I am trying to write a piece of code which asks some of the turtles to
calculate 2 parameters, compare them and then if one is lower than the
other, subtracts a characteristic of turtles by 1.
here is the code I wrote but I receive this error: "this isn't something you
can use set on netlogo"
set monthly-electricity-consumption random-float((monthly-electricity-demand * 1.2) - (monthly-electricity-demand * 0.8)) + (monthly-electricity-demand *
0.8)
ask turtles with [shape = "triangle"] [
if monthly-electricity-consumption > monthly-electricity-demand [
set [company-loyalty-level] of self company-loyalty-level - 1]]
Do you have any advice?
I'm assuming that the line
set [company-loyalty-level] of self company-loyalty-level - 1]]
is the line that generates the error. NetLogo does not allow one turtle to directly set the variables of another turtle by using the of construction. E.g.,
ask turtle 1 [set [company-loyalty-level] of turtle 2 (company-loyalty-level - 1)]
breaks that rule. In your case, by using self, turtle 2 and turtle 1 are the same turtle, but NetLogo will still throw that error. The line that you want is simply,
set company-loyalty-level company-loyalty-level - 1]]
Once you are within an ask, the variable is understood to be the one associated with the turtle being asked. The of self is not necessary.
Related
I have an agent set named "open patches", which I found using the following, in which "number_open_patches" is a specific radius depending on "length":
`ask turtles ['
let my_patch self
set number_open_patches ((50 * (length ^ 2)) / 100) / resolution
let open_patches patches in-radius number_open_patches
I found the distance of these patches to a particular turtle "my_patch" using:
let distance_patch [distance my_patch] of open_patches
I then calculate "patch attractiveness", which is just the distance of each of the patches ("distance_patch") in "open_patches" plugged into the equation below:
let patch_attractiveness ( map [[x] -> exp ((x / open_patches) * (log e (.2 / 1)))] distance_patch)
I would finally like to select the patch within "open_patches" that has the greatest value of "patch_attractiveness". I am using the code below:
let destination max-one-of open_patches [patch_attractiveness]
I get an error as it returns 'nobody'. Why would this be? How do I know that I would actually be selecting the right patch as "distance_patch" is just a list in random order?
EDIT: I also tried the code below, but am getting a "division by zero" error when I try to calculate patch attractiveness.
ask turtles [
let my_patch self
set number_open_patches ((50 * (length ^ 2)) / 100) / resolution
let open_patches patches in-radius number_open_patches
ask open_patches [
set distance_patch [distance self] of my_patch
set patch_attractiveness exp ((distance_patch / number_open_patches) * (log e (.2 / 1)))
]
]
You didn't give enough code to see what the problems are. Your line with the exponential dividing by an agent set doesn't seem syntactically correct. You don't show the code for calculating open_patches_in_radius and I suspect there are none and the rest of the line using that is just a distraction.
That said, I'd suggest going the opposite way.
let the patches own a distance and an attractiveness variable
but "distance" is a reserved word so it needs a different name.
ask the open-patches to determine their distance to your "my_patch" turtle and store that in their distance slot. ( instead of asking your patch )
compute whatever global things you need for your formula
ask open-patches to set their attractiveness based on their distance ( already computed) and some exponential decay formula that apparently includes some global factor.
then finally let your destination be the open-patch with the max attractiveness, which now has a much simpler formula using
"max-one-of agentset [reporter]" syntax, such as
let dest max-one-of open-patch [ attractiveness ]
or possibly you want only open-patches within a given radius, but again depending on the scattering of open-patches and the size of the radius, there may be none. Check for none when setting open-patches-in-radius because that's a valid subcase.
I'm new to NetLogo and I'm trying to create 2 sub-breeds (denoted by different shapes) within each breed for 2 breeds total (i.e. sharks and fishes). The chunks of code work as expected when ran individually, however, when running both chunks the first sub-breed of the fishes does not seem to initialise in the interface tab. For some reason the initialisation of the shark breed seems to interfere with the initiation of the fishes breed.
Any idea what i'm doing wrong? Code below.
; Create the agents
breed [sharks shark]
breed [fishes fish]
; Create the agents' variables
fishes-own
[
x0
y0
]
globals
[
species
species-f
]
to setup
; Always start with this
clear-all
reset-ticks
; Create sharks and species
create-sharks N-sharks ; N-sharks is a slider
[
set color blue
setxy random-xcor random-ycor
set species N-sharks
ask sharks
[
set shape "default"
set size 2.5
]
ask sharks with [who >= (species * (1 / 2))]
[
set shape "square"
set size 2
]
ask sharks with [who < (species * (1 / 6))]
[
set shape "star"
set size 3
]
] ; End create sharks and species
; Create fishes
create-fishes N-fishes
[
setxy random-xcor random-ycor
set x0 xcor
set y0 ycor
set species-f (N-fishes * species-ratio)
ifelse who <= species-f
[
set shape "sheep"
set size 5
]
[
set shape "cow"
set size 3
]
set color white
] ; End create fishes
end
There seem to be a number of misunderstandings about what code is run when by NetLogo, and what some pieces of code do: when you use create-turtles *number* [*commands*] (or with breeds, as in your case), commands is run once by every turtle being created (first the new turtles are created, then they run the commands in turn).
That means that every time you use ask sharks within the create-sharks' command block, every new shark will ask all other sharks to set shape and size. For example, if you create 100 sharks, instead of setting the shape and size only once you are doing it 10,000 times (100 * 100).
So you need to put all those commands out of their respective create-<breed>'s command blocks; for example:
create-sharks N-sharks [
set color blue
setxy random-xcor random-ycor
]
ask sharks [
set size 2.5
]
ask sharks with [who >= (species * (1 / 2))] [
set shape "square"
set size 2
]
ask sharks with [who < (species * (1 / 6))] [
set shape "star"
set size 3
]
This is still improvable code, but it shows how to achieve exactly the same thing by doing it once and not doing it N-sharks ^ 2 times.
A better way to do it is to bring those commands back inside the create-<breed>'s command block, but letting each agent carry out the task for itself only. That is, without using ask sharks but using ifelse, so that each shark will check its own condition:
create-sharks N-sharks [
set color blue
setxy random-xcor random-ycor
(ifelse
who >= (species * (1 / 2)) ; the first condition
[set shape "square" set size 2]
who < (species * (1 / 6)) ; the second condition
[set shape "star" set size 3]
; else
[set size 2.5])
]
All of this applies to the other breed too.
It is generally said that using who is a sign that the code should be improved, but let's not focus on this now.
You will have noticed that I omitted the part where you set species N-sharks. This is because I think there is another misunderstanding here and it is not clear to me what you wanted to do: species (such as species-f for the other breed) is a global variable. You are basically asking each of the 100 sharks (again, 100 for example) to do the same thing: set the value of a global variable to equal the value of another global variable. In this case, you are asking each shark to set species 100. This seems very unnecessary, especially considering that N-sharks is a slider used for setup and thus probably won't be changed during the simulation (which means that there is probably no need to store the current value of N-sharks as a separate global variable).
Why are you basing the repartition of your sub-breeds on the value of species? What do you want species to represent? Is it correct for it to be a separate variable from N-sharks? If yes, then it is not clear what is its point; if no, then it can be eliminated.
You need to make sure that whatever you wanted to do with N-sharks & species, and with N-fishes and species-f, is better reflected in your code.
Also because I think this is the reason why your first fish sub-breed isn't showing. First of all, what is species-ratio? It is not present in your example, but it seems to be relevant for your question. In any case, if your first fish sub-breed isn't showing, it means that there is no fish who satisfies the condition who <= species-f.
This doesn't surprise me. who numbers are progressive for turtles: if you create 15 sharks and later you create 10 fish...
... the oldest shark will have who = 0
... the youngest shark will have who = 14
... the oldest fish will have who = 15
... the youngest fish will have who = 24
As you can see in this example, there is no fish for which who <= N-fishes (where N-fishes = 10). Let alone that in your case you set species-f (N-fishes * species-ratio) and, although you didn't tell what species-ratio is, I imagine it is a value between 0 and 1 - thus making the value of species-f even smaller.
The problem is that you are using who to determine your fishes. To quote the Netlogo programming guide: "Who numbers are assigned irrespective of breeds. If you already have a frog 0, then the first mouse will be mouse 1, not mouse 0, since the who number 0 is already taken."
As such, you need to take the number of sharks into account for determining the cutoff point for your fish species
set species-f (N-fishes * species-ratio) + N-sharks - 1
One other comment I have on your code is that you use the following structure:
create-sharks N-shark [
ask sharks [...]
]
Every shark executes the entire command block of create-sharks. That means that every shark asks every shark to do something. Thus follows that every shark sets their shape to default N-shark times.
It is much more efficient to use one of the following constructions
create-sharks N-shark [...]
ask sharks [...]
or
create-sharks N-shark [
if <condition> [<reshape-self>]
]
I'm new in Netlogo . I'm trying to use the code included in the Many Regions Example of the library. But I get one error in this procedure
to keep-in-region [ which-region ]
if region != which-region [
let region-min-pxcor first item (which-region - 1) region-boundaries
let region-max-pxcor last item (which-region - 1) region-boundaries
let region-width (region-max-pxcor - region-min-pxcor) + 1
ifelse xcor < region-min-pxcor [
set xcor xcor + region-width ]
[if xcor > region-max-pxcor [
set xcor xcor - region-width
]
]
]
The error I get is .... any advices? Thanks in advance
-1 no esta mas que o iqual a cero.
error while a-seller 31 running ITEM
called by procedure KEEP-IN-REGION
called by procedure ADJUST
called by procedure GO
called by Botón 'go'
The error you are getting ("-1 isn't greater than or equal to zero.", once translated to English) is caused by passing -1 as an index for the item primitive.
There are two lines in the code that make use of item:
let region-min-pxcor first item (which-region - 1) region-boundaries
let region-max-pxcor last item (which-region - 1) region-boundaries
As you can see (which-region - 1) is the expression passed to item as index. If you are getting -1, it must be because which-region = 0.
In the "Many Regions Example" model, region 0 is reserved for the patches that are not part of any regions, namely, the patches that act as region boundaries. The regions themselves are numbered from 1.
Look in your code for the place where keep-in-region is called and make sure that you are using a region number that is between 1 and the number of regions you have (inclusively).
Disclaimer: I originally wrote the "Many Regions" code example. Maybe I could have used a value like nobody for the patches outside of any region and number the actual regions starting from 0 instead of one, but I don't remember if I actively decided against it or just didn't think of it.
there is a question that I want to ask that is when I trying to type this code, I got the error that is
The > operator can only be used on two numbers, two strings, or two agents of the same type, but not on a number and a list.
What I want to ask is how can I fix this, the false happen at this line on the code :
if pri-lev > [pri-lev] of oppoint1 and pri-lev > [pri-lev] of oppoint2
I tried to change it into "cars-on" or "cars with" but they are all useless. I also try to find on the Netlogo dictionary but I found no results on the code for directing an agent on a specific path.
What I am trying to do here is when an agent comes to a specific section, it will check if any agents listed as "oppoint1"; "oppoint2"; "oppoint3"; "oppoint4" and then compare a value call pri-lev to others value for setting its decision on keeping on moving or stopping and wait for others.
These are the part of my code:
ask cars
[
let oppoint1 (cars-at (xcor + 2) (ycor + 2))
let oppoint2 (cars-at (xcor - 1) (ycor + 1))
let oppoint3 (cars-at (xcor - 2) (ycor + 1))
let oppoint4 (cars-at (xcor - 3) (ycor + 1))
ifelse oppoint1 != nobody and oppoint2 != nobody
[
if pri-lev > [pri-lev] of oppoint1 and pri-lev > [pri-lev] of oppoint2
[
set pri-lev 4
speed-up
]
]
[
if oppoint2 = nobody and oppoint3 = nobody and oppoint4 = nobody
[
set speed 1
fd speed
if turning = "Rtrue"
[
set heading heading + 90
speed-up
]
]
]
]
Sincerely, Minh
it seems that the reason you are getting this error is that you are comparing the attribute of one (the current car) to the attributes of many (the oppoint agent-sets). Your code now says something like "If my privilege is greater than the privilege of that group, do this thing..." The problem is that [pri-lev] of oppoint1 returns a list of the pri-lev of all members of the oppoint1 agentset, like [ 10 12 13 24 ], and Netlogo won't automatically iterate over that list and compare each item to the attribute of the asking turtle.
There are several ways to deal with this. For example, you could make sure that you only ever compare two turtles- maybe by making sure that you only ever have one turtle per patch at a given time. If you are going to potentially compare one agent to an agent-set, you can use the any? primitive to check if any members of the group you're looking at satisfy your conditional statement. For example, given this setup:
turtles-own [
pri-lev
]
to setup
ca
reset-ticks
crt 10 [
set pri-lev 1 + random 10
]
end
You can ask one-of your turtles to check if not any? of the turtles on the current patch have a higher pri-lev than the asking turtle. If none of them do, the current turtle will move forward. Otherwise, it will print that there is another turtle with a higher pri-lev on the current patch.
to compare-with
ask one-of turtles [
let other-turtles other turtles-here
ifelse not any? other-turtles with [ pri-lev > [pri-lev] of myself ] [
fd 1
]
[
print ("A turtle here has a higher pri-lev than I do." )
]
]
tick
end
I have seen on here how to create an intersection or union of two agentsets, but I am trying to say if any turtle in agentset a is in agentset b, return true. I was trying
ifelse (member? (one-of my-intersections) destination-intersections)
but I am pretty sure this is just testing if one element in my-intersections is in destination -intersections instead of testing every element. Is there some way to use a for each? Or is there another functionality I am unaware of?
Again, I have already referenced
NetLogo two agentsets operations
Thank you!!
The most straightforward way to write such a test is:
to-report test [ as bs ]
report any? as with [ member? self bs ]
end
To try it:
create-turtles 5
show test (turtle-set turtle 0 turtle 1 turtle 2) (turtle-set turtle 3 turtle 4)
show test (turtle-set turtle 0 turtle 1 turtle 2) (turtle-set turtle 0 turtle 3 turtle 4)
will show:
observer: false
observer: true
It's not the most efficient way, however, because the with clause builds an intermediate agentset that's not really needed.
A faster test would be:
to-report test [ as bs ]
let result false
ask as [
if member? self bs [
set result true
stop
]
]
report result
end
Edit:
I spoke too fast. As per the NetLogo wiki page on compiler architecture, the combination of any? and with does get optimized to exit early. Bottom line: you should use the first version (i.e., any? as with [ member? self bs ]).