I am trying to use a foldLeft on an array. Eg:
var x = some array
x.foldLeft(new Array[Int](10))((a, c) => a(c) = a(c)+1)
This refuses to compile with the error found Int(0) required Array[Int].
In order to use foldLeft in what you want to do, and following your style, you can just return the same accumulator array in the computation like this:
val ret = a.foldLeft(new Array[Int](10)) {
(acc, c) => acc(c) += 1; acc
}
Alternatively, since your numbers are from 0 to 9, you can also do this to achieve the same result:
val ret = (0 to 9).map(x => a.count(_ == x))
Assignment in Scala does not return a value (but instead Unit) so your expression that is supposed to return the Array[Int] for the next step returns Unit which does not work.
You would have to use a block and return the array in the end like this:
x.foldLeft(new Array[Int](10)) { (a, c) =>
a(c) = a(c)+1
a
}
Related
I might have something like this:
val found = source.toCharArray.foreach{ c =>
// Process char c
// Sometimes (e.g. on newline) I want to emit a result to be
// captured in 'found'. There may be 0 or more captured results.
}
This shows my intent. I want to iterate over some collection of things. Whenever the need arrises I want to "emit" a result to be captured in found. It's not a direct 1-for-1 like map. collect() is a "pull", applying a partial function over the collection. I want a "push" behavior, where I visit everything but push out something when needed.
Is there a pattern or collection method I'm missing that does this?
Apparently, you have a Collection[Thing], and you want to obtain a new Collection[Event] by emitting a Collection[Event] for each Thing. That is, you want a function
(Collection[Thing], Thing => Collection[Event]) => Collection[Event]
That's exactly what flatMap does.
You can write it down with nested fors where the second generator defines what "events" have to be "emitted" for each input from the source. For example:
val input = "a2ba4b"
val result = (for {
c <- input
emitted <- {
if (c == 'a') List('A')
else if (c.isDigit) List.fill(c.toString.toInt)('|')
else Nil
}
} yield emitted).mkString
println(result)
prints
A||A||||
because each 'a' emits an 'A', each digit emits the right amount of tally marks, and all other symbols are ignored.
There are several other ways to express the same thing, for example, the above expression could also be rewritten with an explicit flatMap and with a pattern match instead of if-else:
println(input.flatMap{
case 'a' => "A"
case d if d.isDigit => "|" * (d.toString.toInt)
case _ => ""
})
I think you are looking for a way to build a Stream for your condition. Streams are lazy and are computed only when required.
val sourceString = "sdfdsdsfssd\ndfgdfgd\nsdfsfsggdfg\ndsgsfgdfgdfg\nsdfsffdg\nersdff\n"
val sourceStream = sourceString.toCharArray.toStream
def foundStreamCreator( source: Stream[Char], emmitBoundaryFunction: Char => Boolean): Stream[String] = {
def loop(sourceStream: Stream[Char], collector: List[Char]): Stream[String] =
sourceStream.isEmpty match {
case true => collector.mkString.reverse #:: Stream.empty[String]
case false => {
val char = sourceStream.head
emmitBoundaryFunction(char) match {
case true =>
collector.mkString.reverse #:: loop(sourceStream.tail, List.empty[Char])
case false =>
loop(sourceStream.tail, char :: collector)
}
}
}
loop(source, List.empty[Char])
}
val foundStream = foundStreamCreator(sourceStream, c => c == '\n')
val foundIterator = foundStream.toIterator
foundIterator.next()
// res0: String = sdfdsdsfssd
foundIterator.next()
// res1: String = dfgdfgd
foundIterator.next()
// res2: String = sdfsfsggdfg
It looks like foldLeft to me:
val found = ((List.empty[String], "") /: source.toCharArray) {case ((agg, tmp), char) =>
if (char == '\n') (tmp :: agg, "") // <- emit
else (agg, tmp + char)
}._1
Where you keep collecting items in a temporary location and then emit it when you run into a character signifying something. Since I used List you'll have to reverse at the end if you want it in order.
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 a map m
val m = Map(1->2, 3->4, 5->6, 7->8, 4->4, 9->9, 10->12, 11->11)
Now i want a map whose keys are equal to the values. So i do this
def eq(k: Int, v: Int) = if (k == v) Some(k->v) else None
m.flatMap((k,v) => eq(k,v))
This gives me the error
error: wrong number of parameters; expected = 1
m.flatMap((k,v) => eq(k,v))
Whats wrong with the above code? flatMap expects a one argument function and here i am passing one argument which is a Pair of integers.
Also this works
m.flatMap {
case (k,v) => eq(k,v)
}
but this does not
m.flatMap {
(k,v) => eq(k,v)
}
Looks like i am missing something. Help?
There is no such syntax:
m.flatMap((k,v) => eq(k,v))
Well, in fact there is such syntax, but actually it is used in functions that accept two arguments (like reduce):
List(1,2,3,4).reduce((acc, x) => acc + x)
The
m.flatMap {
case (k,v) => eq(k,v)
}
syntax works because in fact it is something like this:
val temp: PartialFunction[Tuple2[X,Y], Tuple2[Y,X]] = {
case (k,v) => eq(k,v) // using literal expression to construct function
}
m.flatMap(temp) // with braces ommited
They key thing here is the usage of case word (actually, there is a discussion to enable your very syntax) which turns usual braces expression, like { ... } into full blown anonymous partial function
(If you want to simply fix the error you're getting, see the 2nd solution (with flatMap); if you want a generally nicer solution, read from the beginning.)
What you need instead is filter not flatMap:
def eq(k: Int, v: Int) = k == v
val m = Map(1->2, 3->4, 5->6, 7->8, 4->4, 9->9, 10->12, 11->11)
m.filter((eq _).tupled)
...which of course reduces to just the following, without the need for eq:
m.filter { case (k, v) => k == v }
result:
Map(9 -> 9, 11 -> 11, 4 -> 4)
OR... If you want to stick with flatMap
First you must know that flatMap will pass to your function TUPLES not keys and values as separate arguments.
Additionally, you must change the Option returned by eq to something that can be fed back to flatMap on sequences such as List or Map (actually any GenTraversableOnce to be precise):
def eq(k: Int, v: Int) = if (k == v) List(k -> v) else Nil
m.flatMap { case (k,v) => eq(k,v) } // use pattern matching to unpack the tuple
or the uglier but equivalent:
m.flatMap { x => eq(x._1, x._2) }
alternatively, you can convert eq to take a tuple instead:
m.flatMap((eq _).tupled)
I think that what you want is a single argument that will be a couple, not two arguments. Something like this may work
m.flatMap(k => eq(k._1, k._2))
The code snippet that works uses pattern matching. You give names to both elements of your couple. It's a partial function and can be use here in your flatMap.
You have to do:
m.flatMap { case (k,v) => eq(k,v) }
Note that here I switch to curly braces, which indicates a function block rather than parameters, and the function here is a case statement. This means that the function block I'm passing to flatMap is a partialFunction that is only invoked for items that match the case statement.
Your eq function takes two parameters, that is why you are getting the type error. Try:
def f(p: (Int, Int)) = if (p._1 == p._2) Some(p) else None
m flatMap f
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) =>
var set = TreeSet(5,4,3,2,1)
println(set)
val diffSet: TreeSet[Int] = set
// if I change above code to val diffSet: Set[Int] = set
// the result is unsorted set.
for (i <- diffSet; x = i) {
println(i)
}
println("-" * 20)
// the above code translates to below and print the same result
val temp = diffSet.map(i => (i, i))
for ((i, x) <- temp) {
println(i)
}
My question is if I defined a method like this:
def genSet:Set[Int] = {
TreeSet(5, 4, 3, 2, 1)
}
and when i want to use a for loop with it
for (i <- genSet; x = i + 1) {
println(x)
}
the result is unsorted, how to fix this behavior without change the genSet's return type. if I use for loop like below, it will be fine, but I hope to keep the above code style.
for (i <- genSet) {
val x = i + 1
println(x)
}
Why the map version winds up unsorted
The map method (called with a function that we'll call func) takes an implicit CanBuildFrom parameter that takes into account the type of the collection that map is being called on, in addition to the type that func returns to choose an appropriate return type. This is used to make Map.map[Int] or BitSet.map[String] do the right thing (return general purpose lists) while Map.map[(String,Int)] or BitSet.map[Int] also do the right thing (return a Map and a BitSet) respectively.
The CanBuildFrom is chosen at compile time, so it must be chosen based on the static type of the set that you call map on (the type the compiler knows about at compile time). The static type of set is TreeSet, but the static type of diffset is Set. The dynamic type of both (at runtime) is TreeSet.
When you call map on set (a TreeSet), the compiler chooses immutable.this.SortedSet.canBuildFrom[Int](math.this.Ordering.Int) as the CanBuildFrom.
When you call map on diffset (a Set), the compiler chooses immutable.this.Set.canBuildFrom[Int] as the CanBuildFrom.
Why the for version winds up unsorted
The loop
for (i <- genSet; x = i + 1) {
println(x)
}
desugars into
genSet.map(((i) => {
val x = i.$plus(1);
scala.Tuple2(i, x)
})).foreach(((x$1) => x$1: #scala.unchecked match {
case scala.Tuple2((i # _), (x # _)) => println(x)
}))
The desugared version includes a map function which will use the unsorted CanBuildFrom as I explained above.
On the other hand, the loop
for (i <- genSet) {
val x = i + 1
println(x)
}
desugars into
genSet.foreach(((i) => {
val x = i.$plus(1);
println(x)
}))
Which doesn't use a CanBuildFrom at all, since no new collection is being returned.
Set does not guarantee ordering. Even if the underlying class is a TreeSet, if the expected result is a Set you'll loose the ordering in the first transformation you do.
If you want ordering, do not use Set. I suggest, say, SortedSet.
Change the sig of genSet to return a SortedSet
def genSet:SortedSet[Int] = {
TreeSet(5, 4, 3, 2, 1)
}
This is probably some sort of bug. I would have expected your code to work too.
I think map is the culprit. This results in the same behavior:
for (i <- genSet.map(_ + 1)) { println(i) }
And for(i <- genSet; x = i + 1) equates to for(x <- genSet.map({i => i + 1}))
You can do:
scala> for (i <-genSet.view; x = i + 1) println(x)
2
3
4
5
6
Although, it's the type of trick that when you look at it after a few months, you may wonder why you added .view ...