很抱歉，这个问题可能有一个简单的答案，但我对Netlogo相对较新，并且花了几天的时间来处理这个问题无济于事，所以任何建议都非常受欢迎 - 谢谢。
植绒鱼类和鸟类有很好的植绒方法（参见MacLennan，2007（http://web.eecs.utk.edu/~mclennan/Classes/420-594-F07/NetLogo/Flock.html），基于Huth和Wissel（1992））和我想要为了适应这种植绒方法，这样你就不会蜂拥到最近的三个邻居（他们都是同一群体），而只是涌向你的代理人群中的那些人，，或者更具体地说，是你自己群体中的那些人（忽略那些）在它之外）。
据我所知，一个建议可能是在开始时创建两个不同的品种，例如：鹿和驼鹿，然后你只会涌向那些属于同一品种的人。但是，在我的例子中，我有代理人在每个滴答时重新评估哪个组最适合哪个组，并在组之间翻转（通过代理集定义）。
对于同质人群，原始代码（MacLennan，2007）如下：
TO GO ;--------------------------------------------------------------------------------------------
set i 0 ; initializes variables used in finding neighbors
set distance_1 1000
set distance_2 1000
set distance_3 1000
; (below) determines ID numbers and distances for three closest neighbors; each turtle is
; first checked to be visible (i.e., out of blind spot and in front of obstacles), and then
; its distance is compared with that of neighbor_3. If a bird is closer than neighbor_3,
; it is then compared with neighbor_2 and (if it's closer than neighbor_2) with neighbor_1;
; variables are then re-assigned to accomodate the new neighbor as one of the closest three.
repeat count turtles
[ifelse not ((i = who) or in_blind_spot? or behind_opaque_obstacle?)
[set dist distance turtle i
ifelse dist < distance_3
[ifelse dist < distance_2
[set neighbor_3 neighbor_2
set distance_3 distance_2
ifelse dist < distance_1
[set neighbor_2 neighbor_1
set distance_2 distance_1
set neighbor_1 i
set distance_1 dist]
[set neighbor_2 i
set distance_2 dist
repeat 2 [fd 0]]] ; (see note at end for explanantion of "fd 0")
[set neighbor_3 i
set distance_3 dist
repeat 4 [fd 0]]]
[repeat 6 [fd 0]]]
[repeat 7 [fd 0]]
set i i + 1]
set acceleration 0 ; sets acceleration to zero
set target_1 calc_target neighbor_1 distance_1 ; calculates targets (see procedure below) based on
set target_2 calc_target neighbor_2 distance_2 ; the headings and positions of each neighbor;
set target_3 calc_target neighbor_3 distance_3 ; acceleration is set by the calc_target procedure
; (below) calculates a bird's composite target, weighting each individual target by the inverse
; of distance to that neighbor
set target atan ((sin target_1) * 100 / distance_1 + (sin target_2) * 100 / distance_2 + (sin target_3) * 100 / distance_3)
((cos target_1) * 100 / distance_1 + (cos target_2) * 100 / distance_2 + (cos target_3) * 100 / distance_3)
; (below) turns bird towards target by a fraction specified by "flexibility"; random motion is also
; included to an extent governed by "noise". The "... + 540) mod 360) - 180" clause is necessary to
; ensure that the bird turns in the right direction. For example, if a bird is oriented at 0 degrees
; and its target is 300 degrees, it should consider a 60 [ = ((0 - 300 + 540) mod 360) - 180] degree
; turn to the left instead of a 300 [ = 300 - 0] degree turn to the right.
fd 1 + acceleration ; moves birds forward
end
我认为最简单的方法是保留尽可能多的原始代码，然后查看更改代码行
[set dist distance turtle i
因为乌龟意味着世界上的任何海龟，而不仅仅是我小组中的那些乌龟。
要做到这一点，我考虑过将我的agentset'group-members'临时转换为品种，以便我可以用groupmember i替换单词turtle i ...例如
create-groupies-from group-members
[set dist distance group-member i
但是，当我尝试使用create-<breed>-from代码时，我收到了错误 '已经定义了名为CREATE-GROUPIES-FROM的内容，甚至当这个品种被包括在品种中时[]
因此，我尝试调整代码，只考虑聚集其他群组成员，如下所示：
to flock
set id-within-group 0 ; initializes variables used in finding neighbors
set distance_1 1000
set distance_2 1000
set distance_3 1000
set group-members turtle-set turtles with [group-id = [group-id] of myself] ; set those sharing a group id as my group-members
(foreach (sort group-members) (n-values count group-members [?]) [ ; set each turtle in my agentset (group) with an ID no. 0 - n?
ask ?1 [ set id-within-group ?2 ]
])
repeat count group-members ; repeat the next step as many times as there are group members
[ifelse not (id-within-group = who)
[
set dist distance turtle id-within-group
print "turtle id-within-group"
print turtle id-within-group
print dist
ifelse dist < distance_3
[ifelse dist < distance_2
[set neighbor_3 neighbor_2
set distance_3 distance_2
ifelse dist < distance_1
[set neighbor_2 neighbor_1
set distance_2 distance_1
set neighbor_1 id-within-group
set distance_1 dist]
[set neighbor_2 id-within-group
set distance_2 dist
repeat 2 [fd 0]]]
[set neighbor_3 id-within-group
set distance_3 dist
repeat 4 [fd 0]]]
[repeat 6 [fd 0]]]
[repeat 7 [fd 0]]
print "_____________________"
print "ID" print self
print "group-id" print group-id
]
;]
set acceleration 0 ; sets acceleration to zero
set target_1 calc_target neighbor_1 distance_1 ; calculates targets (see procedure below) based on
set target_2 calc_target neighbor_2 distance_2 ; the headings and positions of each neighbor;
set target_3 calc_target neighbor_3 distance_3 ; acceleration is set by the calc_target procedure
; (below) calculates a bird's composite target, weighting each individual target by the inverse
; of distance to that neighbor
set target atan ((sin target_1) * 100 / distance_1 + (sin target_2) * 100 / distance_2 + (sin target_3) * 100 / distance_3)
((cos target_1) * 100 / distance_1 + (cos target_2) * 100 / distance_2 + (cos target_3) * 100 / distance_3)
; (below) turns bird towards target by a fraction specified by "flexibility"; random motion is also
; included to an extent governed by "noise". The "... + 540) mod 360) - 180" clause is necessary to
; ensure that the bird turns in the right direction. For example, if a bird is oriented at 0 degrees
; and its target is 300 degrees, it should consider a 60 [ = ((0 - 300 + 540) mod 360) - 180] degree
; turn to the left instead of a 300 [ = 300 - 0] degree turn to the right.
lt (flexibility / 100 * (((heading - target + 540) mod 360) - 180)) + (random noise) - (random noise)
fd 1 + acceleration ; moves birds forward
end
在我修改过的示例中，虽然这些组确实正确地分配了一个单独的ID组成员，但他们并没有蜂拥而至。此外，邻居，距离和target_1 - _3都具有相同的值，这是错误的。
我想知道是否有人对我遇到的一些问题有任何见解？任何解决方案或相关文献都会非常感激 - 谢谢！
答案 0 :(得分：0)
首先，只是为了让您知道，海龟可以通过要求相关海龟将其品种属性设置为新品种来改变品种：ask ... [set breed ...]（例如，请参阅NetLogo字典中的品种基元）。 / p>
尽管如此，我认为你和一个＆＃39;组合这样做是对的。属性和使用代理集来做你想要的。这就像你想要的但不正确。希望其他人能够改进它。
turtles-own
[ groupID ]
to setup
create-turtles 20
[ set groupID random 2
set xcor random-xcor
set ycor random-ycor
]
ask turtles
[ let close-friends min-n-of 3 turtles with [groupID = [groupID] of myself] [distance myself]
ask close-friends
[ face myself
forward 1
]
]
end
这段代码（我认为）在同一组中找到三只最接近的海龟。然而，它将这三个移向了要求的乌龟，而不是让要求乌龟移动到那三个。我不知道如何做反过来而不用列表做丑陋的东西。
答案 1 :(得分：0)
turtles-own
[ groupID ]
to setup
clear-all
create-turtles 20
[ set groupID random 2
set xcor random-xcor
set ycor random-ycor
set color 155 - 10 * groupID
]
reset-ticks
end
to go
ask turtles
[ let close-friends min-n-of 3 turtles with [groupID = [groupID] of myself] [distance myself]
let friend-list sort-on [distance myself] close-friends
foreach friend-list
[ face ?
forward 1
]
]
tick
end