I want to iterate over a scala list in an incremental way, i.e. the first pass should yield the head, the second the first 2 elements, the next the first 3, etc...
I can code this myself as a recursive function, but does a pre-existing function exist for this in the standard library?
You can use the .inits method to get there, albeit there may be performance issues for a large list (I haven't played around with making this lazy):
scala> val data = List(0,1,2,3,4)
data: List[Int] = List(0, 1, 2, 3, 4)
scala> data.inits.toList.reverse.flatten
res2: List[Int] = List(0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4)
You can use the take like so:
scala> val myList = 1 to 10 toList
myList: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> for(cnt <- myList.indices) yield myList.take(cnt+1)
res1: scala.collection.immutable.IndexedSeq[List[Int]] = Vector(List(1), List(1, 2), List(1, 2, 3), List(1, 2, 3, 4), List(1, 2, 3, 4, 5), List(1, 2, 3, 4, 5, 6), List(1, 2, 3, 4, 5, 6, 7), List(1, 2, 3, 4, 5, 6, 7, 8), List(1, 2, 3, 4, 5, 6, 7, 8, 9), List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
OK, since I've whined enough, here's an iterator version that tries reasonably hard to not waste space or compute more than is needed at at one point:
class stini[A](xs: List[A]) extends Iterator[List[A]] {
var ys: List[A] = Nil
var remaining = xs
def hasNext = remaining.nonEmpty
def next = {
val e = remaining.head
remaining = remaining.tail
ys = e :: ys
ys.reverse
}
}
val it = new stini(List(1, 2, 3, 4))
it.toList
//> List[List[Int]] =
// List(List(1), List(1, 2), List(1, 2, 3), List(1, 2, 3, 4))
Try: for((x, i) <- l.view.zipWithIndex) println(l.take(i + 1))
if you need something side-effected (I just did println to give you an example)
Related
So my problem is simple. I have come across this many times, and my brain is unable to find a solution.
How can I unpack a list into another list, for an indefinite amount of variables?
Here is what I mean.
val list1 = List(List(7, 4), List(7, 6))
val list2 = List(List(1), List(5), List(8))
val desired_list1 = List(List(1, 7, 4), List(5, 7, 4), List(8, 7, 4))
val desired_list2 = List(List(1, 7, 6), List(5, 7, 6), List(8, 7, 6))
//** The desired_list1 and 2 must be a List[List[Int]] it cannot be List[List[Any]]
//Here's my attempt, which oddly enough completely ignores all elements of list1(0) which are not the first(7).
val attempt = list2.map(i => i +: list1(0)).map(j => j.collect{ case k:Int => k; case l # a :: b => a}).map(m => m.map{ case i:Any => i.toString.toInt})
//The result is
attempt: List[List[Int]] = List(List(1, 7), List(5, 7), List(8, 7))
//while it should be:
val desired_list1 = List(List(1, 7, 4), List(5, 7, 4), List(8, 7, 4))
I need a way to unpack that is not manual, please do not tell me to do this:
val attempt = list2.map(k => k +: list1(0)).map{ case (k, List(x, y)) => (k, x, y)}
Basically, list1 could have any number of elements. e.g
val list1 = List(List(99, 83, 2, 3, 4), List(99, 83, 2, 5 7))
However, these numbers are never repeated, so I guess it could be a set as well. But I don't know much about sets or if it'd help in any way.
This seems to be what you want:
val lists: List[List[Int]] = List(List(7, 4), List(7, 6))
val prefixes: List[List[Int]] = List(List(1), List(5), List(8))
val res: List[List[Int]] = for{
prefix <- prefixes.flatten
rest <- lists
} yield prefix :: res
// res: List[List[Int]] = List(List(1, 7, 4), List(1, 7, 6), List(5, 7, 4), List(5, 7, 6), List(8, 7, 4), List(8, 7, 6))
Unless, you actually want a list that contains your desired_list1 and desired_list2 together, in which case you need:
val res3 = lists.map{ rest =>
prefixes.flatten.map{prefix =>
prefix :: rest
}
}
/// res3: List[List[List[Int]]] =
// List(
// List(
// List(1, 7, 4), List(5, 7, 4), List(8, 7, 4)
// ),
// List(
// List(1, 7, 6), List(5, 7, 6), List(8, 7, 6)
// )
// )
I am not sure exactly why you would want that but try this out:
val list1 = List(List(7, 4), List(7, 6))
val list2 = List(1, 5, 8)
list2.flatMap(element=> list1.map(innerList=> element:: innerList))
output: List[List[Int]] = List(List(1, 7, 4), List(1, 7, 6), List(5, 7, 4), List(5, 7, 6), List(8, 7, 4), List(8, 7, 6))
In Scala, what would be the right way of selecting elements of a list based on the position of two elements? Suppose I have the list below and I would like to select all the elements between 2 and 7, including them (note: not greater than/smaller than, but the elements that come after 2 and before 7 in the list):
scala> val l = List(1, 14, 2, 17, 35, 9, 12, 7, 9, 40)
l: List[Int] = List(1, 14, 2, 17, 35, 9, 12, 7, 9, 40)
scala> def someMethod(l: List[Int], from: Int, to: Int) : List[Int] = {
| // some code here
| }
someMethod: (l: List[Int], from: Int, to: Int)List[Int]
scala> someMethod(l, 2, 7)
res0: List[Int] = List(2, 17, 35, 9, 12, 7)
Expected output:
For lists that don't contain 2 and/or 7: an empty list
Input: (1, 2, 2, 2, 3, 4, 7, 8); Output: (2, 2, 2, 3, 4, 7)
Input: (1, 2, 3, 4, 7, 7, 7, 8); Output: (2, 3, 4, 7)
Input: (1, 2, 3, 4, 7, 1, 2, 3, 5, 7, 8); Output: ((2, 3, 4, 7), (2, 3, 5, 7))
Too bad that the regex-engines work only with strings, not with general lists - would be really nice if you could find all matches for something like L.*?R with two arbitrary delimiters L and R. Since it doesn't work with regex, you have to build a little automaton yourself. Here is one way to do it:
#annotation.tailrec
def findDelimitedSlices[A](
xs: List[A],
l: A,
r: A,
revAcc: List[List[A]] = Nil
): List[List[A]] = {
xs match {
case h :: t => if (h == l) {
val idx = xs.indexOf(r)
if (idx >= 0) {
val (s, rest) = xs.splitAt(idx + 1)
findDelimitedSlices(rest, l, r, s :: revAcc)
} else {
revAcc.reverse
}
} else {
findDelimitedSlices(t, l, r, revAcc)
}
case Nil => revAcc.reverse
}
}
Input:
for (example <- List(
List(1, 2, 2, 2, 3, 4, 7, 8),
List(1, 2, 3, 4, 7, 7, 7, 8),
List(1, 2, 3, 4, 7, 1, 2, 3, 5, 7, 8)
)) {
println(findDelimitedSlices(example, 2, 7))
}
Output:
List(List(2, 2, 2, 3, 4, 7))
List(List(2, 3, 4, 7))
List(List(2, 3, 4, 7), List(2, 3, 5, 7))
You're looking for slice:
# l.slice(2, 7)
res1: List[Int] = List(2, 17, 35, 9, 12)
# l.slice(2, 8)
res2: List[Int] = List(2, 17, 35, 9, 12, 7)
Hi I am new to scala and getting silly doubts, I have a list of lists which looks like this
(4,List(List(2, 4, 0, 2, 4), List(3, 4, 0, 2, 4), List(4, 0, 1, 2, 4)))
I want to get the lists which starts with 4. How to do it.
you use filter to traverse through the List and apply your predicate on each list to check if first elem is 4.
example:
scala> val (data, options) = (4, List(List(2, 4, 0, 2, 4), List(3, 4, 0, 2, 4), List(4, 0, 1, 2, 4)))
data: Int = 4
options: List[List[Int]] = List(List(2, 4, 0, 2, 4), List(3, 4, 0, 2, 4), List(4, 0, 1, 2, 4))
scala> options.filter(_.headOption.contains(data))
res0: List[List[Int]] = List(List(4, 0, 1, 2, 4))
Also see: Scala List.filter with two conditions, applied only once
There are several ways.
Here is another
listOfLists.collect{ case l # 4 :: _ => l}
Potentially more powerful because we can filter on the first n elements, e.g.
listOfLists.collect{ case l # 4 :: 0 :: 1 :: _ => l}
If you have a Tuple (Int, List[List[Int]]), and want to return Lists that start with the Int provided in the start, for this case 4:
I would recommend you do something like this:
val myTuple = (4,List(List(2, 4, 0, 2, 4), List(3, 4, 0, 2, 4), List(4, 0, 1, 2, 4)))
myTuple._2.filter(_.headOption.contains(myTuple._1))
And this will return List(List(4, 0, 1, 2, 4))
What we are doing here is, we are first accessing the List[List[Int]] in the Tuple by doing myTuple._2 then we filter to remove Lists that don't have a head value as 4 - which we passed in as myTuple._1.
Note we are using headOption instead of head to get the first element in a List, this is to handle exceptions where no List contains the value provided in the start, for this case 4 (more details on this can be found here http://www.bks2.com/blog/2012/12/31/head_vs_headOption/)
val t = (4, List(List(2, 4, 0, 2, 4), List(3, 4, 0, 2, 4), List(4, 0, 1, 2, 4)))
t._2.filter(_.head==t._1)
In REPL:
scala> t._2.filter(_.head==t._1)
res5: List[List[Int]] = List(List(4, 0, 1, 2, 4))
This feels like a peculiar problem and I am very new to Scala, so I don't know how to ask the right questions in order to get progress on this problem.
As a demonstration, say I have a list of lists like this:
val data = List(List(1, 2, 3, 4), List(1, 2, 2, 3, 4), List(1, 2, 3, 3, 3, 4), List(1, 2, 3, 4), List(2, 3, 4))
and I want to be able to reduce it down to a List of integers that looks mostly like a distinct set of the multiple lists, with one exception: where each list has more than one of each integer, I want to represent that in the final list. So as a general rule, the list with the most representations of that integer will have "their" repetitions of that integer in the final list. So that would ideally give:
List(1, 2, 2, 3, 3, 3, 4)
I know I can do data.flatten.distinct and get:
List(1, 2, 3, 4)
but that's not what I want and I know there's probably a bit more work to get to the desired result.
I am wondering if there is a good way to achieve the desired result in a functional way in scala.
Try this
val data = List(List(1, 2, 3, 4), List(1, 2, 2, 3, 4), List(1, 2, 3, 3, 3, 4), List(1, 2, 3, 4), List(2, 3, 4))
val map = data.map(_.groupBy(identity)).foldLeft(Map[Int, List[Int]]()) {
case (r, c) => r ++ c.map {
case (k, v) => k -> (if (v.size > r.getOrElse(k, List()).size) v else r(k))
}
}.values.flatten
//> map : Iterable[Int] = List(2, 2, 4, 1, 3, 3, 3)
It does not maintain the ordering. After this you can call to sort this.
Maybe this is cleaner
data.flatMap(_.groupBy(identity)).groupBy(_._1).mapValues(_.sortBy(_._2.size).reverse(0)._2).values.flatten
//> res0: Iterable[Int] = List(2, 2, 4, 1, 3, 3, 3)
I don't quite get it, but you can just order elements
data.flatten.sorted
Which would give you
List(1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4)
if you want them ordered by number of encounters, you can do it like this:
data.flatten.groupBy(k => k).mapValues(_.size).toList.sortBy(_._2).map(_._1)
which would give you
List(1, 4, 2, 3)
I'm trying to get List(0,1,2,...n)
Is there a cleaner/better way than:
scala> List(0 to 9)
res0: List[scala.collection.immutable.Range.Inclusive] = List(Range(0, 1, 2, 3, 4,
5, 6, 7, 8, 9))
scala> List(0 to 9).flatten
res1: List[Int] = List(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
The best way might be:
(0 to 9).toList
scala> List.range(0, 10)
res0: List[Int] = List(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
Also
List(0 to 9: _*)
I suspect though that List.range is the most efficient one.