I have two lists that will be created during runtime. I want to combine the lists that have been made so that the data can be accessed later on within the code , with the end goal of simplifying my code and improving my model efficiency. Can lists be concatenated by the inclusion of one within the other or is there another way?
Thanks.
The usual way to concatenate lists is by using the sentence primitive. This will give you a new list made with the elements of your two original lists, like in Jen's answer.
Alternatively, you could use the list primitive to build a list with your two original lists included as sub-lists.
The following example shows both methods:
to setup
let list1 [ 1 2 3 ]
let list2 [ 4 5 6 ]
print sentence list1 list2 ; will print: [1 2 3 4 5 6]
print list list1 list2 ; will print: [[1 2 3] [4 5 6]]
end
Which one you should prefer depends, of course, on what you want to do with it...
The sentence command can combine two lists without brackets left
to setup
let mylist1 [1 2 3]
let mylist2 [4 5 6]
set mylist1 sentence mylist1 mylist2
show mylist1
end
Related
I want to do multiple where query without effect data. I want to get data that include at least 1 data per array. Pseudo code
data =[1,3]
array1 = [1,2]
array2 = [3,4]
if(data.IsIntersect(array1) and data.IsIntersect(array2))
IsIntersect checks are there a intersection beetween arrays
I did so far
queryBuilder.andWhere(
'properties.id IN (:...sizeIds) AND properties.id IN (:...colorIds)',
{ sizeIds: [1, 2], colorIds: [3, 4] },
);
It returns empty because firstly checks properties for 'sizeIds' then it checks for 'colorIds'. For example
properties includes 1,3
check for sizeIds, returns 1
check for colorIds, return empty
How can I do that with typeORM?
How can properties.id be 1 and 3? And if it is, how could 1 or 3 be in both? You're asking for the impossible.
I assume you mean to ask for when properties.id is 1 or 3, because if it is [1,3] then you should use the postgres array syntax {1,3} & the ANY keyword (some variation on this: Check if value exists in Postgres array).
tldr, I think all you need is brackets and OR instead of AND:
queryBuilder.andWhere(
'(properties.id IN (:...sizeIds) OR properties.id IN (:...colorIds))',
{ sizeIds: [1, 2], colorIds: [3, 4] },
);
If properties.id is in fact an array, then please add the entity definition to your question. If you want to merge the rows where properties.id is in the list you will need a GROUP BY (https://orkhan.gitbook.io/typeorm/docs/select-query-builder).
I am trying something (in netlogo), but it is not working. I want a value of a position from a list of numbers. And I want to use the number that comes out of it to retrieve a name from a list of names.
So if I have a list like [1 2 3 4] en a list with ["chicken" "duck" "monkey" "dog"]
I want my number 2 to correspond with "duck".
So far, my zq is a list of numbers and my usedstrategies is a list of names.
let m precision (max zq) 1
let l position m zq
let p (position l zq) usedstrategies
But when I try this the result will be false, because l is not part of usedstrategies.
Ideas?
You need the item primitive to select from the list after matching on the other list. I am not sure what the precision line is for. However, here is a self contained piece of code that I think demonstrates what you want to do. Note that NetLogo counts positions from 0, not 1. I also used arbitrary numbers in the list so you don't get confused between the number in the list and its position.
to testme
let usedstrategies (list "chicken" "duck" "monkey" "dog")
let zq (list 5 6 7 8)
let strategynum position 7 zq
let thisstrategy item strategynum usedstrategies
type "Selected strategy number " type strategynum
type " which is " print thisstrategy
end
Jen's solution is perfectly fine, but I think this could also be a good use case for the table extension. Here is an example:
extensions [table]
to demo
let usedstrategies ["chicken" "duck" "monkey" "dog"]
let zq [5 6 7 8]
let strategies table:from-list (map list zq usedstrategies)
; get item corresponding with number 7:
print table:get strategies 7
end
A "table", here, is a data structure where a set of keys are associated with values. Here, your numbers are the keys and the strategies are the values.
If you try to get an item for which there is no key in the table (e.g., table:get strategies 9), you'll get the following error:
Extension exception: No value for 9 in table.
Here is a bit more detail about how the code works.
To construct the table, we use the table:from-list reporter, which takes a list of lists as input and gives you back a table where the first item of each sublist is used as a key and the second item is used as a value.
To construct our list of lists, we use the map primitive. This part is a bit more tricky to understand. The map primitive needs two kind of inputs: one or more lists, and a reporter to be applied to elements of these lists. The reporter comes first, and the whole expression needs to be inside parentheses:
(map list zq usedstrategies)
This expression "zips" our two lists together: it takes the first element of zq and the first element of usedstrategies, passes them to the list reporter, which constructs a list with these two elements, and adds that result to a new list. It then takes the second element of zq and the second element of usedstrategies and does the same thing with them, until we have a list that looks like:
[[5 "chicken"] [6 "duck"] [7 "monkey"] [8 "dog"]]
Note that the zipping expression could also have be written:
(map [ [a b] -> list a b ] zq usedstrategies)
...but it's a more roundabout way to do it. The list reporter by itself is already what we want; there is no need to construct a separate anonymous reporter that does the same thing.
Say I have a list
b:1 1 2 3 4
and I want to find the location of the element in list b using another list
a:1 2
When I type in b in\ a, I got
11000b
00000b
where it should be
11000b
00100b
What is going on and how to get the desired answer?
Thanks in advance!
You need to use each-right /:
q)b in/:a
11000b
00100b
With b in\a the first output is getting passed back in as b. Effectively:
q)1 1 2 3 4 in 1
11000b
q)11000b in 2
00000b
You can also use each-both ':
q)in[b]'[a]
11000b
00100b
This question already has answers here:
NumPy selecting specific column index per row by using a list of indexes
(7 answers)
Closed 2 years ago.
Is there a better way to get the "output_array" from the "input_array" and "select_id" ?
Can we get rid of range( input_array.shape[0] ) ?
>>> input_array = numpy.array( [ [3,14], [12, 5], [75, 50] ] )
>>> select_id = [0, 1, 1]
>>> print input_array
[[ 3 14]
[12 5]
[75 50]]
>>> output_array = input_array[ range( input_array.shape[0] ), select_id ]
>>> print output_array
[ 3 5 50]
You can choose from given array using numpy.choose which constructs an array from an index array (in your case select_id) and a set of arrays (in your case input_array) to choose from. However you may first need to transpose input_array to match dimensions. The following shows a small example:
In [101]: input_array
Out[101]:
array([[ 3, 14],
[12, 5],
[75, 50]])
In [102]: input_array.shape
Out[102]: (3, 2)
In [103]: select_id
Out[103]: [0, 1, 1]
In [104]: output_array = np.choose(select_id, input_array.T)
In [105]: output_array
Out[105]: array([ 3, 5, 50])
(because I can't post this as a comment on the accepted answer)
Note that numpy.choose only works if you have 32 or fewer choices (in this case, the dimension of your array along which you're indexing must be of size 32 or smaller). Additionally, the documentation for numpy.choose says
To reduce the chance of misinterpretation, even though the following "abuse" is nominally supported, choices should neither be, nor be thought of as, a single array, i.e., the outermost sequence-like container should be either a list or a tuple.
The OP asks:
Is there a better way to get the output_array from the input_array and select_id?
I would say, the way you originally suggested seems the best out of those presented here. It is easy to understand, scales to large arrays, and is efficient.
Can we get rid of range(input_array.shape[0])?
Yes, as shown by other answers, but the accepted one doesn't work in general so well as what the OP already suggests doing.
I think enumerate is handy.
[input_array[enum, item] for enum, item in enumerate(select_id)]
How about:
[input_array[x,y] for x,y in zip(range(len(input_array[:,0])),select_id)]
I am trying to get the value of a mode of a list. I know I can get the mode of a list by using the mode operator in a way such as "show modes [1 2 2 2 2 3 4]" which will report 2 as the mode, but how would I get the count for that mode, 4 in this case since their are 4 2's. I suppose I could get the mode and iterate through the list checking to see if each number is equal the mode, but I am just wondering if there is a simpler way.
Thanks in advance!
A combination of length and filter should get you what you want:
let xs [1 2 2 3 3 4]
foreach modes xs [
let mode ?
let n length filter [ ? = mode ] xs
print (word "mode: " mode ", occurrences: " n)
]
Will output:
mode: 2, occurrences: 2
mode: 3, occurrences: 2
(Of course, by definition, each mode will have the same number of occurrences.)