I need to group list of tuples in some unique way.
For example, if I have
val l = List((1,2,3),(4,2,5),(2,3,3),(10,3,2))
Then I should group the list with second value and map with the set of first value
So the result should be
Map(2 -> Set(1,4), 3 -> Set(2,10))
By so far, I came up with this
l groupBy { p => p._2 } mapValues { v => (v map { vv => vv._1 }).toSet }
This works, but I believe there should be a much more efficient way...
This is similar to this question. Basically, as #serejja said, your approach is correct and also the most concise one. You could use collection.breakOut as builder factory argument to the last map and thereby save the additional iteration to get the Set type:
l.groupBy(_._2).mapValues(_.map(_._1)(collection.breakOut): Set[Int])
You shouldn't probably go beyond this, unless you really need to squeeze the performance.
Otherwise, this is how a general toMultiMap function could look like which allows you to control the values collection type:
import collection.generic.CanBuildFrom
import collection.mutable
def toMultiMap[A, K, V, Values](xs: TraversableOnce[A])
(key: A => K)(value: A => V)
(implicit cbfv: CanBuildFrom[Nothing, V, Values]): Map[K, Values] = {
val b = mutable.Map.empty[K, mutable.Builder[V, Values]]
xs.foreach { elem =>
b.getOrElseUpdate(key(elem), cbfv()) += value(elem)
}
b.map { case (k, vb) => (k, vb.result()) } (collection.breakOut)
}
What it does is, it uses a mutable Map during building stage, and values gathered in a mutable Builder first (the builder is provided by the CanBuildFrom instance). After the iteration over all input elements has completed, that mutable map of builder values is converted into an immutable map of the values collection type (again using the collection.breakOut trick to get the desired output collection straight away).
Ex:
val l = List((1,2,3),(4,2,5),(2,3,3),(10,3,2))
val v = toMultiMap(l)(_._2)(_._1) // uses Vector for values
val s: Map[Int, Set[Int] = toMultiMap(l)(_._2)(_._1) // uses Set for values
So your annotated result type directs the type inference of the values type. If you do not annotate the result, Scala will pick Vector as default collection type.
Related
I am processing XML using scala, and I am converting the XML into my own data structures. Currently, I am using plain Map instances to hold (sub-)elements, however, the order of elements from the XML gets lost this way, and I cannot reproduce the original XML.
Therefore, I want to use LinkedHashMap instances instead of Map, however I am using groupBy on the list of nodes, which creates a Map:
For example:
def parse(n:Node): Unit =
{
val leaves:Map[String, Seq[XmlItem]] =
n.child
.filter(node => { ... })
.groupBy(_.label)
.map((tuple:Tuple2[String, Seq[Node]]) =>
{
val items = tuple._2.map(node =>
{
val attributes = ...
if (node.text.nonEmpty)
XmlItem(Some(node.text), attributes)
else
XmlItem(None, attributes)
})
(tuple._1, items)
})
...
}
In this example, I want leaves to be of type LinkedHashMap to retain the order of n.child. How can I achieve this?
Note: I am grouping by label/tagname because elements can occur multiple times, and for each label/tagname, I keep a list of elements in my data structures.
Solution
As answered by #jwvh I am using foldLeft as a substitution for groupBy. Also, I decided to go with LinkedHashMap instead of ListMap.
def parse(n:Node): Unit =
{
val leaves:mutable.LinkedHashMap[String, Seq[XmlItem]] =
n.child
.filter(node => { ... })
.foldLeft(mutable.LinkedHashMap.empty[String, Seq[Node]])((m, sn) =>
{
m.update(sn.label, m.getOrElse(sn.label, Seq.empty[Node]) ++ Seq(sn))
m
})
.map((tuple:Tuple2[String, Seq[Node]]) =>
{
val items = tuple._2.map(node =>
{
val attributes = ...
if (node.text.nonEmpty)
XmlItem(Some(node.text), attributes)
else
XmlItem(None, attributes)
})
(tuple._1, items)
})
To get the rough equivalent to .groupBy() in a ListMap you could fold over your collection. The problem is that ListMap preserves the order of elements as they were appended, not as they were encountered.
import collection.immutable.ListMap
List('a','b','a','c').foldLeft(ListMap.empty[Char,Seq[Char]]){
case (lm,c) => lm.updated(c, c +: lm.getOrElse(c, Seq()))
}
//res0: ListMap[Char,Seq[Char]] = ListMap(b -> Seq(b), a -> Seq(a, a), c -> Seq(c))
To fix this you can foldRight instead of foldLeft. The result is the original order of elements as encountered (scanning left to right) but in reverse.
List('a','b','a','c').foldRight(ListMap.empty[Char,Seq[Char]]){
case (c,lm) => lm.updated(c, c +: lm.getOrElse(c, Seq()))
}
//res1: ListMap[Char,Seq[Char]] = ListMap(c -> Seq(c), b -> Seq(b), a -> Seq(a, a))
This isn't necessarily a bad thing since a ListMap is more efficient with last and init ops, O(1), than it is with head and tail ops, O(n).
To process the ListMap in the original left-to-right order you could .toList and .reverse it.
List('a','b','a','c').foldRight(ListMap.empty[Char,Seq[Char]]){
case (c,lm) => lm.updated(c, c +: lm.getOrElse(c, Seq()))
}.toList.reverse
//res2: List[(Char, Seq[Char])] = List((a,Seq(a, a)), (b,Seq(b)), (c,Seq(c)))
Purely immutable solution would be quite slow. So I'd go with
import collection.mutable.{ArrayBuffer, LinkedHashMap}
implicit class ExtraTraversableOps[A](seq: collection.TraversableOnce[A]) {
def orderedGroupBy[B](f: A => B): collection.Map[B, collection.Seq[A]] = {
val map = LinkedHashMap.empty[B, ArrayBuffer[A]]
for (x <- seq) {
val key = f(x)
map.getOrElseUpdate(key, ArrayBuffer.empty) += x
}
map
}
To use, just change .groupBy in your code to .orderedGroupBy.
The returned Map can't be mutated using this type (though it can be cast to mutable.Map or to mutable.LinkedHashMap), so it's safe enough for most purposes (and you could create a ListMap from it at the end if really desired).
I am trying to find an elegant way to do:
val l = List(1,2,3)
val (item, idx) = l.zipWithIndex.find(predicate)
val updatedItem = updating(item)
l.update(idx, updatedItem)
Can I do all in one operation ? Find the item, if it exist replace with updated value and keep it in place.
I could do:
l.map{ i =>
if (predicate(i)) {
updating(i)
} else {
i
}
}
but that's pretty ugly.
The other complexity is the fact that I want to update only the first element which match predicate .
Edit: Attempt:
implicit class UpdateList[A](l: List[A]) {
def filterMap(p: A => Boolean)(update: A => A): List[A] = {
l.map(a => if (p(a)) update(a) else a)
}
def updateFirst(p: A => Boolean)(update: A => A): List[A] = {
val found = l.zipWithIndex.find { case (item, _) => p(item) }
found match {
case Some((item, idx)) => l.updated(idx, update(item))
case None => l
}
}
}
I don't know any way to make this in one pass of the collection without using a mutable variable. With two passes you can do it using foldLeft as in:
def updateFirst[A](list:List[A])(predicate:A => Boolean, newValue:A):List[A] = {
list.foldLeft((List.empty[A], predicate))((acc, it) => {acc match {
case (nl,pr) => if (pr(it)) (newValue::nl, _ => false) else (it::nl, pr)
}})._1.reverse
}
The idea is that foldLeft allows passing additional data through iteration. In this particular implementation I change the predicate to the fixed one that always returns false. Unfortunately you can't build a List from the head in an efficient way so this requires another pass for reverse.
I believe it is obvious how to do it using a combination of map and var
Note: performance of the List.map is the same as of a single pass over the list only because internally the standard library is mutable. Particularly the cons class :: is declared as
final case class ::[B](override val head: B, private[scala] var tl: List[B]) extends List[B] {
so tl is actually a var and this is exploited by the map implementation to build a list from the head in an efficient way. The field is private[scala] so you can't use the same trick from outside of the standard library. Unfortunately I don't see any other API call that allows to use this feature to reduce the complexity of your problem to a single pass.
You can avoid .zipWithIndex() by using .indexWhere().
To improve complexity, use Vector so that l(idx) becomes effectively constant time.
val l = Vector(1,2,3)
val idx = l.indexWhere(predicate)
val updatedItem = updating(l(idx))
l.updated(idx, updatedItem)
Reason for using scala.collection.immutable.Vector rather than List:
Scala's List is a linked list, which means data are access in O(n) time. Scala's Vector is indexed, meaning data can be read from any point in effectively constant time.
You may also consider mutable collections if you're modifying just one element in a very large collection.
https://docs.scala-lang.org/overviews/collections/performance-characteristics.html
I have a collection which I want to map to a new collection, however each resulting value is dependent on the value before it in some way.I could solve this with a leftFold
val result:List[B] = (myList:List[A]).foldLeft(C -> List.empty[B]){
case ((c, list), a) =>
..some function returning something like..
C -> (B :: list)
}
The problem here is I need to iterate through the entire list to retrieve the resultant list. Say I wanted a function that maps TraversableOnce[A] to TraversableOnce[B] and only evaluate members as I call them?
It seems to me to be a fairly conventional problem so Im wondering if there is a common approach to this. What I currently have is:
implicit class TraversableOnceEx[T](val self : TraversableOnce[T]) extends AnyVal {
def foldyMappyFunction[A, U](a:A)(func:(A,T) => (A,U)):TraversableOnce[U] = {
var currentA = a
self.map { t =>
val result = func(currentA, t)
currentA = result._1
result._2
}
}
}
As far as functional purity goes, you couldn't run it in parallel, but otherwise it seems sound.
An example would be;
Return me each element and if it is the first time that element has appeared before.
val elements:TraversableOnce[E]
val result = elements.mappyFoldyFunction(Set.empty[E]) {
(s, e) => (s + e) -> (e -> s.contains(e))
}
result:TraversableOnce[(E,Boolean)]
You might be able to make use of the State Monad. Here is your example re-written using scalaz:
import scalaz._, Scalaz._
def foldyMappy(i: Int) = State[Set[Int], (Int, Boolean)](s => (s + i, (i, s contains(i))))
val r = List(1, 2, 3, 3, 6).traverseS(foldyMappy)(Set.empty[Int])._2
//List((1,false), (2,false), (3,false), (3,true), (6,false))
println(r)
It is look like you need SeqView. Use view or view(from: Int, until: Int) methods for create a non-strict view of list.
I really don't understand your example as your contains check will always result to false.
foldLeft is different. It will result in a single value by aggregating all elements of the list.
You clearly need map (List => List).
Anyway, answering your question about laziness:
you should use Stream instead of List. Stream doesn't evaluate the tail before actually calling it.
Stream API
I'm new to Scala.
I'm using SortedMap in my code, and I wanted to use mapValues to create a new map with some transformation on the values.
Instead of returning a new SortedMap, the mapValues function returns a new Map, which I then have to convert to a SortedMap.
For example
val my_map = SortedMap(1 -> "one", 0 -> "zero", 2 -> "two")
val new_map = my_map.mapValues(name => name.toUpperCase)
// returns scala.collection.immutable.Map[Int,java.lang.String] = Map(0 -> ZERO, 1 -> ONE, 2 -> TWO)
val sorted_new_map = SortedMap(new_map.toArray:_ *)
This looks inefficient - the last convertion probably sorts the keys again, or at least verify that they are sorted.
I could use the normal map function which operates both on the keys and the values, and deliberately not change the keys in my transformation function. This looks inefficient too, since the implementation of Map probably assume that the transformation may change the order of the keys (like in the case: my_map.map(tup => (-tup._1, tup._2)) - so it probably "re-sorts" them too.
Is anyone familiar with the internal implementations of Map and SortedMap, and could tell me if my assumptions are correct? Can the compiler recognize automatically that the keys have not been reordered? Is there an internal reason for why mapValues should not return a SortedMap? Is there a better way to transform the map's values without loosing the order of the keys?
Thanks
You've stumbled upon a tricky feature of Scala's Map implementation. The catch that you are missing is that mapValues does not actually return a new Map: it returns a view of a Map. In other words, it wraps your original map in such a way that whenever you access a value it will compute .toUpperCase before returning the value to you.
The upside to this behavior is that Scala won't compute the function for values that aren't accessed, and it won't spend time copying all the data into a new Map. The downside is that the function is re-computed every time that value is accessed. So you might end up doing extra computation if you access the same values many times.
So why does SortedMap not return a SortedMap? Because it's actually returning a Map-wrapper. The underlying Map, then one that is wrapped, is still a SortedMap, so if you were to iterate through, it would still be in sorted order. You and I know that, but the type-checker doesn't. It certainly seems like they could have written it in such a way that it still maintains the SortedMap trait, but they didn't.
You can see in the code that it's not returning a SortedMap, but that the iteration behavior is still going to be sorted:
// from MapLike
override def mapValues[C](f: B => C): Map[A, C] = new DefaultMap[A, C] {
def iterator = for ((k, v) <- self.iterator) yield (k, f(v))
...
The solution to your problem is the same as the solution to getting around the view issue: use .map{ case (k,v) => (k,f(v)) }, as you mentioned in your question.
If you really want that convenience method though, you can do what I do, and write you own, better, version of mapValues:
class EnrichedWithMapVals[T, U, Repr <: GenTraversable[(T, U)]](self: GenTraversableLike[(T, U), Repr]) {
/**
* In a collection of pairs, map a function over the second item of each
* pair. Ensures that the map is computed at call-time, and not returned
* as a view as 'Map.mapValues' would do.
*
* #param f function to map over the second item of each pair
* #return a collection of pairs
*/
def mapVals[R, That](f: U => R)(implicit bf: CanBuildFrom[Repr, (T, R), That]) = {
val b = bf(self.asInstanceOf[Repr])
b.sizeHint(self.size)
for ((k, v) <- self) b += k -> f(v)
b.result
}
}
implicit def enrichWithMapVals[T, U, Repr <: GenTraversable[(T, U)]](self: GenTraversableLike[(T, U), Repr]): EnrichedWithMapVals[T, U, Repr] =
new EnrichedWithMapVals(self)
Now when you call mapVals on a SortedMap you get back a non-view SortedMap:
scala> val m3 = m1.mapVals(_ + 1)
m3: SortedMap[String,Int] = Map(aardvark -> 2, cow -> 6, dog -> 10)
It actually works on any collection of pairs, not just Map implementations:
scala> List(('a,1),('b,2),('c,3)).mapVals(_+1)
res8: List[(Symbol, Int)] = List(('a,2), ('b,3), ('c,4))
Given a key k in a SortedMap, how can I efficiently find the largest key m that is less than or equal to k, and also the smallest key n that is greater than or equal to k. Thank you.
Looking at the source code for 2.9.0, the following code seems about to be the best you can do
def getLessOrEqual[A,B](sm: SortedMap[A,B], bound: A): B = {
val key = sm.to(x).lastKey
sm(key)
}
I don't know exactly how the splitting of the RedBlack tree works, but I guess it's something like a O(log n) traversal of the tree/construction of new elements and then a balancing, presumable also O(log n). Then you need to go down the new tree again to get the last key. Unfortunately you can't retrieve the value in the same go. So you have to go down again to fetch the value.
In addition the lastKey might throw an exception and there is no similar method that returns an Option.
I'm waiting for corrections.
Edit and personal comment
The SortedMap area of the std lib seems to be a bit neglected. I'm also missing a mutable SortedMap. And looking through the sources, I noticed that there are some important methods missing (like the one the OP asks for or the ones pointed out in my answer) and also some have bad implementation, like 'last' which is defined by TraversableLike and goes through the complete tree from first to last to obtain the last element.
Edit 2
Now the question is reformulated my answer is not valid anymore (well it wasn't before anyway). I think you have to do the thing I'm describing twice for lessOrEqual and greaterOrEqual. Well you can take a shortcut if you find the equal element.
Scala's SortedSet trait has no method that will give you the closest element to some other element.
It is presently implemented with TreeSet, which is based on RedBlack. The RedBlack tree is not visible through methods on TreeSet, but the protected method tree is protected. Unfortunately, it is basically useless. You'd have to override methods returning TreeSet to return your subclass, but most of them are based on newSet, which is private.
So, in the end, you'd have to duplicate most of TreeSet. On the other hand, it isn't all that much code.
Once you have access to RedBlack, you'd have to implement something similar to RedBlack.Tree's lookup, so you'd have O(logn) performance. That's actually the same complexity of range, though it would certainly do less work.
Alternatively, you'd make a zipper for the tree, so that you could actually navigate through the set in constant time. It would be a lot more work, of course.
Using Scala 2.11.7, the following will give what you want:
scala> val set = SortedSet('a', 'f', 'j', 'z')
set: scala.collection.SortedSet[Char] = TreeSet(a, f, j, z)
scala> val beforeH = set.to('h').last
beforeH: Char = f
scala> val afterH = set.from('h').head
afterH: Char = j
Generally you should use lastOption and headOption as the specified elements may not exist. If you are looking to squeeze a little more efficiency out, you can try replacing from(...).head with keysIteratorFrom(...).head
Sadly, the Scala library only allows to make this type of query efficiently:
and also the smallest key n that is greater than or equal to k.
val n = TreeMap(...).keysIteratorFrom(k).next
You can hack this by keeping two structures, one with normal keys, and one with negated keys. Then you can use the other structure to make the second type of query.
val n = - TreeMap(...).keysIteratorFrom(-k).next
Looks like I should file a ticket to add 'fromIterator' and 'toIterator' methods to 'Sorted' trait.
Well, one option is certainly using java.util.TreeMap.
It has lowerKey and higherKey methods, which do excatly what you want.
I had a similar problem: I wanted to find the closest element to a given key in a SortedMap. I remember the answer to this question being, "You have to hack TreeSet," so when I had to implement it for a project, I found a way to wrap TreeSet without getting into its internals.
I didn't see jazmit's answer, which more closely answers the original poster's question with minimum fuss (two method calls). However, those method calls do more work than needed for this application (multiple tree traversals), and my solution provides lots of hooks where other users can modify it to their own needs.
Here it is:
import scala.collection.immutable.TreeSet
import scala.collection.SortedMap
// generalize the idea of an Ordering to metric sets
trait MetricOrdering[T] extends Ordering[T] {
def distance(x: T, y: T): Double
def compare(x: T, y: T) = {
val d = distance(x, y)
if (d > 0.0) 1
else if (d < 0.0) -1
else 0
}
}
class MetricSortedMap[A, B]
(elems: (A, B)*)
(implicit val ordering: MetricOrdering[A])
extends SortedMap[A, B] {
// while TreeSet searches for an element, keep track of the best it finds
// with *thread-safe* mutable state, of course
private val best = new java.lang.ThreadLocal[(Double, A, B)]
best.set((-1.0, null.asInstanceOf[A], null.asInstanceOf[B]))
private val ord = new MetricOrdering[(A, B)] {
def distance(x: (A, B), y: (A, B)) = {
val diff = ordering.distance(x._1, y._1)
val absdiff = Math.abs(diff)
// the "to" position is a key-null pair; the object of interest
// is the other one
if (absdiff < best.get._1)
(x, y) match {
// in practice, TreeSet always picks this first case, but that's
// insider knowledge
case ((to, null), (pos, obj)) =>
best.set((absdiff, pos, obj))
case ((pos, obj), (to, null)) =>
best.set((absdiff, pos, obj))
case _ =>
}
diff
}
}
// use a TreeSet as a backing (not TreeMap because we need to get
// the whole pair back when we query it)
private val treeSet = TreeSet[(A, B)](elems: _*)(ord)
// find the closest key and return:
// (distance to key, the key, its associated value)
def closest(to: A): (Double, A, B) = {
treeSet.headOption match {
case Some((pos, obj)) =>
best.set((ordering.distance(to, pos), pos, obj))
case None =>
throw new java.util.NoSuchElementException(
"SortedMap has no elements, and hence no closest element")
}
treeSet((to, null.asInstanceOf[B])) // called for side effects
best.get
}
// satisfy the contract (or throw UnsupportedOperationException)
def +[B1 >: B](kv: (A, B1)): SortedMap[A, B1] =
new MetricSortedMap[A, B](
elems :+ (kv._1, kv._2.asInstanceOf[B]): _*)
def -(key: A): SortedMap[A, B] =
new MetricSortedMap[A, B](elems.filter(_._1 != key): _*)
def get(key: A): Option[B] = treeSet.find(_._1 == key).map(_._2)
def iterator: Iterator[(A, B)] = treeSet.iterator
def rangeImpl(from: Option[A], until: Option[A]): SortedMap[A, B] =
new MetricSortedMap[A, B](treeSet.rangeImpl(
from.map((_, null.asInstanceOf[B])),
until.map((_, null.asInstanceOf[B]))).toSeq: _*)
}
// test it with A = Double
implicit val doubleOrdering =
new MetricOrdering[Double] {
def distance(x: Double, y: Double) = x - y
}
// and B = String
val stuff = new MetricSortedMap[Double, String](
3.3 -> "three",
1.1 -> "one",
5.5 -> "five",
4.4 -> "four",
2.2 -> "two")
println(stuff.iterator.toList)
println(stuff.closest(1.5))
println(stuff.closest(1000))
println(stuff.closest(-1000))
println(stuff.closest(3.3))
println(stuff.closest(3.4))
println(stuff.closest(3.2))
I've been doing:
val m = SortedMap(myMap.toSeq:_*)
val offsetMap = (m.toSeq zip m.keys.toSeq.drop(1)).map {
case ( (k,v),newKey) => (newKey,v)
}.toMap
When I want the results of my map off-set by one key. I'm also looking for a better way, preferably without storing an extra map.