I wanted to write it functionally, and the best I could do was:
list.zipWithIndex.filter((tt:Tuple2[Thing,Int])=>(tt._2%3==0)).unzip._1
to get elements 0, 3, 6,...
Is there a more readable Scala idiom for this?
If efficiency is not an issue, you could do the following:
list.grouped(3).map(_.head)
Note that this constructs intermediate lists.
Alternatively you can use a for-comprehension:
for {
(x,i) <- list zipWithIndex
if i % 3 == 0
} yield x
This is of course almost identical to your original solution, just written differently.
My last alternative for you is the use of collect on the zipped list:
list.zipWithIndex.collect {
case (x,i) if i % 3 == 0 => x
}
Not much clear, but still:
xs.indices.collect { case i if i % 3 == 0 => xs(i) }
A nice, functional solution, without creating temporary vectors, lists, and so on:
def everyNth[T](xs: List[T], n:Int): List[T] = xs match {
case hd::tl => hd::everyNth(tl.drop(n-1), n)
case Nil => Nil
}
Clojure has a take-nth function that does what you want, but I was surprised to find that there's not an equivalent method in Scala. You could code up a similar recursive solution based off the Clojure code, or you could read this blog post:
Scala collections: Filtering each n-th element
The author actually has a nice graph at the end showing the relative performance of each of his solutions.
I would do it like in Octave mathematical program.
val indices = 0 until n by 3 // Range 0,3,6,9 ...
and then I needed some way to select the indices from a collection. Obviously I had to have a collection with random-access O(1). Like Array or Vector. For example here I use Vector. To wrap the access into a nice DSL I'd add an implicit class:
implicit class VectorEnrichedWithIndices[T](v:Vector[T]) {
def apply(indices:TraversableOnce[Int]):Vector[T] = {
// some implementation
indices.toVector.map(v)
}
}
The usage would look like:
val vector = list.toVector
val every3rdElement = vector(0 until vector.size by 3)
Ah, how about this?
val l = List(10,9,8,7,6,5,4,3,2,1,0)
for (i <- (0 to l.size - 1 by 3).toList) yield l(i)
//res0: List[Int] = List(10, 7, 4, 1)
which can be made more general by
def seqByN[A](xs: Seq[A], n: Int): Seq[A] = for (i <- 0 to xs.size - 1 by n) yield xs(i)
scala> seqByN(List(10,9,8,7,6,5,4,3,2,1,0), 3)
res1: Seq[Int] = Vector(10,7,4,1)
scala> seqByN(List(10,9,8,7,6,5,4,3,2,1,0), 3).toList
res2: Seq[Int] = List(10,7,4,1)
scala> seqByN(List[Int](), 3)
res1: Seq[Int] = Vector()
But by functional do you mean only using the various List combinator functions? Otherwise, are Streams functional enough?
def fromByN[A](xs: List[A], n: Int): Stream[A] = if (xs.isEmpty) Stream.empty else
xs.head #:: fromByN(xs drop n, n)
scala> fromByN(List(10,9,8,7,6,5,4,3,2,1,0), 3).toList
res17: List[Int] = List(10, 7, 4, 1)
Related
This question already has answers here:
Scala - Combine two lists in an alternating fashion
(4 answers)
Closed 3 years ago.
The elements of the resulting list should alternate between the elements of the arguments. Assume that the two arguments have the same length.
USE RECURSION
My code as follows
val finalString = new ListBuffer[Int]
val buff2= new ListBuffer[Int]
def alternate(xs:List[Int], ys:List[Int]):List[Int] = {
while (xs.nonEmpty) {
finalString += xs.head + ys.head
alternate(xs.tail,ys.tail)
}
return finalString.toList
}
EXPECTED RESULT:
alternate ( List (1 , 3, 5) , List (2 , 4, 6)) = List (1 , 2, 3, 4, 6)
As far for the output, I don't get any output. The program is still running and cannot be executed.
Are there any Scala experts?
There are a few problems with the recursive solutions suggested so far (including yours, which would actually work, if you replace while with if): appending to end of list is a linear operation, making the whole thing quadratic (taking a .length of a list too, as well ас accessing elements by index), don't do that; also, if the lists are long, a recursion may require a lot of space on the stack, you should be using tail-recursion whenever possible.
Here is a solution that is free of both those problems: it builds the output backwards, by prepending elements to the list (constant time operation) rather than appending them, and reverses the result at the end. It is also tail-recursive: the recursive call is the last operation in the function, which allows the compiler to convert it into a loop, so that it will only use a single stack frame for execution regardless of the size of the lists.
#tailrec
def alternate(
a: List[Int],
b: List[Int],
result: List[Int] = Nil
): List[Int] = (a,b) match {
case (Nil, _) | (_, Nil) => result.reversed
case (ah :: at, bh :: bt) => alternate(at, bt, bh :: ah :: result)
}
(if the lists are of different lengths, the whole thing stops when the shortest one ends, and whatever is left in the longer one is thrown out. You may want to modify the first case (split it into two, perhaps) if you desire a different behavior).
BTW, your own solution is actually better than most suggested here: it is actually tail recursive (or rather can be made one if you add else after your if, which is now while), and appending to ListBuffer isn't actually as bad as to a List. But using mutable state is generally considered "code smell" in scala, and should be avoided (that's one of the main ideas behind using recursion instead of loops in the first place).
Condition xs.nonEmpty is true always so you have infinite while loop.
Maybe you meant if instead of while.
A more Scala-ish approach would be something like:
def alternate(xs: List[Int], ys: List[Int]): List[Int] = {
xs.zip(ys).flatMap{case (x, y) => List(x, y)}
}
alternate(List(1,3,5), List(2,4,6))
// List(1, 2, 3, 4, 5, 6)
A recursive solution using match
def alternate[T](a: List[T], b: List[T]): List[T] =
(a, b) match {
case (h1::t1, h2::t2) =>
h1 +: h2 +: alternate(t1, t2)
case _ =>
a ++ b
}
This could be more efficient at the cost of clarity.
Update
This is the more efficient solution:
def alternate[T](a: List[T], b: List[T]): List[T] = {
#annotation.tailrec
def loop(a: List[T], b: List[T], res: List[T]): List[T] =
(a, b) match {
case (h1 :: t1, h2 :: t2) =>
loop(t1, t2, h2 +: h1 +: res)
case _ =>
a ++ b ++ res.reverse
}
loop(a, b, Nil)
}
This retains the original function signature but uses an inner function that is an efficient, tail-recursive implementation of the algorithm.
You're accessing variables from outside the method, which is bad. I would suggest something like the following:
object Main extends App {
val l1 = List(1, 3, 5)
val l2 = List(2, 4, 6)
def alternate[A](l1: List[A], l2: List[A]): List[A] = {
if (l1.isEmpty || l2.isEmpty) List()
else List(l1.head,l2.head) ++ alternate(l1.tail, l2.tail)
}
println(alternate(l1, l2))
}
Still recursive but without accessing state from outside the method.
Assuming both lists are of the same length, you can use a ListBuffer to build up the alternating list. alternate is a pure function:
import scala.collection.mutable.ListBuffer
object Alternate extends App {
def alternate[T](xs: List[T], ys: List[T]): List[T] = {
val buffer = new ListBuffer[T]
for ((x, y) <- xs.zip(ys)) {
buffer += x
buffer += y
}
buffer.toList
}
alternate(List(1, 3, 5), List(2, 4, 6)).foreach(println)
}
Suppose that I use a sequence of various maps and/or flatMaps to generate a sequence of collections. Is it possible to access information about the "current" collection from within any of those methods? For example, without knowing anything specific about the functions used in the previous maps or flatMaps, and without using any intermediate declarations, how can I get the maximum value (or length, or first element, etc.) of the collection upon which the last map acts?
List(1, 2, 3)
.flatMap(x => f(x) /* some unknown function */)
.map(x => x + ??? /* what is the max element of the collection? */)
Edit for clarification:
In the example, I'm not looking for the max (or whatever) of the initial List. I'm looking for the max of the collection after the flatMap has been applied.
By "without using any intermediate declarations" I mean that I do not want to use any temporary collections en route to the final result. So, the example by Steve Waldman below, while giving the desired result, is not what I am seeking. (I include this condition is mostly for aesthetic reasons.)
Edit for clarification, part 2:
The ideal solution would be some magic keyword or syntactic sugar that lets me reference the current collection:
List(1, 2, 3)
.flatMap(x => f(x))
.map(x => x + theCurrentList.max)
I'm prepared to accept the fact, however, that this simply is not possible.
Maybe just define the list as a val, so you can name it? I don't know of any facility built into map(...) or flatMap(...) that would help.
val myList = List(1, 2, 3)
myList
.flatMap(x => f(x) /* some unknown function */)
.map(x => x + myList.max /* what is the max element of the List? */)
Update: By this approach at least, if you have multiple transformations and want to see the transformed version, you'd have to name that. You could get away with
val myList = List(1, 2, 3).flatMap(x => f(x) /* some unknown function */)
myList.map(x => x + myList.max /* what is the max element of the List? */)
Or, if there will be multiple transformations, get in the habit of naming the stages.
val rawList = List(1, 2, 3)
val smordified = rawList.flatMap(x => f(x) /* some unknown function */)
val maxified = smordified.map(x => x + smordified.max /* what is the max element of the List? */)
maxified
Update 2: Watch it work in the REPL even with heterogenous types:
scala> def f( x : Int ) : Vector[Double] = Vector(x * math.random, x * math.random )
f: (x: Int)Vector[Double]
scala> val rawList = List(1, 2, 3)
rawList: List[Int] = List(1, 2, 3)
scala> val smordified = rawList.flatMap(x => f(x) /* some unknown function */)
smordified: List[Double] = List(0.40730853571901315, 0.15151641399798665, 1.5305929709857609, 0.35211231420067435, 0.644241939254793, 0.15530230501048903)
scala> val maxified = smordified.map(x => x + smordified.max /* what is the max element of the List? */)
maxified: List[Double] = List(1.937901506704774, 1.6821093849837476, 3.0611859419715217, 1.8827052851864352, 2.1748349102405538, 1.6858952759962498)
scala> maxified
res3: List[Double] = List(1.937901506704774, 1.6821093849837476, 3.0611859419715217, 1.8827052851864352, 2.1748349102405538, 1.6858952759962498)
It is possible, but not pretty, and not likely something you want if you are doing it for "aesthetic reasons."
import scala.math.max
def f(x: Int): Seq[Int] = ???
List(1, 2, 3).
flatMap(x => f(x) /* some unknown function */).
foldRight((List[Int](),List[Int]())) {
case (x, (xs, Nil)) => ((x :: xs), List.fill(xs.size + 1)(x))
case (x, (xs, xMax :: _)) => ((x :: xs), List.fill(xs.size + 1)(max(x, xMax)))
}.
zipped.
map {
case (x, xMax) => x + xMax
}
// Or alternately, a slightly more efficient version using Streams.
List(1, 2, 3).
flatMap(x => f(x) /* some unknown function */).
foldRight((List[Int](),Stream[Int]())) {
case (x, (xs, Stream())) =>
((x :: xs), Stream.continually(x))
case (x, (xs, curXMax #:: _)) =>
val newXMax = max(x, curXMax)
((x :: xs), Stream.continually(newXMax))
}.
zipped.
map {
case (x, xMax) => x + xMax
}
Seriously though, I just took this on to see if I could do it. While the code didn't turn out as bad as I expected, I still don't think it's particularly readable. I'd discourage using this over something similar to Steve Waldman's answer. Sometimes, it's simply better to just introduce a val, rather than being dogmatic about it.
You could define a mapWithSelf (resp. flatMapWithSelf) operation along these lines and add it as an implicit enrichment to the collection. For List it might look like:
// Scala 2.13 APIs
object Enrichments {
implicit class WithSelfOps[A](val lst: List[A]) extends AnyVal {
def mapWithSelf[B](f: (A, List[A]) => B): List[B] =
lst.map(f(_, lst))
def flatMapWithSelf[B](f: (A, List[A]) => IterableOnce[B]): List[B] =
lst.flatMap(f(_, lst))
}
}
The enrichment basically fixes the value of the collection before the operation and threads it through. It should be possible to generify this (at least for the strict collections), though it would look a little different in 2.12 vs. 2.13+.
Usage would look like
import Enrichments._
val someF: Int => IterableOnce[Int] = ???
List(1, 2, 3)
.flatMap(someF)
.mapWithSelf { (x, lst) =>
x + lst.max
}
So at the usage site, it's aesthetically pleasant. Note that if you're computing something which traverses the list, you'll be traversing the list every time (leading to a quadratic runtime). You can get around that with some mutability or by just saving the intermediate list after the flatMap.
One somewhat-simple way of referencing prior output within the current map/collect operation is to use a named reference outside the map, then reference it from within the map block:
var prevOutput = ... // starting value of whatever is referenced within the map
myValues.map {
prevOutput = ... // expression that references prior `prevOutput`
prevOutput // return above computed value for the map to collect
}
This draws attention to the fact that we're referencing prior elements while building the new sequence.
This would be more messy, though, if you wanted to reference arbitrarily previous values, not just the previous one.
ScalaCheck generators work with the syntactic sugar in Scala's for expressions:
for( s1 <- Gen.choose(1, 10); s2 <- Gen.choose(10, 100) ) yield ( s1, s2 )
I'd like to be able to mix "traditional" Scala for expressions with ScalaCheck expressions. For example:
for( s0 <- 0 until 10; s1 <- Gen.choose(1, 10); s2 <- Gen.choose(10, 100) ) yield ( s0, s1, s2 )
However, this won't compile, since the 0 until 10 expressions isn't a type of Gen.
How can I achieve this (and Seq and/or Traversable expressions in general) within the same for loop?
EDIT: The consensus in the answers seems to be that the kind of syntactic sugar I'm looking for is not possible.
However, could this not be done with some form of stateful Gen? In particular, one which used the state monad?
I'd like to be able to mix "traditional" Scala for expressions with ScalaCheck expressions.
You cannot do that; at least, not as you suggest. What you can do is to define a generator that produces 10-long lists (i.e. lists of length 10) of triples, in which
the first element is not random (but instead ranges from 0 to 10),
the second element is randomly chosen between 1 and 10,
the third element is randomly chosen between 10 and 100.
Generator implementation
I'm assuming that org.scalacheck.Gen is in scope.
Define a generator for a pair composed of the second and third elements of a triple:
val pairGen: Gen[(Int, Int)] = for {
s1 <- Gen.choose(1, 10)
s2 <- Gen.choose(10, 100)
} yield (s1, s2)
Define a generator for a 10-long list of such pairs:
val listOfPairsGen: Gen[List[(Int, Int)]] = Gen.listOfN(10, pairGen)
Define
val intList: List[Int] = (0 until 10).toList
Zip intList with the result of listOfPairsGen, and "flatten each element to a triple":
val myGen: Gen[List[(Int, Int, Int)]] =
listOfPairsGen map { list: List[(Int, Int)] =>
(intList zip list) map { case (a, (b, c)) => (a, b, c) }
}
Examples
scala> myGen.sample.head
res0: List[(Int, Int, Int)] = List((0,2,58), (1,10,34), (2,3,94), (3,2,91), (4,6,15), (5,7,99), (6,4,82), (7,10,69), (8,8,78), (9,10,27))
scala> myGen.sample.get
res1: List[(Int, Int, Int)] = List((0,2,56), (1,2,83), (2,4,76), (3,4,87), (4,4,55), (5,6,80), (6,4,94), (7,7,67), (8,10,92), (9,4,84))
scala> myGen.sample.get
res2: List[(Int, Int, Int)] = List((0,10,40), (1,9,48), (2,10,63), (3,5,100), (4,5,67), (5,4,73), (6,8,56), (7,6,58), (8,6,82), (9,10,86))
scala> myGen.sample.get
res3: List[(Int, Int, Int)] = List((0,6,56), (1,7,94), (2,4,40), (3,7,27), (4,1,91), (5,3,50), (6,1,70), (7,6,90), (8,7,23), (9,7,49))
As Jubobs already explained, you cannot
mix "traditional" Scala for expressions with ScalaCheck expressions
However, you could achieve the described result with a somewhat less idiomatic approach using an index:
def indexedGenerator = {
val index = new AtomicInteger(0)
for (s1 <- Gen.choose(1, 10); s2 <- Gen.choose(10, 100)) yield (index.getAndIncrement(), s1, s2)
}
val gen = indexedGenerator
println(gen.sample) |-> Some((0,1,60))
println(gen.sample) |-> Some((1,8,82))
println(gen.sample) |-> Some((2,9,29))
println(gen.sample) |-> Some((3,6,76))
println(gen.sample) |-> Some((4,5,32))
Though you would have to be cautious about when and where to instantiate a Generator like this, as the count var will not be reset for every property check. Thus, you need to create a new local instance every time you need the index to start at 0.
I have a list with assorted keywords that may repeat. I need to generate a list with distinct keywords but sorted by the frequency of which they appeared on the original list.
How would be the idiomatic Scala for that? Here is a working but ugly implementation:
val keys = List("c","a","b","b","a","a")
keys.groupBy(p => p).toList.sortWith( (a,b) => a._2.size > b._2.size ).map(_._1)
// List("a","b","c")
Shorter version:
keys.distinct.sortBy(keys count _.==).reverse
That is not particular efficient, however. The groupBy version ought to perform better, though it can be improved:
keys.groupBy(identity).toSeq.sortBy(_._2.size).map(_._1)
One can also get rid of the reverse in the first version by declaring an Ordering:
val ord = Ordering by (keys count (_: String).==)
keys.distinct.sorted(ord.reverse)
Note that reverse in this version just produces a new Ordering that works in the opposite manner of the original. This version also suggests a way to get better performance:
val freq = collection.mutable.Map.empty[String, Int] withDefaultValue 0
keys foreach (k => freq(k) += 1)
val ord = Ordering by freq
keys.distinct.sorted(ord.reverse)
Nothing wrong with that implementation that comments can't fix!
Seriously, break it down a bit and describe what & why you're taking each step.
Not as "concise" perhaps, but the purpose of concise code in scala is to make code more readable. When concise code is not clear it's time to back up, break up (introduce well named local variables), and comment.
Here's my take, don't know if it's less "ugly":
scala> keys.groupBy(p => p).values.toList.sortBy(_.size).reverse.map(_.head)
res39: List[String] = List(a, b, c)
fold version:
val keys = List("c","a","b","b","a","a")
val keysCounts =
(Map.empty[String, Int] /: keys) { case (counts, k) =>
counts updated (k, (counts getOrElse (k, 0)) + 1)
}
keysCounts.toList sortBy { case (_, count) => -count } map { case (w, _) => w }
Perhaps,
val mapCount = keys.map(x => (x,keys.count(_ == x))).distinct
// mapCount : List[(java.lang.String, Int)] = List((c,1), (a,3), (b,2))
val sortedList = mapCount.sortWith(_._2 > _._2).map(_._1)
// sortedList : List[java.lang.String] = List(a, b, c)
How about:
keys.distinct.sorted
Newbie didn't read the question carefully. Let me try again:
keys.foldLeft (Map[String,Int]()) { (counts, elem) => counts + (elem -> (counts.getOrElse(elem, 0) - 1))}
.toList.sortBy(_._2).map(_._1)
Could use a mutable Map if you prefer. Negative frequency counts are stored in the map. If that bothers you, you can make them positive and negate the sortBy argument.
Just a little change from #Daniel 's 4th version, may have a better performance:
scala> def sortByFreq[T](xs: List[T]): List[T] = {
| val freq = collection.mutable.Map.empty[T, Int] withDefaultValue 0
| xs foreach (k => freq(k) -= 1)
| xs.distinct sortBy freq
| }
sortByFreq: [T](xs: List[T])List[T]
scala> sortByFreq(keys)
res2: List[String] = List(a, b, c)
My prefered versions would be:
Most canonical / expressive?
keys.groupBy(identity).toList.map{ case (k,v) => (-v.size,k) }.sorted.map(_._2)
Shortest and probably most efficient?
keys.groupBy(identity).toList.sortBy(-_._2.size).map(_._1)
Straight forward
keys.groupBy(identity).values.toList.sortBy(-_.size).map(_.head)
As far as I understand, the Scala for-comprehension notation relies on the first generator to define how elements are to be combined. Namely, for (i <- list) yield i returns a list and for (i <- set) yield i returns a set.
I was wondering if there was a way to specify how elements are combined independently of the properties of the first generator. For instance, I would like to get "the set of all elements from a given list", or "the sum of all elements from a given set". The only way I have found is to first build a list or a set as prescribed by the for-comprehension notation, then apply a transformation function to it - building a useless data structure in the process.
What I have in mind is a general "algebraic" comprehension notation as it exists for instance in Ateji PX:
`+ { i | int i : set } // the sum of all elements from a given set
set() { i | int i : list } // the set of all elements from a given list
concat(",") { s | String s : list } // string concatenation with a separator symbol
Here the first element (`+, set(), concat(",")) is a so-called "monoid" that defines how elements are combined, independently of the structure of the first generator (there can be multiple generators and filters, I just tried to keep the examples concise).
Any idea about how to achieve a similar result in Scala while keeping a nice and concise notation ? As far as I understand, the for-comprehension notation is hard-wired in the compiler and cannot be upgraded.
Thanks for your feedback.
About the for comprehension
The for comprehension in scala is syntactic sugar for calls to flatMap, filter, map and foreach. In exactly the same way as calls to those methods, the type of the target collection leads to the type of the returned collection. That is:
list map f //is a List
vector map f // is a Vector
This property is one of the underlying design goals of the scala collections library and would be seen as desirable in most situations.
Answering the question
You do not need to construct any intermediate collection of course:
(list.view map (_.prop)).toSet //uses list.view
(list.iterator map (_.prop)).toSet //uses iterator
(for { l <- list.view} yield l.prop).toSet //uses view
(Set.empty[Prop] /: coll) { _ + _.prop } //uses foldLeft
Will all yield Sets without generating unnecessary collections. My personal preference is for the first. In terms of idiomatic scala collection manipulation, each "collection" comes with these methods:
//Conversions
toSeq
toSet
toArray
toList
toIndexedSeq
iterator
toStream
//Strings
mkString
//accumulation
sum
The last is used where the element type of a collection has an implicit Numeric instance in scope; such as:
Set(1, 2, 3, 4).sum //10
Set('a, 'b).sum //does not compile
Note that the String concatenation example in scala looks like:
list.mkString(",")
And in the scalaz FP library might look something like (which uses Monoid to sum Strings):
list.intercalate(",").asMA.sum
Your suggestions do not look anything like Scala; I'm not sure whether they are inspired by another language.
foldLeft? That's what you're describing.
The sum of all elements from a given set:
(0 /: Set(1,2,3))(_ + _)
the set of all elements from a given list
(Set[Int]() /: List(1,2,3,2,1))((acc,x) => acc + x)
String concatenation with a separator symbol:
("" /: List("a", "b"))(_ + _) // (edit - ok concat a bit more verbose:
("" /: List("a", "b"))((acc,x) => acc + (if (acc == "") "" else ",") + x)
You can also force the result type of the for comprehension by explicitly supplying the implicit CanBuildFrom parameter as scala.collection.breakout and specifying the result type.
Consider this REPL session:
scala> val list = List(1, 1, 2, 2, 3, 3)
list: List[Int] = List(1, 1, 2, 2, 3, 3)
scala> val res = for(i <- list) yield i
res: List[Int] = List(1, 1, 2, 2, 3, 3)
scala> val res: Set[Int] = (for(i <- list) yield i)(collection.breakOut)
res: Set[Int] = Set(1, 2, 3)
It results in a type error when not specifying the CanBuildFrom explicitly:
scala> val res: Set[Int] = for(i <- list) yield i
<console>:8: error: type mismatch;
found : List[Int]
required: Set[Int]
val res: Set[Int] = for(i <- list) yield i
^
For a deeper understanding of this I suggest the following read:
http://www.scala-lang.org/docu/files/collections-api/collections-impl.html
If you want to use for comprehensions and still be able to combine your values in some result value you could do the following.
case class WithCollector[B, A](init: B)(p: (B, A) => B) {
var x: B = init
val collect = { (y: A) => { x = p(x, y) } }
def apply(pr: (A => Unit) => Unit) = {
pr(collect)
x
}
}
// Some examples
object Test {
def main(args: Array[String]): Unit = {
// It's still functional
val r1 = WithCollector[Int, Int](0)(_ + _) { collect =>
for (i <- 1 to 10; if i % 2 == 0; j <- 1 to 3) collect(i + j)
}
println(r1) // 120
import collection.mutable.Set
val r2 = WithCollector[Set[Int], Int](Set[Int]())(_ += _) { collect =>
for (i <- 1 to 10; if i % 2 == 0; j <- 1 to 3) collect(i + j)
}
println(r2) // Set(9, 10, 11, 6, 13, 4, 12, 3, 7, 8, 5)
}
}