Is there any way to convert a nested arrayBuffer to a nested array in scala ? I tried using the toArray function, but it did not convert the nested arrayBuffers
I have an array buffer of type Any and following is my sample nested
ArrayBuffer(
ArrayBuffer(ArrayBuffer(ArrayBuffer(1, b), 5)))
The code below works for your specific test case. It converts nested ArrayBuffers that are the first element of the enclosing ArrayBuffer:
def convert(a: Any): Array[Any] = a match {
case ArrayBuffer(inner # ArrayBuffer(_*)) => Array(convert(inner))
case ArrayBuffer(head # ArrayBuffer(_*), tail # _*) => (convert(head) +: tail).toArray
case arr # ArrayBuffer(_*) => arr.toArray
}
val result = convert(ArrayBuffer(ArrayBuffer(ArrayBuffer(ArrayBuffer(1, "b"), 5))))
// result is Array(Array(Array(Array(1, "b"), 5)))
Perhaps the above will help you come up with a more general solution.
I have created a generic function to convert arraybuffer to array.This takes care of nested arrayBuffers also
def convert(a: ArrayBuffer[Any]): Array[Any] = {
val checkExistance = (x:ArrayBuffer[Any]) => x.zipWithIndex.collect{ case(x,y) if x.isInstanceOf[ArrayBuffer[Any]] => y}.toArray
//filter(x => x.isInstanceOf[ArrayBuffer[Any]])
val arr = checkExistance(a)
for(i <- arr) {
//if (a(i).isInstanceOf[ArrayBuffer[Any]]) {
val m = a(i).asInstanceOf[ArrayBuffer[Any]]
if(checkExistance(m).length > 0) {
a.update(i,(convert(m)).toArray)
}
else {
val n = m.toArray
a.update(i,n)
}
}
a.toArray
}
Related
I have an Iterator of Strings and would like to concatenate each element preceding one that matches a predicate, e.g. for an Iterator of
Iterator("a", "b", "c break", "d break", "e")
and a predicate of
!line.endsWith("break")
I would like to print out
(Group: 0): a-b-c break
(Group: 1): d break
(Group: 2): e
(without needing to hold in memory more than a single group at a time)
I know I can achieve this with an iterator like below, but there has to be a more "Scala" way of writing this, right?
import scala.collection.mutable.ListBuffer
object IteratingAndAccumulating extends App {
class AccumulatingIterator(lines: Iterator[String])extends Iterator[ListBuffer[String]] {
override def hasNext: Boolean = lines.hasNext
override def next(): ListBuffer[String] = getNextLine(lines, new ListBuffer[String])
def getNextLine(lines: Iterator[String], accumulator: ListBuffer[String]): ListBuffer[String] = {
val line = lines.next
accumulator += line
if (line.endsWith("break") || !lines.hasNext) accumulator
else getNextLine(lines, accumulator)
}
}
new AccumulatingIterator(Iterator("a", "b", "c break", "d break", "e"))
.map(_.mkString("-")).zipWithIndex.foreach{
case (conc, i) =>
println(s"(Group: $i): $conc")
}
}
many thanks,
Fil
Here is a simple solution if you don't mind loading the entire contents into memory at once:
val lines: List[List[String]] = it.foldLeft(List(List.empty[String])) {
case (head::tail, x) if predicate(x) => Nil :: (x::head) :: tail
case (head::tail, x) => (x::head ) :: tail
}.dropWhile(_.isEmpty).map(_.reverse).reverse
If you would rather iterate through the strings and groups one-by-one, it gets a little bit more involved:
// first "instrument" the iterator, by "demarcating" group boundaries with None:
val instrumented: Iterator[Option[String]] = it.flatMap {
case x if predicate(x) => Seq(Some(x), None)
case x => Seq(Some(x))
}
// And now, wrap it around into another iterator, constructing groups:
val lines: Iterator[Iterator[String]] = Iterator.continually {
instrumented.takeWhile(_.nonEmpty).flatten
}.takeWhile(_.nonEmpty)
I have a val assume
val a: Future[List[(List[Case_Class], List[Case_Class])]] = //some method call
In my tuple, there are two lists, say
listA = List[Case_Class] and listB = List[Case_Class].
Both the lists are having two elements at a time. What I need is applying some logic(ex: fold) to my val 'a' so as to achieve
Future[(List(Case_Class), List(Case_Class))]
What should be the logic that I must apply?
Something like
a.map(_.unzip match { case (x, y) => (x.flatten, y.flatten) })
Please excuse the unruly number of brackets in the below example but this seems to work:
case class Thing()
val a: Future[List[(List[Thing], List[Thing])]] = {
Future.successful(List((List(Thing()), List(Thing()))))
}
val b: Future[(List[Thing], List[Thing])] = {
a.map(x => (x.unzip._1.flatten, x.unzip._2.flatten))
}
Await.ready(b, Duration.Inf)
// scala.concurrent.Future[(List[Thing], List[Thing])] =
// Future(Success((List(Thing()),List(Thing()))))
More on unzip here.
I'm trying to 'group' a string into segments, I guess this example would explain it more succintly
scala> val str: String = "aaaabbcddeeeeeeffg"
... (do something)
res0: List("aaaa","bb","c","dd","eeeee","ff","g")
I can thnk of a few ways to do this in an imperative style (with vars and stepping through the string to find groups) but I was wondering if any better functional solution could
be attained? I've been looking through the Scala API but there doesn't seem to be something that fits my needs.
Any help would be appreciated
You can split the string recursively with span:
def s(x : String) : List[String] = if(x.size == 0) Nil else {
val (l,r) = x.span(_ == x(0))
l :: s(r)
}
Tail recursive:
#annotation.tailrec def s(x : String, y : List[String] = Nil) : List[String] = {
if(x.size == 0) y.reverse
else {
val (l,r) = x.span(_ == x(0))
s(r, l :: y)
}
}
Seems that all other answers are very concentrated on collection operations. But pure string + regex solution is much simpler:
str split """(?<=(\w))(?!\1)""" toList
In this regex I use positive lookbehind and negative lookahead for the captured char
def group(s: String): List[String] = s match {
case "" => Nil
case s => s.takeWhile(_==s.head) :: group(s.dropWhile(_==s.head))
}
Edit: Tail recursive version:
def group(s: String, result: List[String] = Nil): List[String] = s match {
case "" => result reverse
case s => group(s.dropWhile(_==s.head), s.takeWhile(_==s.head) :: result)
}
can be used just like the other because the second parameter has a default value and thus doesnt have to be supplied.
Make it one-liner:
scala> val str = "aaaabbcddddeeeeefff"
str: java.lang.String = aaaabbcddddeeeeefff
scala> str.groupBy(identity).map(_._2)
res: scala.collection.immutable.Iterable[String] = List(eeeee, fff, aaaa, bb, c, dddd)
UPDATE:
As #Paul mentioned about the order here is updated version:
scala> str.groupBy(identity).toList.sortBy(_._1).map(_._2)
res: List[String] = List(aaaa, bb, c, dddd, eeeee, fff)
You could use some helper functions like this:
val str = "aaaabbcddddeeeeefff"
def zame(chars:List[Char]) = chars.partition(_==chars.head)
def q(chars:List[Char]):List[List[Char]] = chars match {
case Nil => Nil
case rest =>
val (thesame,others) = zame(rest)
thesame :: q(others)
}
q(str.toList) map (_.mkString)
This should do the trick, right? No doubt it can be cleaned up into one-liners even further
A functional* solution using fold:
def group(s : String) : Seq[String] = {
s.tail.foldLeft(Seq(s.head.toString)) { case (carry, elem) =>
if ( carry.last(0) == elem ) {
carry.init :+ (carry.last + elem)
}
else {
carry :+ elem.toString
}
}
}
There is a lot of cost hidden in all those sequence operations performed on strings (via implicit conversion). I guess the real complexity heavily depends on the kind of Seq strings are converted to.
(*) Afaik all/most operations in the collection library depend in iterators, an imho inherently unfunctional concept. But the code looks functional, at least.
Starting Scala 2.13, List is now provided with the unfold builder which can be combined with String::span:
List.unfold("aaaabbaaacdeeffg") {
case "" => None
case rest => Some(rest.span(_ == rest.head))
}
// List[String] = List("aaaa", "bb", "aaa", "c", "d", "ee", "ff", "g")
or alternatively, coupled with Scala 2.13's Option#unless builder:
List.unfold("aaaabbaaacdeeffg") {
rest => Option.unless(rest.isEmpty)(rest.span(_ == rest.head))
}
// List[String] = List("aaaa", "bb", "aaa", "c", "d", "ee", "ff", "g")
Details:
Unfold (def unfold[A, S](init: S)(f: (S) => Option[(A, S)]): List[A]) is based on an internal state (init) which is initialized in our case with "aaaabbaaacdeeffg".
For each iteration, we span (def span(p: (Char) => Boolean): (String, String)) this internal state in order to find the prefix containing the same symbol and produce a (String, String) tuple which contains the prefix and the rest of the string. span is very fortunate in this context as it produces exactly what unfold expects: a tuple containing the next element of the list and the new internal state.
The unfolding stops when the internal state is "" in which case we produce None as expected by unfold to exit.
Edit: Have to read more carefully. Below is no functional code.
Sometimes, a little mutable state helps:
def group(s : String) = {
var tmp = ""
val b = Seq.newBuilder[String]
s.foreach { c =>
if ( tmp != "" && tmp.head != c ) {
b += tmp
tmp = ""
}
tmp += c
}
b += tmp
b.result
}
Runtime O(n) (if segments have at most constant length) and tmp.+= probably creates the most overhead. Use a string builder instead for strict runtime in O(n).
group("aaaabbcddeeeeeeffg")
> Seq[String] = List(aaaa, bb, c, dd, eeeeee, ff, g)
If you want to use scala API you can use the built in function for that:
str.groupBy(c => c).values
Or if you mind it being sorted and in a list:
str.groupBy(c => c).values.toList.sorted
I am looking for an approach to join multiple Lists in the following manner:
ListA a b c
ListB 1 2 3 4
ListC + # * § %
..
..
..
Resulting List: a 1 + b 2 # c 3 * 4 § %
In Words: The elements in sequential order, starting at first list combined into the resulting list. An arbitrary amount of input lists could be there varying in length.
I used multiple approaches with variants of zip, sliding iterators but none worked and especially took care of varying list lengths. There has to be an elegant way in scala ;)
val lists = List(ListA, ListB, ListC)
lists.flatMap(_.zipWithIndex).sortBy(_._2).map(_._1)
It's pretty self-explanatory. It just zips each value with its position on its respective list, sorts by index, then pulls the values back out.
Here's how I would do it:
class ListTests extends FunSuite {
test("The three lists from his example") {
val l1 = List("a", "b", "c")
val l2 = List(1, 2, 3, 4)
val l3 = List("+", "#", "*", "§", "%")
// All lists together
val l = List(l1, l2, l3)
// Max length of a list (to pad the shorter ones)
val maxLen = l.map(_.size).max
// Wrap the elements in Option and pad with None
val padded = l.map { list => list.map(Some(_)) ++ Stream.continually(None).take(maxLen - list.size) }
// Transpose
val trans = padded.transpose
// Flatten the lists then flatten the options
val result = trans.flatten.flatten
// Viola
assert(List("a", 1, "+", "b", 2, "#", "c", 3, "*", 4, "§", "%") === result)
}
}
Here's an imperative solution if efficiency is paramount:
def combine[T](xss: List[List[T]]): List[T] = {
val b = List.newBuilder[T]
var its = xss.map(_.iterator)
while (!its.isEmpty) {
its = its.filter(_.hasNext)
its.foreach(b += _.next)
}
b.result
}
You can use padTo, transpose, and flatten to good effect here:
lists.map(_.map(Some(_)).padTo(lists.map(_.length).max, None)).transpose.flatten.flatten
Here's a small recursive solution.
def flatList(lists: List[List[Any]]) = {
def loop(output: List[Any], xss: List[List[Any]]): List[Any] = (xss collect { case x :: xs => x }) match {
case Nil => output
case heads => loop(output ::: heads, xss.collect({ case x :: xs => xs }))
}
loop(List[Any](), lists)
}
And here is a simple streams approach which can cope with an arbitrary sequence of sequences, each of potentially infinite length.
def flatSeqs[A](ssa: Seq[Seq[A]]): Stream[A] = {
def seqs(xss: Seq[Seq[A]]): Stream[Seq[A]] = xss collect { case xs if !xs.isEmpty => xs } match {
case Nil => Stream.empty
case heads => heads #:: seqs(xss collect { case xs if !xs.isEmpty => xs.tail })
}
seqs(ssa).flatten
}
Here's something short but not exceedingly efficient:
def heads[A](xss: List[List[A]]) = xss.map(_.splitAt(1)).unzip
def interleave[A](xss: List[List[A]]) = Iterator.
iterate(heads(xss)){ case (_, tails) => heads(tails) }.
map(_._1.flatten).
takeWhile(! _.isEmpty).
flatten.toList
Here's a recursive solution that's O(n). The accepted solution (using sort) is O(nlog(n)). Some testing I've done suggests the second solution using transpose is also O(nlog(n)) due to the implementation of transpose. The use of reverse below looks suspicious (since it's an O(n) operation itself) but convince yourself that it either can't be called too often or on too-large lists.
def intercalate[T](lists: List[List[T]]) : List[T] = {
def intercalateHelper(newLists: List[List[T]], oldLists: List[List[T]], merged: List[T]): List[T] = {
(newLists, oldLists) match {
case (Nil, Nil) => merged
case (Nil, zss) => intercalateHelper(zss.reverse, Nil, merged)
case (Nil::xss, zss) => intercalateHelper(xss, zss, merged)
case ( (y::ys)::xss, zss) => intercalateHelper(xss, ys::zss, y::merged)
}
}
intercalateHelper(lists, List.empty, List.empty).reverse
}
I was wondering if I can tune the following Scala code :
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] = {
var listNoDuplicates: List[(Class1, Class2)] = Nil
for (outerIndex <- 0 until listOfTuple.size) {
if (outerIndex != listOfTuple.size - 1)
for (innerIndex <- outerIndex + 1 until listOfTuple.size) {
if (listOfTuple(i)._1.flag.equals(listOfTuple(j)._1.flag))
listNoDuplicates = listOfTuple(i) :: listNoDuplicates
}
}
listNoDuplicates
}
Usually if you have someting looking like:
var accumulator: A = new A
for( b <- collection ) {
accumulator = update(accumulator, b)
}
val result = accumulator
can be converted in something like:
val result = collection.foldLeft( new A ){ (acc,b) => update( acc, b ) }
So here we can first use a map to force the unicity of flags. Supposing the flag has a type F:
val result = listOfTuples.foldLeft( Map[F,(ClassA,ClassB)] ){
( map, tuple ) => map + ( tuple._1.flag -> tuple )
}
Then the remaining tuples can be extracted from the map and converted to a list:
val uniqList = map.values.toList
It will keep the last tuple encoutered, if you want to keep the first one, replace foldLeft by foldRight, and invert the argument of the lambda.
Example:
case class ClassA( flag: Int )
case class ClassB( value: Int )
val listOfTuples =
List( (ClassA(1),ClassB(2)), (ClassA(3),ClassB(4)), (ClassA(1),ClassB(-1)) )
val result = listOfTuples.foldRight( Map[Int,(ClassA,ClassB)]() ) {
( tuple, map ) => map + ( tuple._1.flag -> tuple )
}
val uniqList = result.values.toList
//uniqList: List((ClassA(1),ClassB(2)), (ClassA(3),ClassB(4)))
Edit: If you need to retain the order of the initial list, use instead:
val uniqList = listOfTuples.filter( result.values.toSet )
This compiles, but as I can't test it it's hard to say if it does "The Right Thing" (tm):
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] =
(for {outerIndex <- 0 until listOfTuple.size
if outerIndex != listOfTuple.size - 1
innerIndex <- outerIndex + 1 until listOfTuple.size
if listOfTuple(i)._1.flag == listOfTuple(j)._1.flag
} yield listOfTuple(i)).reverse.toList
Note that you can use == instead of equals (use eq if you need reference equality).
BTW: https://codereview.stackexchange.com/ is better suited for this type of question.
Do not use index with lists (like listOfTuple(i)). Index on lists have very lousy performance. So, some ways...
The easiest:
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] =
SortedSet(listOfTuple: _*)(Ordering by (_._1.flag)).toList
This will preserve the last element of the list. If you want it to preserve the first element, pass listOfTuple.reverse instead. Because of the sorting, performance is, at best, O(nlogn). So, here's a faster way, using a mutable HashSet:
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] = {
// Produce a hash map to find the duplicates
import scala.collection.mutable.HashSet
val seen = HashSet[Flag]()
// now fold
listOfTuple.foldLeft(Nil: List[(Class1,Class2)]) {
case (acc, el) =>
val result = if (seen(el._1.flag)) acc else el :: acc
seen += el._1.flag
result
}.reverse
}
One can avoid using a mutable HashSet in two ways:
Make seen a var, so that it can be updated.
Pass the set along with the list being created in the fold. The case then becomes:
case ((seen, acc), el) =>