If I wanted to pattern match on a basic option type in Scala, I would run something along the lines of
val opt = Option(5)
val lessThanTen = opt match {
case Some(e) => if (e < 10) true else false
case None => None
}
But suppose that opt comes as a result of one of Slick's Queries, and therefore has the Lifted Embedding Type of Rep[Option[Int]]
How can I carry out the same pattern matching in a way that allows us the to see inside of the the lifted type? I.e. something along the lines of
val opt = Rep(Option(5))
val lessThanTen = opt match {
case Rep[Some(e)] => Rep[if (e < 10) true else false]
case Rep[None] => Rep[None]
}
But of course, one that compiles ;)
You can use the map method to apply some operation on the content of a Rep.
val rep: Rep[Option[Int]] = ???
val boolRep = rep.map {
case Some(i) => Some(i < 10)
case None => None
}
Even better: Option, like many other collection types in Scala, also has a similar map method, so you can write
val boolRep = rep.map(_.map(_ < 10))
In that expression, the first _ is the Option[Int], and the second one is the Int itself. In cases where the Option[Int] is None, the map method has nothing to apply the given function to, so it returns None by definition.
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 get first argument for main method that is optional, something like this:
val all = args(0) == "all"
However, this would fail with exception if no argument is provided.
Is there any one-liner simple method to set all to false when args[0] is missing; and not doing the common if-no-args-set-false-else... thingy?
In general case you can use lifting:
args.lift(0).map(_ == "all").getOrElse(false)
Or even (thanks to #enzyme):
args.lift(0).contains("all")
You can use headOption and fold (on Option):
val all = args.headOption.fold(false)(_ == "all")
Of course, as #mohit pointed out, map followed by getOrElse will work as well.
If you really need indexed access, you could pimp a get method on any Seq:
implicit class RichIndexedSeq[V, T <% Seq[V]](seq: T) {
def get(i: Int): Option[V] =
if (i < 0 || i >= seq.length) None
else Some(seq(i))
}
However, if this is really about arguments, you'll be probably better off, handling arguments in a fold:
case class MyArgs(n: Int = 1, debug: Boolean = false,
file: Option[String] = None)
val myArgs = args.foldLeft(MyArgs()) {
case (args, "-debug") =>
args.copy(debug = true)
case (args, str) if str.startsWith("-n") =>
args.copy(n = ???) // parse string
case (args, str) if str.startsWith("-f") =>
args.copy(file = Some(???) // parse string
case _ =>
sys.error("Unknown arg")
}
if (myArgs.file.isEmpty)
sys.error("Need file")
You can use foldLeft with initial false value:
val all = (false /: args)(_ | _ == "all")
But be careful, One Liners can be difficult to read.
Something like this will work assuming args(0) returns Some or None:
val all = args(0).map(_ == "all").getOrElse(false)
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")
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) =>