I have an Option of a string. I want to update the contained value:
if(x.isEmpty) {
...another calculation
} else {
x.map(val => ...update val)
}
Is this an idiomatic way?
x.fold(another calculation)(v => ...update v)
e.g.
x.fold("no value")("Value is " + _)
Note that this extracts the value from the Option so if you want to have the result as an Option you need to wrap it in Some.
Note that if your inner computation gets too long or unreadable for a fold, there's always good old-fashioned pattern matching.
x match {
case None => {
// None case ...
}
case Some(y) => {
// Some case (y is the inside) ...
}
}
Like everything in Scala, this is an expression, so it can be assigned to a variable or used in another other expression you like.
Alternatively, using the lazy keyword:
// T is the type of the value
val valueOpt: Option[T] = ???
lazy val backupValue: T = ??? // Other calculation, but only evaluated if needed
val value: T = valueOpt.map( v => /*Update v*/ ).getOrElse( backupValue )
// do something with the value you want to manipulate
I have a configuration value that matches to one of the values in a map and depending on to which it matches i take an action. Here is some sample code of what i am trying to do
val x = 1 // or 2 or 3
val config = Map("c1"-> 1, "c2"-> 2, "c3"-> 3)
x match {
case config("c1") =>
println("1")
case config("c2") =>
println("2")
case config("c3") =>
println("3")
}
Now this should print 1 because config("c1") evaluates to 1 but it gives error
error: value config is not a case class, nor does it have an unapply/unapplySeq member
case config("c1") =>
Similarly for the other 2 cases. Why should i have an unapply here? Any pointers?
An expression like that looks like an extractor, hence the message about unapply/unapplySeq methods. If you don't want to use an extractor but just want to match against a plain value, you need to store that value in a stable identifier - you can't use an arbitrary expression as a match case:
val case1 = config("c1")
x match {
case case1 => println("1")
...
}
To the best of my knowledge, in Scala, x match {case config("c1") gets translated to config.unapply(x) with the branching dependent on the result of the unapply method. As Imm already mentioned in his answer, this isn't the case for stable identifiers (literals and val), and I'd encourage you to use his solution.
Nevertheless, to show you how you could solve the problem using extractors, I'd like to post a different solution:
def main(args: Array[String]): Unit = {
object config {
val configData = Map("c1" -> 1, "c2" -> 2, "c3" -> 3)
def unapply(value: Int): Option[String] = configData find (_._2 == value) map (_._1)
}
1 to 4 foreach {
case config("c1") => println("1")
case config("c2") => println("2")
case config("c3") => println("3")
case _ => println("no match")
}
}
I changed the match for a foreach to show the different results, but this has no effect on the implementation. This would print:
1
2
3
no match
As you can see, case config("c1") now calls the unapply method and checks whether the result is Some("c1"). Note that this is inverse to how you'd use a map: The key is searched according to the value. However, this makes sense: If in the map, "c1" and "c2" both map to 1, then 1 matches both, the same way _ matches everything, in our case even 4 which is not configured.
Here's also a very brief tutorial on extractors. I don't find it particularly good, because both, the returned type and the argument type are Int, but it might help you understand what's going on.
As others have stated, with x match { case config("c1") => ..., scala looks for an extractor by the name of config (something with an unapply method that takes a single value and returns an Optional value); Making pattern matching work this way seems like an abuse of the pattern, and I would not use an extractor for this.
Personally, I would recommend one of the following:
if (x == config("c1"))
println("1")
else if (x == config("c2"))
println("2")
else ...
Or, if you're set on using a match statement, you can use conditionals like this:
x match {
case _ if x == config("c1") =>
println("1")
case _ if x == config("c2") =>
println("2")
case _ if x == config("c3") =>
println("3")
}
Not as clean; unfortunately, there isn't a way to invoke a method call literally where the extractor goes. You can use back-ticks to tell scala "match against the value of this variable" (rather than default behavior, which would yield the value named as that variable):
val (c1,c2,c3) = (config("c1"), config("c2"), config("c3"))
x match {
case `c1` =>
println("1")
case `c2` =>
println("2")
case `c3` =>
println("3")
}
Finally, if your goal is to reverse-apply a map, maybe try this instead?
scala> Map("a" -> 1).map { case (k,v) => (v,k) }
res0: scala.collection.immutable.Map[Int,String] = Map(1 -> a)
I have some expensive computation in a loop, and I need to find max value produced by the calculations, though if, say, it will equal to LIMIT I'd like to stop the calculation and return my accumulator.
It may easily be done by recursion:
val list: List[Int] = ???
val UpperBound = ???
def findMax(ls: List[Int], max: Int): Int = ls match {
case h :: rest =>
val v = expensiveComputation(h)
if (v == UpperBound) v
else findMax(rest, math.max(max, v))
case _ => max
}
findMax(list, 0)
My question: whether this behaviour template has a name and reflected in scala collection library?
Update: Do something up to N times or until condition is met in Scala - There is an interesting idea (using laziness and find or exists at the end) but it is not directly applicable to my particular case or requires mutable var to track accumulator.
I think your recursive function is quite nice, so honestly I wouldn't change that, but here's a way to use the collections library:
list.foldLeft(0) {
case (max, next) =>
if(max == UpperBound)
max
else
math.max(expensiveComputation(next), max)
}
It will iterate over the whole list, but after it has hit the upper bound it won't perform the expensive computation.
Update
Based on your comment I tried adapting foldLeft a bit, based on LinearSeqOptimized's foldLeft implementation.
def foldLeftWithExit[A, B](list: Seq[A])(z: B)(exit: B => Boolean)(f: (B, A) => B): B = {
var acc = z
var remaining = list
while (!remaining.isEmpty && !exit(acc)) {
acc = f(acc, list.head)
remaining = remaining.tail
}
acc
}
Calling it:
foldLeftWithExit(list)(0)(UpperBound==){
case (max, next) => math.max(expensiveComputation(next), max)
}
You could potentially use implicits to omit the first parameter of list.
Hope this helps.
Scenario:
val col: IndexedSeq[Array[Char]] = for (i <- 1 to n) yield {
val x = for (j <- 1 to m) yield 'x'
x.toArray
}
This is a fairly simple char matrix. toArray used to allow updating.
var west = last.x - 1
while (west >= 0 && arr(last.y)(west) == '.') {
arr(last.y)(west) = ch;
west -= 1;
}
This is updating all . to ch until a non-dot char is found.
Generically, update until stop condition is met, unknown number of steps.
What is the idiomatic equivalent of it?
Conclusion
It's doable, but the trade-off isn't worth it, a lot of performance is lost to expressive syntax when the collection allows updating.
Your wish for a "cleaner, more idiomatic" solution is of course a little fuzzy, because it leaves a lot of room for subjectivity. In general, I'd consider a tail-recursive updating routine more idiomatic, but it might not be "cleaner" if you're more familiar with a non-functional programming style. I came up with this:
#tailrec
def update(arr:List[Char], replace:Char, replacement:Char, result:List[Char] = Nil):List[Char] = arr match {
case `replace` :: tail =>
update(tail, replace, replacement, replacement :: result)
case _ => result.reverse ::: arr
}
This takes one of the inner sequences (assuming a List for easier pattern matching, since Arrays are trivially convertible to lists), and replaces the replace char with the replacement recursively.
You can then use map to update the outer sequence, like so:
col.map { x => update(x, '.', ch) }
Another more reusable alternative is writing your own mapUntil, or using one which is implemented in a supplemental library (Scalaz probably has something like it). The one I came up with looks like this:
def mapUntil[T](input:List[T])(f:(T => Option[T])) = {
#tailrec
def inner(xs:List[T], result:List[T]):List[T] = xs match {
case Nil => Nil
case head :: tail => f(head) match {
case None => (head :: result).reverse ::: tail
case Some(x) => inner(tail, x :: result)
}
}
inner(input, Nil)
}
It does the same as a regular map invocation, except that it stops as soon as the passed function returns None, e.g.
mapUntil(List(1,2,3,4)) {
case x if x >= 3 => None
case x => Some(x-1)
}
Will result in
List[Int] = List(0, 1, 3, 4)
If you want to look at Scalaz, this answer might be a good place to start.
x3ro's answer is the right answer, esp. if you care about performance or are going to be using this operation in multiple places. I would like to add simple solution using only what you find in the collections API:
col.map { a =>
val (l, r) = a.span(_ == '.')
l.map {
case '.' => ch
case x => x
} ++ r
}
What's the best way to terminate a fold early? As a simplified example, imagine I want to sum up the numbers in an Iterable, but if I encounter something I'm not expecting (say an odd number) I might want to terminate. This is a first approximation
def sumEvenNumbers(nums: Iterable[Int]): Option[Int] = {
nums.foldLeft (Some(0): Option[Int]) {
case (Some(s), n) if n % 2 == 0 => Some(s + n)
case _ => None
}
}
However, this solution is pretty ugly (as in, if I did a .foreach and a return -- it'd be much cleaner and clearer) and worst of all, it traverses the entire iterable even if it encounters a non-even number.
So what would be the best way to write a fold like this, that terminates early? Should I just go and write this recursively, or is there a more accepted way?
My first choice would usually be to use recursion. It is only moderately less compact, is potentially faster (certainly no slower), and in early termination can make the logic more clear. In this case you need nested defs which is a little awkward:
def sumEvenNumbers(nums: Iterable[Int]) = {
def sumEven(it: Iterator[Int], n: Int): Option[Int] = {
if (it.hasNext) {
val x = it.next
if ((x % 2) == 0) sumEven(it, n+x) else None
}
else Some(n)
}
sumEven(nums.iterator, 0)
}
My second choice would be to use return, as it keeps everything else intact and you only need to wrap the fold in a def so you have something to return from--in this case, you already have a method, so:
def sumEvenNumbers(nums: Iterable[Int]): Option[Int] = {
Some(nums.foldLeft(0){ (n,x) =>
if ((n % 2) != 0) return None
n+x
})
}
which in this particular case is a lot more compact than recursion (though we got especially unlucky with recursion since we had to do an iterable/iterator transformation). The jumpy control flow is something to avoid when all else is equal, but here it's not. No harm in using it in cases where it's valuable.
If I was doing this often and wanted it within the middle of a method somewhere (so I couldn't just use return), I would probably use exception-handling to generate non-local control flow. That is, after all, what it is good at, and error handling is not the only time it's useful. The only trick is to avoid generating a stack trace (which is really slow), and that's easy because the trait NoStackTrace and its child trait ControlThrowable already do that for you. Scala already uses this internally (in fact, that's how it implements the return from inside the fold!). Let's make our own (can't be nested, though one could fix that):
import scala.util.control.ControlThrowable
case class Returned[A](value: A) extends ControlThrowable {}
def shortcut[A](a: => A) = try { a } catch { case Returned(v) => v }
def sumEvenNumbers(nums: Iterable[Int]) = shortcut{
Option(nums.foldLeft(0){ (n,x) =>
if ((x % 2) != 0) throw Returned(None)
n+x
})
}
Here of course using return is better, but note that you could put shortcut anywhere, not just wrapping an entire method.
Next in line for me would be to re-implement fold (either myself or to find a library that does it) so that it could signal early termination. The two natural ways of doing this are to not propagate the value but an Option containing the value, where None signifies termination; or to use a second indicator function that signals completion. The Scalaz lazy fold shown by Kim Stebel already covers the first case, so I'll show the second (with a mutable implementation):
def foldOrFail[A,B](it: Iterable[A])(zero: B)(fail: A => Boolean)(f: (B,A) => B): Option[B] = {
val ii = it.iterator
var b = zero
while (ii.hasNext) {
val x = ii.next
if (fail(x)) return None
b = f(b,x)
}
Some(b)
}
def sumEvenNumbers(nums: Iterable[Int]) = foldOrFail(nums)(0)(_ % 2 != 0)(_ + _)
(Whether you implement the termination by recursion, return, laziness, etc. is up to you.)
I think that covers the main reasonable variants; there are some other options also, but I'm not sure why one would use them in this case. (Iterator itself would work well if it had a findOrPrevious, but it doesn't, and the extra work it takes to do that by hand makes it a silly option to use here.)
The scenario you describe (exit upon some unwanted condition) seems like a good use case for the takeWhile method. It is essentially filter, but should end upon encountering an element that doesn't meet the condition.
For example:
val list = List(2,4,6,8,6,4,2,5,3,2)
list.takeWhile(_ % 2 == 0) //result is List(2,4,6,8,6,4,2)
This will work just fine for Iterators/Iterables too. The solution I suggest for your "sum of even numbers, but break on odd" is:
list.iterator.takeWhile(_ % 2 == 0).foldLeft(...)
And just to prove that it's not wasting your time once it hits an odd number...
scala> val list = List(2,4,5,6,8)
list: List[Int] = List(2, 4, 5, 6, 8)
scala> def condition(i: Int) = {
| println("processing " + i)
| i % 2 == 0
| }
condition: (i: Int)Boolean
scala> list.iterator.takeWhile(condition _).sum
processing 2
processing 4
processing 5
res4: Int = 6
You can do what you want in a functional style using the lazy version of foldRight in scalaz. For a more in depth explanation, see this blog post. While this solution uses a Stream, you can convert an Iterable into a Stream efficiently with iterable.toStream.
import scalaz._
import Scalaz._
val str = Stream(2,1,2,2,2,2,2,2,2)
var i = 0 //only here for testing
val r = str.foldr(Some(0):Option[Int])((n,s) => {
println(i)
i+=1
if (n % 2 == 0) s.map(n+) else None
})
This only prints
0
1
which clearly shows that the anonymous function is only called twice (i.e. until it encounters the odd number). That is due to the definition of foldr, whose signature (in case of Stream) is def foldr[B](b: B)(f: (Int, => B) => B)(implicit r: scalaz.Foldable[Stream]): B. Note that the anonymous function takes a by name parameter as its second argument, so it need no be evaluated.
Btw, you can still write this with the OP's pattern matching solution, but I find if/else and map more elegant.
Well, Scala does allow non local returns. There are differing opinions on whether or not this is a good style.
scala> def sumEvenNumbers(nums: Iterable[Int]): Option[Int] = {
| nums.foldLeft (Some(0): Option[Int]) {
| case (None, _) => return None
| case (Some(s), n) if n % 2 == 0 => Some(s + n)
| case (Some(_), _) => None
| }
| }
sumEvenNumbers: (nums: Iterable[Int])Option[Int]
scala> sumEvenNumbers(2 to 10)
res8: Option[Int] = None
scala> sumEvenNumbers(2 to 10 by 2)
res9: Option[Int] = Some(30)
EDIT:
In this particular case, as #Arjan suggested, you can also do:
def sumEvenNumbers(nums: Iterable[Int]): Option[Int] = {
nums.foldLeft (Some(0): Option[Int]) {
case (Some(s), n) if n % 2 == 0 => Some(s + n)
case _ => return None
}
}
You can use foldM from cats lib (as suggested by #Didac) but I suggest to use Either instead of Option if you want to get actual sum out.
bifoldMap is used to extract the result from Either.
import cats.implicits._
def sumEven(nums: Stream[Int]): Either[Int, Int] = {
nums.foldM(0) {
case (acc, n) if n % 2 == 0 => Either.right(acc + n)
case (acc, n) => {
println(s"Stopping on number: $n")
Either.left(acc)
}
}
}
examples:
println("Result: " + sumEven(Stream(2, 2, 3, 11)).bifoldMap(identity, identity))
> Stopping on number: 3
> Result: 4
println("Result: " + sumEven(Stream(2, 7, 2, 3)).bifoldMap(identity, identity))
> Stopping on number: 7
> Result: 2
Cats has a method called foldM which does short-circuiting (for Vector, List, Stream, ...).
It works as follows:
def sumEvenNumbers(nums: Stream[Int]): Option[Long] = {
import cats.implicits._
nums.foldM(0L) {
case (acc, c) if c % 2 == 0 => Some(acc + c)
case _ => None
}
}
If it finds a not even element it returns None without computing the rest, otherwise it returns the sum of the even entries.
If you want to keep count until an even entry is found, you should use an Either[Long, Long]
#Rex Kerr your answer helped me, but I needed to tweak it to use Either
def foldOrFail[A,B,C,D](map: B => Either[D, C])(merge: (A, C) => A)(initial: A)(it: Iterable[B]): Either[D, A] = {
val ii= it.iterator
var b= initial
while (ii.hasNext) {
val x= ii.next
map(x) match {
case Left(error) => return Left(error)
case Right(d) => b= merge(b, d)
}
}
Right(b)
}
You could try using a temporary var and using takeWhile. Here is a version.
var continue = true
// sample stream of 2's and then a stream of 3's.
val evenSum = (Stream.fill(10)(2) ++ Stream.fill(10)(3)).takeWhile(_ => continue)
.foldLeft(Option[Int](0)){
case (result,i) if i%2 != 0 =>
continue = false;
// return whatever is appropriate either the accumulated sum or None.
result
case (optionSum,i) => optionSum.map( _ + i)
}
The evenSum should be Some(20) in this case.
You can throw a well-chosen exception upon encountering your termination criterion, handling it in the calling code.
A more beutiful solution would be using span:
val (l, r) = numbers.span(_ % 2 == 0)
if(r.isEmpty) Some(l.sum)
else None
... but it traverses the list two times if all the numbers are even
Just for an "academic" reasons (:
var headers = Source.fromFile(file).getLines().next().split(",")
var closeHeaderIdx = headers.takeWhile { s => !"Close".equals(s) }.foldLeft(0)((i, S) => i+1)
Takes twice then it should but it is a nice one liner.
If "Close" not found it will return
headers.size
Another (better) is this one:
var headers = Source.fromFile(file).getLines().next().split(",").toList
var closeHeaderIdx = headers.indexOf("Close")