NetLogo高效地创建具有任意程度分布的网络

问题描述

这是对> NetLogo高效方法的后续问题创建固定数量的链接.专注于避免嵌套的询问"之后，我现在有了这段代码.它效率更高，但是创建了太多链接.显然是逻辑错误，但我看不到.

This is a follow up question to NetLogo Efficient way to create fixed number of links. Having focussed on avoiding the nested `ask', I now have this code. It is much more efficient, but is creating too many links. Clearly a logic error but I can't see it.

globals
[ candidates
  friends
]

to setup
  clear-all
  set friends 2
  create-turtles 5000
  set candidates turtles
  make-network
end

to make-network
  ask turtles
  [ let new-links friends - count my-links
    if new-links > 0
    [ let chosen n-of min (list new-links count other candidates) other candidates
      create-links-with chosen [ hide-link ]
      set candidates other candidates
      ask chosen [ if my-links = friends [ set candidates other candidates ] ]
    ]
  ]
end

推荐答案

很好的解决方案！请注意，other candidates实际上遍历代理集中的每个代理，因此在许多代理中它仍然会变慢，但是要比，因为它没有这些代理运行代码.它的运行速度让我印象深刻！

Nice solution! Note that other candidates actually iterates through every agent in the agentset, so it will still get slow with lots of agents, but less so than in your other question since it's not having those agents run code. I'm really impressed by how quickly it runs!

该错误.在这一部分:

if my-links = friends [ set candidates other candidates ]

我认为您忘记了my-links前面的count.

I think you forgot a count in front of my-links.

代码仍然可以以小于friends的某些代理结束，因为在到达代理之前，世界可能已经不存在了.不知道您对此有多关心.您只需清除链接，然后重试，直到获得正确的号码为止.只要friends很小，就可以了.

The code can still end up with some agents with less than friends, since the world can be out of candidates by the time it gets to them. Not sure how much you care about this. You could just clear the links and retry until you have the right number. That should be okay as long as friends is pretty small.

请注意，您可以通过将set candidates other candidates放在开头这样来加快代码的速度:

Note that you can speed the code up a little bit by putting the set candidates other candidates at the beginning like so:

set candidates other candidates
if new-links > 0
[ let chosen n-of min (list new-links count candidates) candidates
  create-links-with chosen [ hide-link ]
  ask chosen [ if my-links = friends [ set candidates other candidates ] ]
]

那样，您不必多次计算other candidates.

That way, you avoid having to calculate other candidates several times.

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