Question

玩具示例。有两组人： A 和 B 。只有 A 可以说＆＃34;你好＆＃34;对人 B 。人们走遍世界各地，相遇。当人A 遇到人B 时，他们会向他们问好。每个人 A 记录，当发生这种情况时，他们都会打招呼。他们不能向同一个人打招呼，直到五个新的蜱发生。以下程序仅适用于人A 。

每当人A 向人B 打招呼时，我定义：

set tick-last-greeting lput ticks tick-last-greeting
set previous-person-b-greeted lput selected-person-b previous-person-b-greeted

再次发出say-hello程序之前：

if (length  tick-last-greeting != [] and previous-person-b-greeted != []) [
  ; wait 5 ticks
  set temp (map [ticks - ? > 5] tick-last-greeting)
  ; filter the list, I don't know if there is a better way to do this
  set previous-person-b-greeted (map last filter [first ? = false] (map list temp previous-person-b-greeted)) 
  set tick-last-greeting (map last filter [first ? = false] (map list temp tick-last-greeting))
]

所以，我得到一个人B 的列表，不应该被人A 打招呼，但直到五个滴答声发生。这是我的关键问题：如何定义排除列表previous-person-b-greeted的代理的代理集。

set potential-persons-b targets-on (patch-set neighbors patch-here)
if (previous-person-b-greeted > 0) [

; Here, I get an error as expected
let who-previous-person-b [who] of previous-person-b-greeted 
set potential-persons potential-persons with [who != who-previous-person-b]
]

一种可能的解决方案：将列表previous-person-b-greeted转换为代理集（我不知道是否有简单的方法可以执行此操作）。

有什么想法吗？

Answer 1

要将代理列表转换为代理集，请使用turtle-set或patch-set或link-set。例如：

observer> create-turtles 5
observer> let mylist (list turtle 0 turtle 2 turtle 4)  print turtle-set mylist
(agentset, 3 turtles)

Answer 2

我会假设您没有为A人或B人使用特定品种。

也许您可以尝试使用品种，例如：

breed [personA peopleA]
breed [personB peopleB]

将定义2个不同的代理集，然后您可以使用<breeds>-own语句来定义最近受欢迎的人员列表。

peopleA-own [recently-greeted-people recently-greeted-people-time]

然后每当一个人必须问候某人时，你的程序可能如下：

to greet [personB-who]
    if (not (and (member? personB-who recently-greeted-people)
                  (procedure-that-checks-ticks-less-than-5))
       ...ADD HERE OTHER LOGICAL CHECKS DEPENDING ON YOUR PROBLEM
       )
       [ 
         fput personB-who recently-greeted-people
         fput current-tick recently-greeted-people-time
       ]
end

请注意，对于每个personB，who和id会被添加到不同的列表中，然后必须同时将其删除以保持一致。

您可以阅读有关breeds in the NLogo dictionary.

的更多信息

Answer 3

最后，根据您的建议，我最终得到了这个解决方案：

set potential-persons-b sort (targets-on (patch-set neighbors patch-here))

if (previous-person-b-greeted != []) 
[

foreach previous-victimized-target
[ set potential-persons-b remove ? potential-persons-b]

set potential-persons-b turtle-set potential-persons-b
]

这是使用to-report的更通用的解决方案：

to-report subsetting-agents [agent-set1 agent-set2]
  set agent-set1 sort agent-set1
  set agent-set2 sort agent-set2
  foreach agent-set2
  [ set agent-set1 remove ? agent-set1]
  set agent-set1 turtle-set agent-set1
  report agent-set1
end

子集化代理集将代理列表转换为代理集

3 个答案: