Netlogo: is ´map´ limited to two lists? - netlogo

When I try the following command on NL 6.1.1:
(map + [1 2 3] [2 4 6] [7 8 9])
I get [3 6 9]as a result instead of [10 14 18], which means that only the first two lists are considered in the sum.
Anyone knows how to correct this and get all three lists to be considered ?


Per the documentation in the NetLogo dictionary, the map reporter does support multiple lists, but map expects the provided anonymous reporter to take a number of arguments equal to the number of lists you provide. In your case you cannot use the + shortcut since + takes 2 arguments, and we have 3 lists.
So instead you can do this:
(map [ [a b c] -> a + b + c ] [1 2 3] [2 4 6] [7 8 9])
If you need to handle adding an arbitrary number of lists (as in a list variable that could have 2, 3, 4, or more lists inside of it depending on your program), you can try this out using the reduce reporter:
set vals [[1 2 3] [2 4 6] [7 8 9] [10 11 12]]
reduce [ [a b] -> (map + a b) ] vals

Related

Matlab Matrixaddress [duplicate]

This question already has answers here:
How to obtain the mirror image of an array (MATLAB)?
(4 answers)
Closed 2 years ago.
During my course I came across that expression:
A(:,end:-1:1)
I have trouble to understand and read the morphemic structure of the 2nd Operand "end;-1;1"
Lets take the example:
A=[1 2 3; 4 5 6; 7 8 9]
I am aware of:
A(:).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is :.
A(1,:).. Outputs [1 2 3; 4 5 6; 7 8 9] as columns Operator is , then , .
A(:,1).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is , beforehand : .
A(:,end:-1:1)
Output in Matlab show me: 3x3 Matrix.
How am I supposed to read the structure?
Graphem: : ..show me the rows,
Graphem: end:-1 .. ??
Graphem: :1 ..
Somehow ":" was for me the operator for show all of the elements.
It makes sense to me that the "Operand1 , Operand2" shows me the 2 Dimensional Matrix.
First Idea:
The end:-1:1 expression seemed to me like a loop. So -1, 0, 1 => **3x Elements** ?
But when I type
A(1,end:3)
it only shows me the 3rd row.
Second Idea:
A(end:-1:1,1)
It shows me the inverted Matrix..
My background:
I am a undergraduate student from the field of language.
I build in my freetime the 8-Bit Sap1 according Ben Eater.
So I am familiar with program memory or instruction memory.
I understand only the result, but not how it is achieved by the MATLAB compiler.
Someone said to me that the "Matrixaddressing is somehow optimized".
Looking forward to a helpful answer in each step. :)
Thanks in advance!

The end keyword in matrix indexing indicates index of last element in the corresponding dimension. So, A(:,end:-1:1) simply means A(:, size(A, 2):-1:1), which in you example (A=[1 2 3; 4 5 6; 7 8 9]), is equivalent to A(:, 3:-1:1).
But to understand what it does, you need to know what 3:-1:1 does. It creates a subrange. You already know 1:3 creates [1, 2, 3]. 1:3 is simplified form of 1:1:3: rangeStrart:increment:rangeEnd. Now, 3:1 or 3:1:1 creates an empty vector, because rangeStart is greater than rangeEnd. To create [3, 2, 1] you need to use a negative step: 3:-1:1.
So, A(:,end:-1:1) means A(:, [3, 2, 1]), which inverts order of rows of A. Also, A(:,end:3) means A(:, 3:3) and eventually A(:, 3), which returns 3rd row of A.
Edit: about your misunderstandings, addressed by #CrisLuengo
I am aware of:
A(:).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is :.
A(1,:).. Outputs [1 2 3; 4 5 6; 7 8 9] as columns Operator is , then , .
A(:,1).. Outputs [1 2 3; 4 5 6; 7 8 9] as rows. Operator is ,beforehand : .
A(3, 2) is the element in the 3,2 position (third row, second column) of A
A(1, :) is equivalent to A(1, 1:size(A, 2)) and A(1, 1:end) and is the first row of A
A(:, 1) is equivalent to A(1:size(A, 1), 1) and A(1:end, 1) and is the first column of A
A(:) is equivalent to A(1:numel(A)) and is a single column containing all elements of A

In MATLAB, when accessing an array, it is accessed as A(row#,col#). row# and col# can either be integers or a vector of integers. If they are integers then one spot in the matrix is accessed. If they are vectors then MATLAB will loop through the vector and choose spots in A which correspond to the integers in the vector.
end:-1:1 creates a vector which contains the integers ranging from the number of columns (in this case because you put this vector in the col section: A(row#,col#)) to 1. Ex: 4x5 matrix, end:-1:1 would be [5 4 3 2 1].
When you put : in the row part of the matrix, that means you access all rows of the matrix.
Here's an example of A(:,end:-1:1)
The col# vector (:) is [1 2 3] and the row# vector (end:-1:1) is [3 2 1]
A = [1 2 3;
4 5 6;
7 8 9]
A(:,end:-1:1)
[3, 5, 7]

I think you're overthinking this slightly.
If we have a vector
A = [1 2 3]
and we call A(end:-1:1), then we get a vector [3 2 1]. The indexing has returned the same vector, with the values reversed. If we now have a matrix
A = [1 2 3; 4 5 6; 7 8 9]
and call A(:, end:-1:1), we get the matrix with the same values in each row, but now the columns have been reversed to give
A(:, end:-1:1) = [3 2 1; 6 5 4; 9 8 7].
Recall what the colon means in this context.
If we define a vector, v = (1:10), we get a vector with the first element being 1, the last element being 10 and each value in between being integers in steps of 1. If we instead define v = (1:2:10), we get the same, but the elements are separated by 2, not 1.
end:-1:1 is a vector made in just the same way. The first number is the final element in the row of A, and the final number is the first element in the row. Each number is separated by a value of -1. If we try
v = 10:-1:1
we get [10 9 8 7 6 5 4 3 2 1]. If we call v(2:4) we get the second, third and fourth elements of v. If we call v(1:end), we simply get v. If we call v(end:-1:1), we return v, with the elements in the reversed order.
Edit A typo.

How to remove several items from a unsorted list in Netlogo

so i'm a bit struggling with Lists in Netlogo, so basically i've two lists and i want to remove the items that are in List 1 from List 2, for example:
List 1 : [8 6 9 7 1 3]
List 2: [5 9 8]
Resulting List : [6 7 1 3]
I've tried the following code but it returns an empty list:
if List 2 != []
[
foreach List 2
[
let p position ? List 1
if p = true
[
set List 1 remove-item p List 1
]
]
]
Any ideas ?

A combination of member? and filter will get you there:
let list1 [8 6 9 7 1 3]
let list2 [5 9 8]
let result filter [ x -> not member? x list2 ] list1
print result
Will print the desired:
[6 7 1 3]
Tip: whenever you find yourself trying to use an index for anything in NetLogo, you are probably not doing things in the optimal way. NetLogo has tons of functions (like filter, in this case) that operate on lists as a whole. There is rarely a need to explicitly loop through them.

How to form subset form a set of numbers in Netlogo

Dear Netlogo community,
I am looking to generate subset of a set of numbers.
for example if a set is [1 2 3 4 5] then subset would be [1 2] [1 3] [1 4] [1 5] [ 1 2 3] [1 2 4]....... I know we can generate very easily by using bit manipulation in java. But I have no idea how to implement in Netlogo. I am doomed. Any help would be really appreciated. Thanks

This is easiest to solve using recursion:
to-report subsets [xs]
if empty? xs [ report [[]] ]
let recurse subsets butfirst xs
report sentence recurse
map [fput first xs ?] recurse
end
The basic idea is that if you want the subsets of [1 2 3], first you find all the subsets of [2 3]. All of them are themselves subsets of [1 2 3], plus if you stick the 1 on the front of each of them, the resulting lists are part of the answer too.
sample run:
observer> print subsets [1 2 3]
[[] [3] [2] [2 3] [1] [1 3] [1 2] [1 2 3]]

How do I assign a cell array to multiple matrices?

I am relatively new to matlab, and I was wondering if there was a simpler way to do the following:
Given mycellarray = {[1 2 3 4] [5 6 7 8] [9 10 11 12] [13 14 15 16]}, I would like to assign each matrix inside mycellarray to a separate variable. Is there any faster/better/shorter way to do it than this?
a = cell2mat(mycellarray(1,1))
b = cell2mat(mycellarray(1,2))
c = cell2mat(mycellarray(1,3))
d = cell2mat(mycellarray(1,4))
Thanks in advance!

[a,b,c,d]=mycellarray{:}
The {:} makes a comma seperated list of the cell array, which an be assigned to individual variables.
relevant documentation pages:
colon operator
comma seperated list

Here a short example on how to do what you want with the MATLAB function deal.
a = {[1 2 3] [4 5 6] [7 8 9]}
[aa bb cc] = deal(a{:})

How can I append to a vector in Octave?

When ever I have to append to a vector I am doing this.
A = [2 3 4]
A = [A; 3 4 5]
I was wondering if there are any inbuilt functions for this or more elegant ways of doing this in Octave.

The builtin functions are cat, vertcat, and horzcat, found on pages 380-381 of the Octave documentation (v 3.8). They are essentially equivalent to what you have though.
octave:5> A = [2 3 4];
octave:6> A = [A; 3 4 5]
A =
2 3 4
3 4 5
octave:7> B = [4 5 6];
octave:8> B = vertcat(B,[5 6 7])
B =
4 5 6
5 6 7
Another (again equivalent) way would be to directly use matrix indexing (see page 132)
octave:9> C = [6 7 8];
octave:10> C(end+1,:) = [7 8 9]
C =
6 7 8
7 8 9

I think that the most efficient is to use this built in function that you have posted in the question(I'm relying on other experts in octave i did not check it completely; The standard is that matrix operations are generally faster than iterative ones, I don't know what the inner mechanism that allows this to be enabled yet).
Because a vector is a type of matrice, this solution will work for concatinating vectors (of any type) too:
vector = [vector ; value]