组三只海龟都是唯一的，我认为您不希望它成为一个全局变量。您可能希望它也是turtles自己的
变量

下面是一个包含上述思想的示例。使用此设置：

turtles-own [ myteamset teamID myID teammatesID ]

to setup 
  ca
  crt 10 [
    set myID who
    set myteamset nobody 
    set teammatesID [ ]
  ]
  setup-groups
  print remove-duplicates [teammatesID] of turtles
  print sort [teamID] of turtles
  reset-ticks
end

以及
设置组
程序（更多详细信息请参见备注）：

如果这是你想要的，请告诉我，希望有帮助

编辑-根据您的评论：

要获得连续的团队编号，您可以使用一个简单的计数器，该计数器分配给一个团队，然后为下一个团队递增-修改后的
设置组
如下所示：

to setup-groups
  ; Create a temporary variable to use as a counter
  let teamCounter 1
  ask turtles [
    if myteamset = nobody [
      let possible-teammates other turtles with [ myteamset = nobody ]
      ifelse count possible-teammates > 1 [
        set myteamset ( turtle-set self n-of 2 possible-teammates )
        let ids sort [myID] of myteamset
        let teammmembers myteamset

        ; Assign the current teamCounter as team number, then 
        ; increment it by one for the next team
        let tempteam teamCounter
        set teamCounter teamCounter + 1
        ask myteamset [
          set teammatesID ids
          set myteamset teammmembers
          set teamID tempteam
        ]
      ] [
        show "Not enough turtles to make a new team!"
      ]
    ]
  ]
end

第二个问题本身可能值得提出一个新问题，这取决于您希望获得的深度，但这里有一个非常简单的方法，您只需获得所有独特的团队ID，并拥有一个具有该ID的海龟，让他们的团队一起移动：

to move-in-groups
  ; Get the unique team IDs
  let teamNumbers remove-duplicates [teamID] of turtles 

  ; Get one member of each team to ask all members
  ; of its team (itself included) to move 
  foreach teamNumbers [
    tn ->
    ask one-of turtles with [ teamID = tn ] [
      let newHeading heading + random 60 - 30
      if myteamset != nobody [
        ask myteamset [
          set heading newHeading
          fd 1
        ]
      ]
    ]
  ]
end

编辑2:NetLogo 5友好版

to move-in-groups
  ; Get the unique team IDs
  let teamNumbers remove-duplicates [teamID] of turtles

  ; Get one member of each team to ask all members
  ; of its team (itself included) to move
  foreach teamNumbers [
    ask one-of turtles with [ teamID = ? ] [
      let newHeading heading + random 60 - 30
      if myteamset != nobody [
        ask myteamset [
          set heading newHeading
          fd 1
        ]
      ]
    ]
  ]
end

我同意Luke的观点，与团队打交道最合适的方式是作为代理集，而不是复杂的标识符分配。这就是代码执行此操作的方式。请注意，我没有提到你问题中的动作部分

globals [max-teamsize]

turtles-own
[ teamID
  teammates
]

to testme
  clear-all
  create-turtles 100
  [ setxy random-xcor random-ycor
    set teamID 0
    set teammates nobody
  ]
  set max-teamsize 3
  setup-groups
  reset-ticks
end

to setup-groups
  ; assign turtles to team by allocating a team number
  let counter 1
  ask turtles
  [ if teamID = 0
    [ set teamID counter
      let potential-teammates turtles with [teamID = 0]
      ask n-of min (list (max-teamsize - 1) (count potential-teammates)) potential-teammates
      [ set teamID counter
      ]
      set counter counter + 1
    ]
  ]
  ; store teammates as agentset of other turtles in team
  ask turtles
  [ set teammates other turtles with [teamID = [teamID] of myself]
  ]
end

我同意Luke的观点，与团队打交道最合适的方式是作为代理集，而不是复杂的标识符分配。这就是代码执行此操作的方式。请注意，我没有提到你问题中的动作部分

globals [max-teamsize]

turtles-own
[ teamID
  teammates
]

to testme
  clear-all
  create-turtles 100
  [ setxy random-xcor random-ycor
    set teamID 0
    set teammates nobody
  ]
  set max-teamsize 3
  setup-groups
  reset-ticks
end

to setup-groups
  ; assign turtles to team by allocating a team number
  let counter 1
  ask turtles
  [ if teamID = 0
    [ set teamID counter
      let potential-teammates turtles with [teamID = 0]
      ask n-of min (list (max-teamsize - 1) (count potential-teammates)) potential-teammates
      [ set teamID counter
      ]
      set counter counter + 1
    ]
  ]
  ; store teammates as agentset of other turtles in team
  ask turtles
  [ set teammates other turtles with [teamID = [teamID] of myself]
  ]
end

你能澄清一下你所说的“这些队友不应该来自同一个团队吗？这里的团队和团队是不同的吗？根据我的编码，条件是”海龟少于3名成员需要找到其他成员“。例如，乌龟1和乌龟2创建了一个团队，因此在乌龟1的队友中，SID应该有他的ID和乌龟2的ID，乌龟2也有，它有自己的ID和乌龟1的ID。但是，这两个团队仍然需要再找到一个成员，并且