How to create a friendship network between different Breeds - netlogo

Question

Dear NetLogo Specialists,
Please can you give me some advice? This is the first NetLogo Model that I have put together and the first time I have posted a query on this forum.
I am currently adapting the Wilensky Voting model from the NetLogo library.
I am interesting in creating two breeds. The first are agents who are part of an “Echo Chamber”, the second Breed are “not in Echo Chamber”.
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breed [ Echos Echo ]
breed [ NotEchos NotEcho ]
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As with the Wilensky model, I would like the agents to base their voting decision on the voting choices of their eight friends. I can set up the a network where “Echo Chamber” agents are friends with “Echo Chamber” and “not in Echo Chamber” agents are friends with “not in Echo Chamber” agents . I achieved this adapting code from JenB. The current code is below.
The limitation of this method is that friendships are not across Breeds.
Ideally, I would like a piece of code that allows me to select of the eight friends a breed has and how many are from their own breed and how many other breed. For example, I would like agents in the “Echo Chamber” to have six “Echo Chamber” agent friends and two “not in Echo Chamber” agent friends.
Any help would be much appreciated.
Many thanks for your time,
Paul
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to make-network
;; THE PURPOSE OF THIS IS TO CREATE A NETWORK WHERE EACH AGENT IN THE ECHO CHAMBER IS FRIENDS WITH
;; 8 OTHER AGENTS IN THE ECHO CHAMBER
ask NotEchos
[ let needed 8 - count my-links
if needed > 0
[ let NotEchoscandidates other NotEchos with [ count my-links < 8]
create-links-with n-of min (list needed count NotEchoscandidates) NotEchoscandidates
[ hide-link]
]
]
;; THE PURPOSE OF THIS IS TO CREATE A NETWORK WHERE EACH AGENT NOT IN THE ECHO CHAMBER IS FRIENDS WITH
;; 8 OTHER AGENTS NOT IN THE ECHO CHAMBER
ask Echos
[ let needed 8 - count my-links
if needed > 0
[ let Echoscandidates other Echos with [ count my-links < 8]
create-links-with n-of min (list needed count Echoscandidates) Echoscandidates
[ hide-link]
]
]
;; THE PURPOSE OF THESE IS TO CREATE MEASURES TO VALIDATE AND QUALITY ASSURE THE ABM
ask NotEchos [ set num_NotEchos_friends count my-links ]
ask NotEchos [ set total_NotEchos_vote (sum [vote] of link-neighbors) ]
ask Echos [ set num_Echos_friends count my-links ]
ask Echos [ set total_Echos_Vote (sum [vote] of link-neighbors) ]
ask Echos [show total_Echos_vote]
ask NotEchos [show total_NotEchos_vote]
End
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Answer 1

You have NotEcho to NotEcho, Echo to Echo and Echo to NotEcho links and will need separate constructions for each. So far you are limiting your candidates to the right breed, but you haven't adjusted your counts of whether they still have slots available. You need something like (not tested):
ask NotEchos
[ let needed 6 - count my-link-neighbors with [breed = NotEchos]
if needed > 0
[ let NotEchoscandidates other NotEchos with [ count my-neighbors with [breed = NotEchos] < 6]
create-links-with n-of min (list needed count NotEchoscandidates) NotEchoscandidates
[ hide-link]
]
]
Note that I switched to my-link-neighbors instead of my-links because it's the turtles at the end that have a breed, not the links. I also changed 8 to 6 because you want 6 of the same breed.
Also, you will need to think about how many of each breed you have, so that the numbers all come out and there are sufficient to make links.
Finally, are Echo and NotEcho really different turtles? Think about two different opinion groups A and B. A talking to A is likely to be echoing similar sentiments and B talking to B as well, but A and B talking to each other are not echoing. In your model, what do you mean when a NotEcho is linked to a NotEcho. It might be that you really want two opinion groups, with preferential attachment to the same group, and then whether they are same or different at each end is a function of the link (relationship) rather than the turtle.