I have a function that perform a calc but i'm using a var to receive the value of a recursive function and i would like to avoid mutable variables.
def scoreNode(node:Option[Tree], score:Double = 0, depth:Int = 0):Double = {
node.map(n => {
var points = score
n.children.filter(n => n.valid == Some(true)).foreach(h => {
points = scoreNode(Some(h), 10, depth+1)
})
points
}).getOrElse(score)
}
How can i rewrite this piece of code without a mutable variable? I've tried
What you are essentially doing is summing something over all the nodes in a tree. Try to write a more idiomatic code, like this.
def scoreNode(node:Option[Tree], depth:Int = 0):Double =
(for {
n <- node
h <- n.children
if h.valid == Some(true)
res = scoreNode(Some(h), depth + 1) + scala.math.pow(0.8, depth)
} yield res).sum
I do not guarantee this works completely. It is your homework to make it right.
You can use fold:
def scoreNode(node:Option[Tree], score:Double = 0, depth:Int = 0):Double =
node
.map(_.children.filter(n => n.valid == Some(true)).fold(score)((acc, h) => scoreNode(Some(h), acc + scala.math.pow(0.8, depth), depth + 1)))
.getOrElse(score)
Related
I want to count up the number of times that a function f returns each value in it's range (0 to f_max, inclusive) when applied to a given list l, and return the result as an array, in Scala.
Currently, I accomplish as follows:
def count (l: List): Array[Int] = {
val arr = new Array[Int](f_max + 1)
l.foreach {
el => arr(f(el)) += 1
}
return arr
}
So arr(n) is the number of times that f returns n when applied to each element of l. This works however, it is imperative style, and I am wondering if there is a clean way to do this purely functionally.
Thank you
how about a more general approach:
def count[InType, ResultType](l: Seq[InType], f: InType => ResultType): Map[ResultType, Int] = {
l.view // create a view so we don't create new collections after each step
.map(f) // apply your function to every item in the original sequence
.groupBy(x => x) // group the returned values
.map(x => x._1 -> x._2.size) // count returned values
}
val f = (i:Int) => i
count(Seq(1,2,3,4,5,6,6,6,4,2), f)
l.foldLeft(Vector.fill(f_max + 1)(0)) { (acc, el) =>
val result = f(el)
acc.updated(result, acc(result) + 1)
}
Alternatively, a good balance of performance and external purity would be:
def count(l: List[???]): Vector[Int] = {
val arr = l.foldLeft(Array.fill(f_max + 1)(0)) { (acc, el) =>
val result = f(el)
acc(result) += 1
}
arr.toVector
}
I have implemented a calculation to obtain the node score of each nodes.
The formula to obtain the value is:
The children list can not be empty or a flag must be true;
The iterative way works pretty well:
class TreeManager {
def scoreNo(nodes:List[Node]): List[(String, Double)] = {
nodes.headOption.map(node => {
val ranking = node.key.toString -> scoreNode(Some(node)) :: scoreNo(nodes.tail)
ranking ::: scoreNo(node.children)
}).getOrElse(Nil)
}
def scoreNode(node:Option[Node], score:Double = 0, depth:Int = 0):Double = {
node.map(n => {
var nodeScore = score
for(child <- n.children){
if(!child.children.isEmpty || child.hasInvitedSomeone == Some(true)){
nodeScore = scoreNode(Some(child), (nodeScore + scala.math.pow(0.5, depth)), depth+1)
}
}
nodeScore
}).getOrElse(score)
}
}
But after i've refactored this piece of code to use recursion, the results are totally wrong:
class TreeManager {
def scoreRecursive(nodes:List[Node]): List[(Int, Double)] = {
def scoreRec(nodes:List[Node], score:Double = 0, depth:Int = 0): Double = nodes match {
case Nil => score
case n =>
if(!n.head.children.isEmpty || n.head.hasInvitedSomeone == Some(true)){
score + scoreRec(n.tail, score + scala.math.pow(0.5, depth), depth + 1)
} else {
score
}
}
nodes.headOption.map(node => {
val ranking = node.key -> scoreRec(node.children) :: scoreRecursive(nodes.tail)
ranking ::: scoreRecursive(node.children)
}).getOrElse(Nil).sortWith(_._2 > _._2)
}
}
The Node is an object of a tree and it's represented by the following class:
case class Node(key:Int,
children:List[Node] = Nil,
hasInvitedSomeone:Option[Boolean] = Some(false))
And here is the part that i'm running to check results:
object Main {
def main(bla:Array[String]) = {
val xx = new TreeManager
val values = List(
Node(10, List(Node(11, List(Node(13))),
Node(12,
List(
Node(14, List(
Node(15, List(Node(18))), Node(17, hasInvitedSomeone = Some(true)),
Node(16, List(Node(19, List(Node(20)))),
hasInvitedSomeone = Some(true))),
hasInvitedSomeone = Some(true))),
hasInvitedSomeone = Some(true))),
hasInvitedSomeone = Some(true)))
val resIterative = xx.scoreNo(values)
//val resRecursive = xx.scoreRec(values)
println("a")
}
}
The iterative way is working because i've checked it but i didn't get why recursive return wrong values.
Any idea?
Thank in advance.
The recursive version never recurses on children of the nodes, just on the tail. Whereas the iterative version correctly both recurse on the children and iterate on the tail.
You'll notice your "iterative" version is also recursive btw.
I am refactoring some scala code and I am having problems with a while loop. The old code was:
for (s <- sentences){
// ...
while (/*Some condition*/){
// ...
function(trees, ...)
}
}
I've have translated that code into this one, using foldLeft to transverse sentences:
sentences./:(initialSeed){
(seed, s) =>
// ...
// Here I've replaced the while with other foldleft
trees./:(seed){
(v, n) =>
// ....
val updatedVariable = function(...., v)
}
}
Now, It may be the case that I need to stop transversing trees (The inner foldLeft before it is transverse entirely, for that I've found this question:
Abort early in a fold
But I also have the following problem:
As I transverse trees, I need to accumulate values to the variable v, function takes v and returns an updated v, called here updatedVariable. The problem is that I have the feeling that this is not a proper way of coding this functionality.
Could you recommended me a functional/immutable way of doing this?
NOTE: I've simplified the code to show the actual problem, the complete code is this:
val trainVocabulart = sentences./:(Vocabulary()){
(vocab, s) =>
var trees = s.tree
var i = 0
var noConstruction = false
trees./:(vocab){
(v, n) =>
if (i == trees.size - 1) {
if (noConstruction) return v
noConstruction = true
i = 0
} else {
// Build vocabulary
val updatedVocab = buildVocabulary(trees, v, i, Config.LeftCtx, Config.RightCtx)
val y = estimateTrainAction(trees, i)
val (newI, newTrees) = takeAction(trees, i, y)
i = newI
trees = newTrees
// Execute the action and modify the trees
if (y != Shift)
noConstruction = false
Vocabulary(v.positionVocab ++ updatedVocab.positionVocab,
v.positionTag ++ updatedVocab.positionTag,
v.chLVocab ++ updatedVocab.chLVocab,
v.chLTag ++ updatedVocab.chLTag,
v.chRVocab ++ updatedVocab.chRVocab,
v.chRTag ++ updatedVocab.chRTag)
}
v
}
}
And the old one:
for (s <- sentences) {
var trees = s.tree
var i = 0
var noConstruction = false
var exit = false
while (trees.nonEmpty && !exit) {
if (i == trees.size - 1) {
if (noConstruction) exit = true
noConstruction = true
i = 0
} else {
// Build vocabulary
buildVocabulary(trees, i, LeftCtx, RightCtx)
val y = estimateTrainAction(trees, i)
val (newI, newTrees) = takeAction(trees, i, y)
i = newI
trees = newTrees
// Execute the action and modify the trees
if (y != Shift)
noConstruction = false
}
}
}
1st - You don't make this easy. Neither your simplified or complete examples are complete enough to compile.
2nd - You include a link to some reasonable solutions to the break-out-early problem. Is there a reason why none of them look workable for your situation?
3rd - Does that complete example actually work? You're folding over a var ...
trees./:(vocab){
... and inside that operation you modify/update that var ...
trees = newTrees
According to my tests that's a meaningless statement. The original iteration is unchanged by updating the collection.
4th - I'm not convinced that fold is what you want here. fold iterates over a collection and reduces it to a single value, but your aim here doesn't appear to be finding that single value. The result of your /: is thrown away. There is no val result = trees./:(vocab){...
One solution you might look at is: trees.forall{ ... At the end of each iteration you just return true if the next iteration should proceed.
I have a LinkedHashMap which I've been using in a typical way: adding new key-value
pairs to the end, and accessing them in order of insertion. However, now I have a
special case where I need to add pairs to the "head" of the map. I think there's
some functionality inside the LinkedHashMap source for doing this, but it has private
accessibility.
I have a solution where I create a new map, add the pair, then add all the old mappings.
In Java syntax:
newMap.put(newKey, newValue)
newMap.putAll(this.map)
this.map = newMap
It works. But the problem here is that I then need to make my main data structure
(this.map) a var rather than a val.
Can anyone think of a nicer solution? Note that I definitely need the fast lookup
functionality provided by a Map collection. The performance of a prepending is not
such a big deal.
More generally, as a Scala developer how hard would you fight to avoid a var
in a case like this, assuming there's no foreseeable need for concurrency?
Would you create your own version of LinkedHashMap? Looks like a hassle frankly.
This will work but is not especially nice either:
import scala.collection.mutable.LinkedHashMap
def prepend[K,V](map: LinkedHashMap[K,V], kv: (K, V)) = {
val copy = map.toMap
map.clear
map += kv
map ++= copy
}
val map = LinkedHashMap('b -> 2)
prepend(map, 'a -> 1)
map == LinkedHashMap('a -> 1, 'b -> 2)
Have you taken a look at the code of LinkedHashMap? The class has a field firstEntry, and just by taking a quick peek at updateLinkedEntries, it should be relatively easy to create a subclass of LinkedHashMap which only adds a new method prepend and updateLinkedEntriesPrepend resulting in the behavior you need, e.g. (not tested):
private def updateLinkedEntriesPrepend(e: Entry) {
if (firstEntry == null) { firstEntry = e; lastEntry = e }
else {
val oldFirstEntry = firstEntry
firstEntry = e
firstEntry.later = oldFirstEntry
oldFirstEntry.earlier = e
}
}
Here is a sample implementation I threw together real quick (that is, not thoroughly tested!):
class MyLinkedHashMap[A, B] extends LinkedHashMap[A,B] {
def prepend(key: A, value: B): Option[B] = {
val e = findEntry(key)
if (e == null) {
val e = new Entry(key, value)
addEntry(e)
updateLinkedEntriesPrepend(e)
None
} else {
// The key already exists, so we might as well call LinkedHashMap#put
put(key, value)
}
}
private def updateLinkedEntriesPrepend(e: Entry) {
if (firstEntry == null) { firstEntry = e; lastEntry = e }
else {
val oldFirstEntry = firstEntry
firstEntry = e
firstEntry.later = oldFirstEntry
oldFirstEntry.earlier = firstEntry
}
}
}
Tested like this:
object Main {
def main(args:Array[String]) {
val x = new MyLinkedHashMap[String, Int]();
x.prepend("foo", 5)
x.prepend("bar", 6)
x.prepend("olol", 12)
x.foreach(x => println("x:" + x._1 + " y: " + x._2 ));
}
}
Which, on Scala 2.9.0 (yeah, need to update) results in
x:olol y: 12
x:bar y: 6
x:foo y: 5
A quick benchmark shows order of magnitude in performance difference between the extended built-in class and the "map rewrite" approach (I used the code from Debilski's answer in "ExternalMethod" and mine in "BuiltIn"):
benchmark length us linear runtime
ExternalMethod 10 1218.44 =
ExternalMethod 100 1250.28 =
ExternalMethod 1000 19453.59 =
ExternalMethod 10000 349297.25 ==============================
BuiltIn 10 3.10 =
BuiltIn 100 2.48 =
BuiltIn 1000 2.38 =
BuiltIn 10000 3.28 =
The benchmark code:
def timeExternalMethod(reps: Int) = {
var r = reps
while(r > 0) {
for(i <- 1 to 100) prepend(map, (i, i))
r -= 1
}
}
def timeBuiltIn(reps: Int) = {
var r = reps
while(r > 0) {
for(i <- 1 to 100) map.prepend(i, i)
r -= 1
}
}
Using a scala benchmarking template.
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) =>