I have a list that looks like this List(List(0, 2), List(0, 3), List(2, 3), List(3, 2), List(3, 0), List(2, 0))), note this list will only contain pairs and will not contain duplicate pairs. I want to sort the list in descending order by the second item in each sub list in this larger list. If there are duplicate values I don't really which comes first.
For this instance the answer could look like List(List(0,3), List(2,3), List(0,2), List(3,2), List(3,0), List(2,0))
My idea was looping through the larger list and get a list containing each second item in each pair and sort those but I am having trouble keeping track of which second item in each pair belong to which pair afterwards. Perhaps there is a more clever way?
A simple solution would be:
list.sortBy(-_.last)
You can simply do:
list.sortBy(-_(1))
If the lists are always length 2, I would use tuples instead of lists. Then it is just a matter of using sortBy
scala> val l1 = List(List(0, 2), List(0, 3), List(2, 3), List(3, 2), List(3, 0), List(2, 0))
l1: List[List[Int]] = List(List(0, 2), List(0, 3), List(2, 3), List(3, 2), List(3, 0), List(2, 0))
scala> val l2 = l1.map(x => (x(0), x(1)))
l2: List[(Int, Int)] = List((0,2), (0,3), (2,3), (3,2), (3,0), (2,0))
scala> l2.sortBy(-_._2)
res1: List[(Int, Int)] = List((0,3), (2,3), (0,2), (3,2), (3,0), (2,0))
list.sortBy( sublist => -1 * sublist.tail.head )
Related
In Scala, what is the best way to add an element to a list while making sure that the list always contains latest n elements.
So if list is (1, 2, 3, 4, 5) and n = 5, then adding 6 should result in (2, 3, 4, 5, 6).
One possible way could be:
list = list ::: List(6)
list = list.takeRight(5)
Are there any better ways? Also, is there a better data-structure for maintaining such frequently changing collection?
It sounds like a fixed size Circular Buffer would satisfy your need. I think apache commons provides some implementation.
A solution in scala using List could be:
scala> def dropHeadAndAddToList[T](obj: T, list: List[T]):List[T] = {
list.drop(1) :+ obj
}
dropHeadAndAddToList: [T](obj: T, list: List[T])List[T]
scala> val a = List(1,2,3,4,5)
a: List[Int] = List(1, 2, 3, 4, 5)
scala> dropHeadAndAddToList(6, a)
res0: List[Int] = List(2, 3, 4, 5, 6)
I have a list of lists with the following data
val list = List(List("a","b","c"),List("d","e","f"),List("a","a","a"))
I want to disicover how many different data do I have in each position of the sublists
1 -> List("a","d","a")
2 -> List("b","e","a")
3 -> List("c","f","a")
Is there a way to do that? It doesn't need to be indexed, but I need the amount of different values per sublist index, the result could also be
2 // different values a and d
3 // different values b, e and a
3 // different values c, f and a
As I noted in a comment on Jhonny Everson's (almost right but now deleted) answer, you can use transpose to get a list of the columns, and then distinct to remove duplicates:
scala> list.transpose.map(_.distinct.size)
res0: List[Int] = List(2, 3, 3)
This will throw an exception if all the lists don't have the same size, but that's probably what you want.
scala> list.transpose.map(_.toSet.size)
res0: List[Int] = List(2, 3, 3)
The output of list.transpose is List(List(a, d, a), List(b, e, a), List(c, f, a)) which gives you the structure you want then map each List to a Set to get the unique elements and count them.
If you want unique elements per position, then you can use zipWithIndex and reverse the outputs.
scala> list.transpose.map(_.distinct).zipWithIndex.map(t => t._2 -> t._1)
res1: List[(Int, List[String])] = List((0,List(a, d)), (1,List(b, e, a)), (2,List(c, f, a)))
Here is the code you wanted:
var lOfl = List(List.range(1,5), List(2,2,2,3), Nil)
//> lOfl : List[List[Int]] = List(List(1, 2, 3, 4), List(2, 2, 2, 3), List())
for {
l <- lOfl
} yield l.toSet.count(_ => true) //> res1: List[Int] = List(4, 2, 0)
A sample list of list of Int is created as lOfl; for loop iterates through each list of Int; toSet converts list to set eliminating duplicate elements. then count function does as the name says it all.
OR use the Scala's most used collection function (:P :P), map, like below,
lOfl.map(l => l.toSet.count(_ => true) ) //> res2: List[Int] = List(4, 2, 0)
Suppose you are given the following list: {1,0,0,3,4,0,8,0,5,6,0}. Is there any way I can assign a particular index to all the 0s in the list in SCALA? This index must then be used as a parameter to another function.
Not exactly sure what you mean, but perhaps this will give you some ideas:
scala> val list = List(3, 4, 0, 0, 3, 0, 2)
list: List[Int] = List(3, 4, 0, 0, 3, 0, 2)
scala> val indexed = list.zipWithIndex
indexed: List[(Int, Int)] = List((3,0), (4,1), (0,2), (0,3), (3,4), (0,5), (2,6))
scala> val zeroIndices = indexed collect { case (value, index) if value == 0 => index }
zeroIndices: List[Int] = List(2, 3, 5)
Bonus:
scala> zeroIndices map list
res1: List[Int] = List(0, 0, 0)
I'm brand new to Scala, having had very limited experience with functional programming through Haskell.
I'd like to try composing a list of all possible pairs constructed from a single input list. Example:
val nums = List[Int](1, 2, 3, 4, 5) // Create an input list
val pairs = composePairs(nums) // Function I'd like to create
// pairs == List[Int, Int]((1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1) ... etc)
I tried using zip on each element with the whole list, hoping that it would duplicate the one item across the whole. It didn't work (only matched the first possible pair). I'm not sure how to repeat an element (Haskell does it with cycle and take I believe), and I've had trouble following the documentation on Scala.
This leaves me thinking that there's probably a more concise, functional way to get the results I want. Does anybody have a good solution?
How about this:
val pairs = for(x <- nums; y <- nums) yield (x, y)
For those of you who don't want duplicates:
val uniquePairs = for {
(x, idxX) <- nums.zipWithIndex
(y, idxY) <- nums.zipWithIndex
if idxX < idxY
} yield (x, y)
val nums = List(1,2,3,4,5)
uniquePairs: List[(Int, Int)] = List((1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5))
Here's another version using map and flatten
val pairs = nums.flatMap(x => nums.map(y => (x,y)))
List[(Int, Int)] = List((1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2) (5,3), (5,4), (5,5))
This can then be easily wrapped into a composePairs function if you like:
def composePairs(nums: Seq[Int]) =
nums.flatMap(x => nums.map(y => (x,y)))
I tried googling for this but all I got were stories about minor celebrities. Given the lack of documentation, what is a DList?
It's a Difference List, along the lines of "Difference List as functions"
scala> val (l1, l2, l3) = (List(1, 2, 3), List(4, 5, 6), List(7, 8, 9))
l1: List[Int] = List(1, 2, 3)
l2: List[Int] = List(4, 5, 6)
l3: List[Int] = List(7, 8, 9)
Efficient prepending:
scala> l1 ::: l2 ::: l3
res8: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
Inefficient appending. This creates an intermediate list (l1 ++ l2), then ((l1 ++ l2) ++ l3)
scala> l1 ++ l2 ++ l3 // inefficient
res9: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
DList stores up the appends, and only needs to create one complete list, effectively invoking:
scala> List(l1, l2, l3) reduceRight ( _ ::: _)
res10: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9)
Difference lists are a list-like data structure that supports O(1) append operations.
Append, and other operations that modify a list are represented via function composition of the modification functions, rather than directly copying the list.
An example, from Haskell's dlist library:
-- Lists as functions
newtype DList a = DL { unDL :: [a] -> [a] }
-- The empty list
empty = DL id
-- The append case: composition, a la Hughes
append xs ys = DL (unDL xs . unDL ys)
-- Converting to a regular list, linear time.
toList = ($[]) . unDL
The technique goes back at least to Hughes 84, A novel representation of lists and its application to the function "reverse", R. John Hughes, 1984., where, he proposes representing lists as functions, and append as function composition, allowing e.g. reverse to run in linear time. From the paper:
It's a data type in the non-canonical scalaz package, useful for typed lists with constant-time access at both ends. (The trick is to google for "scala" AND "dlist".)
From the project page of scalaz:
DList, a data type for representing elements of the same type with constant time append/prepend operations.