i want this answer
if Input list is >> List( List("A","B"),List("A","C"),List("B","C"),List("B","D"))
Output should be >> List(List("A", "B","C"),List("B","C","D"))
i think it should be done as following>> all indices having first element same are grouped e.g if first element is "A" then group will be ("A","B","B","C").distinct = ("A","B","C")
Related
I thought this was an easy one but I cannot find any solution for it.
I have this vector called cues= ["R" "B" "C" "P" "Y" "G"];
from which I want to randomly select one value, but excluding one (or two) of the values each time.
For example, I would like to get a random value from the vector excluding the "R" value, or in a second condition I would like "R" and "Y" not to be selected from the sample.
I have tried using randsample and randperm but neither of them seems to include this option.
One intuitive way to achieve this would be taking your lists of all values and exclusions, and making a list of inclusions instead, then you can select from that list.
See the comments for each step:
cues= ["R" "B" "C" "P" "Y" "G"]; % All options
exclude = ["R" "Y"]; % Exclusions (could change in a loop or whatever)
include = setdiff( cues, exclude ); % Actual options without exclusions
selection = include( randi(numel(include)) ); % Random selection from options
It seems I found a way to do it.
For a string vector, it can be done using erase function. For example:
cues_minusR = erase (cues,'R');
For a number vector (e.g., cues = [1 2 3 4 5 6]), it works this way:
c1Col=1;
c2Col=2;
cues(c1Col, : ) = [];
cues([c1Col, c2Col], : ) = [];
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.
I have RDD of the following structure (RDD[(String,Map[String,List[Product with Serializable]])]):
This is a sample data:
(600,Map(base_data -> List((10:00 01-08-2016,600,111,1,1), (10:15 01-08-2016,615,111,1,5)), additional_data -> List((1,2)))
(601,Map(base_data -> List((10:01 01-08-2016,600,111,1,2), (10:02 01-08-2016,619,111,1,2), (10:01 01-08-2016,600,111,1,4)), additional_data -> List((5,6)))
I want to calculate the number of unique values of the 4th fields in sub-lists.
For instance let's take the first entry. The list is List((10:00 01-08-2016,600,111,1,1), (10:15 01-08-2016,615,111,1,5)). It contains 2 unique values (1 and 5) in the 4th field of sub-lists.
As to the second entry, it also contains 2 unique values (2 and 4), because 2 is repeated twice.
The resulting RDD should be of the format RDD[Map[String,Any]].
I tried to solve this task as follows:
val result = myRDD.map({
line => Map(("id",line._1),
("unique_count",line._2.get("base_data").groupBy(l => l).count(_))))
})
However this code does not do what I need. In fact, I don't know how to properly indicate that I want to group by 4th field...
You are quite close to the solution. There is no need to call groupBy, but you can access the item of the tuples by index, transform the resulting List into a Set and then just return the size of the Set, which corresponds to the number of unique elements:
("unique_count", line._2("base_data").map(bd => bd.productElement(4)).toSet.size)
I have following scala list-
List((192.168.1.1,8590298237), (192.168.1.1,8590122837), (192.168.1.1,4016236988),
(192.168.1.1,1018539117), (192.168.1.1,2733649135), (192.168.1.2,16755417009),
(192.168.1.1,3315423529), (192.168.1.2,1523080027), (192.168.1.1,1982762904),
(192.168.1.2,6148851261), (192.168.1.1,1070935897), (192.168.1.2,276531515092),
(192.168.1.1,17180030107), (192.168.1.1,8352532280), (192.168.1.3,8590120563),
(192.168.1.3,24651063), (192.168.1.3,4431959144), (192.168.1.3,8232349877),
(192.168.1.2,17493253102), (192.168.1.2,4073818556), (192.168.1.2,42951186251))
I want following output-
List((192.168.1.1, sum of all values of 192.168.1.1),
(192.168.1.2, sum of all values of 192.168.1.2),
(192.168.1.3, sum of all values of 192.168.1.3))
How do I get sum of second elements from list by grouping on first element using scala??
Here you can use the groupBy function in Scala. You do, in your input data however have some issues, the ip numbers or whatever that is must be Strings and the numbers Longs. Here is an example of the groupBy function:
val data = ??? // Your list
val sumList = data.groupBy(_._1).map(x => (x._1, x._2.map(_._2).sum)).toList
If the answer is correct accept it or comment and I'll explain some more.
I have a question : I know merge two list in SML but i can not do the total number of elements of the first list and the second list is less than n, append them fully and return the resulting list appended with 0’s, totaling n elements.
Sample :
f2([1,4,5],[3,6],7);
val it = [1,4,5,3,6,0,0] : int list // 7 elements
Thank you in advance..
Get the length of the two lists with LIST.length and compare the sum with the third argument.
I am not sure what to do if this sum is larger than the third argument but you get the idea.
if sum < n then list1 # list2 # 0.....