I have a list as follows :
[[178 440] [175 440] [160 468] [160 440]]
The values in each sub-list correspond to coordinates x, y.
I would like to obtain a list that contains only unique values of x and y.
From the list above, the results will be:
[[178 440] [160 468]]
Thanks in advance for your help.
Given your example, I assume that by "unique", you mean "not encountered so far".
Here is one way to do it, using the oh so flexible reduce:
to go
let pairs [[178 440] [175 440] [160 468] [160 440]]
print reduce [
ifelse-value (is-unique? ?2 ?1)
[ lput ?2 ?1 ]
[ ?1 ]
] fput [] pairs
end
to-report is-unique? [ pair other-pairs ]
report
(not member? (first pair) (map first other-pairs)) and
(not member? (last pair) (map last other-pairs))
end
The is-unique? reporter should be fairly easy to understand: it uses map to extract the first item (the x) of each of the other-pairs, and check that the first item of pair is not a member? of that. It performs the same check for the last item (the y).
The reduce part is a bit trickier, but not that hard either. We add an empty list at the front of your list of pairs (fput [] pairs) that we are going to use as an "accumulator" for the pairs that meet our criteria. This accumulator is going to be ?1 inside the reporter passed to reduce, and ?2 is going to be each of our pairs in turn. So what we do inside it is just check if ?2 is unique relative to the pairs accumulated so far. If it is, we add it to our list of accumulated pairs. If it's not, we leave the list as it was and just move on to the next ?2.
Related
"nw:weak-component-clusters" in the Networks extension will return a list of weakly connected agentsets. I would like to output the number of turtles in the biggest of these.
So
show nw:weak-component-clusters
observer: [(agentset, 15 turtles) (agentset, 20 turtles) (agentset, 16 turtles)]
would return 20.
Is there an easy way to do this please?
This isn't pretty but it will work:
to find_max
let my_list []
let my_max 0
let turt_list nw:weak-component-clusters
foreach turt_list [x -> ask x [set my_list lput count x my_list]]
set my_max max my_list
show my_max
end
There is a simpler approach using map:
to-report count-of-largest-cluster
report max (map count nw:weak-component-clusters)
end
map takes a reporter and a list as inputs, and reports a list whose items are the result of the input reporter being run for every item of the input list.
nw:weak-component-clusters is a list of agentsets, therefore map count nw:weak-component-clusters is a list of each agentset's count. Note that the parentheses in my solution are optional and only there for readability.
I would like to make the sum = the total of pollen recieved by a plant from other plants (Donnors) which is stored in a list of a list (own by each turtle = plant).
The following code make an error (when computing the sum):
OF expected input to be an agent or agentset but got the list
[[119.05593 50 50] [301.25853 50 50] [30.23906 50 50] [460.525845 50
50] [55.16717 50 50] [301.25853 50 50]] instead.
Does any one could help me about the mistake in the line "set Tot_pol sum ..." ?
Many thanks for your help.
to check-pol [m] ;; we check the pollen recieved by the two morphs
set Donnors [] ;; empty list of pollen donnors
ask zsps with [morph = m] ;; morph of the pollen reciever
[
set totpol 0
;; check for pollen donnors and morph for compatiblity within a radius :
ask zsps with[distance myself <= 20 and morph != m]
[
set totpol (NMaleFlowers * 100 * item round (distance myself) pollination-list) ;; the farther the less pollen
set Donnors lput [ (list totpol NMaleFlowers NFemFlowers)] of myself Donnors
]
set Tot_pol sum [ item (position 0 Donnors) Donnors ] of Donnors ;; total of pollen recieved
]
end
Luke's answer is good and should fix your problem. I suspect, however, that you are going to be doing lots of these types of sums. You may wish to set up a to-report that you can use for whichever item you want to sum over, just by passing the item number and the name of the list of lists. It would look like this:
to-report sum-item [#pos #listoflists ]
let items map [ x -> item #pos x ] #listoflists
report reduce [ [a b] -> a + b] items
end
The first line extracts the relevant item (remember index from 0) into a new list which the second line sums.
You would then use it with set Tot_pol sum-item 0 Donnors
Here's an answer that is not actually responding to your question. Instead, it is a more NetLogo-ish way of doing what I think you are trying to do with your code.
to check-pol [m]
ask zsps with [morph = m]
[ let senders zsps with [distance myself <= 20 and morph != m]
set totpol sum [NMaleFlowers * 100 * round (distance myself)] of senders
]
end
Your code gets into levels of ask that I think are unnecessary. What I think you are doing with your list is keeping track of the pollen donors. But an agentset is a cleaner approach and then you can simply pull out the information you want from the agentset using of.
Further, when you ask zsps with[distance myself <= 20 and morph != m] to set variable values in your code, then THOSE agents (not the receiving agent) are the ones having their variables changed. I think you are trying to take the perspective of the receiver of pollen, who looks around and received pollen from the other agents that are close enough. So the receiving agent should have the value changed.
This is not tested.
I'm not 100% sure what you're after here (you may want to look at the Minimum, Complete, and Verifiable Example guidelines), but if I'm reading you right you want the sum of the first item for each entry in the Donners list.
As to why your approach didn't work- NetLogo is telling you with that error that you've used of with a list, but of only works with agents or agentsets. Instead, you have to use a list processing approach. The simplest way might be to use sum in conjunction with map first in order to get what you need:
to sum-first-item
let example-list [ [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ] ]
let sum-of-firsts sum map first example-list
print sum-of-firsts
end
To translate to Donnors, try:
set Tot_pol sum map first Donnors
That should work, but without reproducible a code example I can't check.
I want to add multiple items from one list to another list, that is organized in one big list, like a matrix.
let targetlists (list firstlist seccondlist thirdlist)
So in my double while loop, I added this code;
set (item x targetlists) lput (item y sourcelist) (item x targetlists)
Sadly it gives me the following error:
This isn't something you can use "set" on.
I found out that it has to do with how I select the target list, as the following code does work, but doesn't do what I want:
set firstlist lput (item y sourcelist) firstlist
JenB is right that, in general, you use replace-item. However, replacing your while loops with map will be much more effective.
I'm not entirely sure what you're trying to do, but it looks like you're trying to put the elements of sourcelist onto the end of the lists in targetlists. Even if that's not what you're doing, this should point you in the right direction:
set targetlists (map [ [ source-item target-row ] ->
lput source-item target-row
] sourcelist targetlists)
This will iterate through the items of sourcelist and targetlists together, calling lput on the pairs.
Also, there's a handy shortcut where, if a reporter already does what you want, you can pass it directly to map. So you can condense this to:
set targetlists (map lput sourcelist targetlists)
Now, given that you mentioned nested whiles and you're indexing into the two lists with two different indices, you might be trying to put the entire contents of sourcelist onto the ends of each of the targetlists. If that's the case, you can just do
set targetlists map [ l -> (sentence l sourcelist) ] targetlists
If I'm totally off, and you're trying to do something completely different, just let me know in the comments and I'll update my answer.
I hope this is a simple solution, but I'm having a difficult time with it.
Problem:
I would like to weight the probability of something occurring by an variable not a constant
Setup
My agent is a farm.
Farms own four variables that represent the
number of cows, goats, pigs, and sheep on it.
When a farm wants to
remove an animal, I'd like the likelihood to remove a member of a
particular species to be directly proportional to quantity of each
species on the farm (i.e. if there are 7 goats, 2 cows, and 1 pig,
there is a 70% probability of taking a goat and a zero percent
probability of taking a sheep)
I have found formula like this for when you know the exact numerical weight that each value will have:
to-report random-weighted [values weights]
let selector (random-float sum weights)
let running-sum 0
(foreach values weights [
set running-sum (running-sum + ?2) ; Random-Weighted Created by NickBenn
if (running-sum > selector) [
report ?1
]
])
end
and the methods described in the rnd extension. But both of these throw the "expected a constant" error when i put "Cow" in instead of a constant.
Something like:
to example1
let values ["Cow" "Sheep" "Goat" "Pig"]
let probabilities [2 0 7 1]
let indices n-values length values [ ? ] ; Made by Nicolas Payette
let index rnd:weighted-one-of indices [ item ? probabilities ]
let loca item index values
end
works well, but if I were to replace it with:
to example1
let values ["Cow" "Sheep" "Goat" "Pig"]
let probabilities [Num-Cows Num-Sheep Num-Goats Num-Pigs]
let indices n-values length values [ ? ] ; Made by Nicolas Payette
let index rnd:weighted-one-of indices [ item ? probabilities ]
let loca item index values
end
it fails.
Alan is right: you need to use the list primitive (as opposed to just brackets) when you want to construct a list from anything else than constants.
I would add two things to that:
The latest version of the rnd extension has two sets of primitives: one for agentsets, and one for lists. So you should probably update and use the rnd:weighted-one-of-list primitive.
Your code is based around using indices to pick an item. That's fine, but that's not the only way to do it.
You could also have something like:
to example1
let values ["Cow" "Sheep" "Goat" "Pig"]
let probabilities (list Num-Cows Num-Sheep Num-Goats Num-Pigs)
let loca first rnd:weighted-one-of-list (map list values probabilities) last
end
This may be a bit trickier to understand, but here is the gist of it:
The (map list values probabilities) expression takes both your values list and your probabilities list and "zips" them together using the list primitive, resulting in a list of pairs: [["Cow" 2] ["Sheep" 0] ["Goat" 7] ["Pig" 1]].
We pass the last reporter to the rnd:weighted-one-of-list primitive to tell it that the last (i.e., second) item of each of these pairs should be used as the probability.
Since rnd:weighted-one-of-list operates on a list of pairs, the item it returns will be a pair (e.g., ["Goat" 7]). We are only interested in the first item of the pair, so we extract it with the first reporter.
Note that we use the NetLogo's concise syntax for tasks when passing list as an argument to map and last as an argument to rnd:weighted-n-of. You could replace list with [ (list ?1 ?2) ] and last with [ last ? ], but it would be uglier.
In my code all turtles own n-features represented by a n-tuple (a1,a2,...,an). where each ai can take values 0 or 1.
I have created some links between turtles. If two turtles share k-features (coordinate-wise matching) and there is a link between them then we call the link as k-link.
How can I find for each k (between 0 to n) how many k-links are there in total?
You don't tell us much about how you have structured your code, so I'm going to assume that your n-tuples are implemented as lists (which would make the most sense in NetLogo).
Here is a full example:
turtles-own [ a ]
links-own [ k ]
globals [ n ]
to setup
ca
set n 5
crt 10 [ ; create turtles with random feature lists
set a n-values n [ random 2 ]
]
ask turtles [ ; make a full network
create-links-with other turtles
]
ask links [ ; calculate k for all links
set k k-of-feature-lists ([a] of end1) ([a] of end2)
]
foreach n-values (n + 1) [ ? ] [ ; display number of k-links
show (word ? "-links: " count links with [ k = ? ])
]
end
to-report k-of-feature-lists [ a1 a2 ]
report length filter [?] (map = a1 a2)
end
Apart from k-of-feature-lists, this is fairly trivial code. What k-of-feature-lists does is to:
transform two lists of features into a single list of booleans containing a true value if the corresponding element is equal in both feature lists and false if it is not. This is accomplished using map and the concise task syntax for =;
filter the list of booleans to keep only the true values;
report the length of that filtered list, which is equal to the number of features that where the same in a1 and a2;
There are plenty of other ways to do that (some more efficient) but this one is nice and concise.