Question

我很难让“会员”正确地加入“家庭”。每个成员都有预定义的变量，'mbzone''abshid''abspid'和'absfid'，有些成员可以拥有与这些预定义变量完全相同的值集。从同样的意义上说，家庭代理人可以为他们预先定义的'hhzone'和'abshid'拥有相同的价值观。

我想让会员代理通过以下代码中的给定条件加入家庭代理。但事实证明，一些家庭的[成员列表]并不仅包含具有独特“abspid”和“absfid”的成员代理。在家庭代理人的成员名单中的含义可能有成员具有相同的abspid和absfid。我的问题是如何防止这种情况发生？

提前谢谢你。对不起，如果我写的不清楚，我很乐意在必要时更详细地解释。

    breed [members member]
    breed [households household]

    members-own [mbzone abshid abspid absfid]
    households-own [hhzone abshid]

    to members-join-households
          ask members [
            set hhid one-of households with [hhzone = [mbzone] of myself and abshid = [abshid] of myself and [abspid] of hhmemberlist != [abspid] of myself and [absfid] of hhmemberlist != [absfid] of myself]
            ask hhid [
              let newmember myself
              set hhmemberlist (turtle-set hhmemberlist newmember)
            ]
          ]
          output-print "member-join-households is done"
        end

Answer 1

@JenB说的是什么。额外的考虑因素：测试你找到了这样一个家庭。

breed [members member]
breed [households household]

members-own [mbzone abshid abspid absfid hhid]
households-own [hhzone abshid hhmemberlist]

to members-join-households
  ask members [
    let _pid abspid
    let _fid absfid
    set hhid one-of households with [
      hhzone = [mbzone] of myself 
      and abshid = [abshid] of myself 
      and not member? _pid ([abspid] of hhmemberlist)
      and not member? _fid ([absfid] of hhmemberlist)
    ]
    if (hhid != nobody) [ ;did such a household exist?
      ask hhid [
        set hhmemberlist (turtle-set hhmemberlist myself)
      ]
    ]
  ]
  output-print "member-join-households is done"
end

但与使用定向链接相比，以这种方式保留成员列表非常笨拙，因此我建议您切换到该列表。如果您的成员形成了指向家庭的有向链接，那么一个家庭的成员就是其中的邻居，而成员的hhid就是它所链接的家庭。

如果代理集中的代理属性不匹配，请求代理加入代理集

1 个答案: