Iterate over Option instances until the first non-empty is found - scala

I have a number of functions that return Option values, like this
case class A()
case class B()
case class C()
def optionA(): Option[A] = None
def optionB(): Option[B] = Some(B())
def optionC(): Option[C] = Some(C())
What I want to do is, I want to run these functions in sequence, but only until one of the functions returns an Option with a value (a Some). Then I want to have that value returned, without running the remaining functions.
This is my current implementation
val res:Option[Any] = Stream(
() => optionA(),
() => optionB(),
() => optionC()
) .map(f => f())
.filter(opt => opt.isDefined)
.head
For the function implementations above, this applies optionA and optionB, gives me a Some(B()), and it never runs optionC, which is what I want.
But I'd like to know if there is is a better/simple/alternative implementation.
Something like val findFirst = optionA compose optionB compose optionC?

optionA().orElse(optionB()).orElse(optionC())
orElse will not evaluate its argument if this is defined.
Or if you have already the options in a collection/stream, you might do
options.find(_.isDefined).flatten

Say you now have a collection of Options, you can then do this:
coll.foldLeft[Option[Int]](None)(_ orElse _)
Which will return you the first non-None value in the collection
Note that I explicitly mention the type of the collection, because scala can't infer what orElse should do without it... (None is of type Option[Any] by default)
If you have a giant list of options, it might be helpful to write
coll.view.foldLeft[Option[Int]](None)(_ orElse _)

Related

Scala - function map is a pattern matching or an interation

I had spent weeks on trying to understand the idea behind "lifting" in scala.
Originally, it was from the example related to Chapter 4 of book "Functional Programming in Scala"
Then I found below topic "How map work on Options in Scala?"
The selected answer specify that:
def map[B](f: A => B): Option[B] = this match (Let's considered this as (*) )
So, from above code, I assume that function "map" is derived from function match. Hence, the mechanism behind "map"
is a kind of pattern matching to provide a case selection between Some, and None
Then, I created below examples by using function map for Seq, Option, and Map (Let's considered below examples as (**) )
Example 1: map for Seq
val xs = Seq(1, 2, 3)
xs.map(println)
Example 2: map for Option
val a:Option[Int] = Some(5)
a.map(println)
val b:Option[Int] = None
b.map(println)
Example 3: map for Map
val capitals = Map("France" -> "Paris", "Japan" -> "Tokyo")
capitals.map(println)
From (*) and (**), I could not know whether "map" is a pattern matching or an iteration, or both.
Thank you for helping me to understand this.
#Jwvh provided a more programming based answer but I want to dig a little bit deeper.
I certainly appreciate you trying to understand how things work in Scala, however if you really want to dig that deep, I am afraid you will need to obtain some basic knowledge of Category Theory since there is no "idea behind lifting in scala" but just the "idea behind lifting"
This is also why functions like "map" can be very confusing. Inherently, programmers are taught map etc. as operations on collections, where as they are actually operations that come with Functors and Natural Transformations (this is normally referred to as fmap in Category Theory and also Haskell).
Before I move on, the short answer is it is a pattern matching in the examples you gave and in some of them it is both. Map is defined specifically to the case, the only condition is that it maintains functoriality
Attention: I will not be defining every single term below, since I would need to write a book to build up to some of the following definitions, interested readers are welcome to research them on their own. You should be able to get some basic understanding by following the types
Let's consider these as Functors, the definition will be something along the lines of this:
In (very very) short, we consider types as objects in the category of our language. The functions between these types (type constructors) are morphisms between types in this category. The set of these transformations are called Endo-Functors (take us from the category of Scala and drop us back in the category of Scala). Functors have to have a polymorphic (which actually has a whole different (extra) definition in category theory) map function, that will take some object A, through some type constructor turn it into object B.
implicit val option: Functor[Option] = new Functor[Option] {
override def map[A,B](optA: Option[A])(f: (A) => B): Option[B] = optA match{
case Some(a) => Some(f(a))
case _ => None
}
}
implicit val seq: Functor[Seq[_]] = new Functor[Seq[_]] {
override def map[A,B](sA: Seq[A])(f: (A) => B): Seq[B] = sA match{
case a :: tail => Seq(f(a), map(tail)(f))
case Nil => Nil
}
}
As you can see in the second case, there is a little bit of both (more of a recursion than iteration but still).
Now before the internet blows up on me, I will say you cant pattern match on Seq in Scala. It works here because the default Seq is also a List. I just provided this example because it is simpler to understand. The underlying definition something along the lines of that.
Now hold on a second. If you look at these types, you see that they also have flatMap defined on them. This means they are something more special than plain Functors. They are Monads. So beyond satisfying functoriality, they obey the monadic laws.
Turns out Monad has a different kind of meaning in the core scala, more on that here: What exactly makes Option a monad in Scala?
But again very very short, this means that we are now in a category where the endofunctors from our previous category are the objects and the mappings between them are morphisms (natural transformations), this is slightly more accurate because if you think about it when you take a type and transform it, you take (carry over) all of it's internal type constructors (2-cell or internal morphisms) with it, you do not only take this sole idea of a type without it's functions.
implicit val optionMonad: Monad[Option] = new Monad[Option] {
override def flatMap[A, B](optA: Option[A])(f: (A) => Option[B]): Option[B] = optA match{
case Some(a) => f(a)
case _ => None
}
def pure[A](a: A): Option[A] = Some(a)
//You can define map using pure and flatmap
}
implicit val seqMonad: Monad[Seq[_]] = new Monad[Seq[_]] {
override def flatMap[A, B](sA: Seq[A])(f: (A) => Seq[B]): Seq[B] = sA match{
case x :: xs => f(a).append(flatMap(tail)(f))
case Nil => Nil
}
override def pure[A](a: A): Seq[A] = Seq(a)
//Same warning as above, also you can implement map with the above 2 funcs
}
One thing you can always count on is map being having pattern match (or some if statement). Why?
In order to satisfy the identity laws, we need to have some sort of a "base case", a unit object and in many cases (such as Lists) those types are gonna be what we call either a product or coproduct.
Hopefully, this did not confuse you further. I wish I could get into every detail of this but it would simply take pages, I highly recommend getting into categories to fully understand where these come from.
From the ScalaDocs page we can see that the type profile for the Standard Library map() method is a little different.
def map[B](f: (A) => B): Seq[B]
So the Standard Library map() is the means to transition from a collection of elements of type A to the same collection but the elements are type B. (A and B might be the same type. They aren't required to be different.)
So, yes, it iterates through the collection applying function f() to each element A to create each new element B. And function f() might use pattern matching in its code, but it doesn't have to.
Now consider a.map(println). Every element of a is sent to println which returns Unit. So if a is List[Int] then the result of a.map(println) is List[Unit], which isn't terribly useful.
When all we want is the side effect of sending information to StdOut then we use foreach() which doesn't create a new collection: a.foreach(println)
Function map for Option isn't about pattern matching. The match/case used in your referred link is just one of the many ways to define the function. It could've been defined using if/else. In fact, that's how it's defined in Scala 2.13 source of class Option:
sealed abstract class Option[+A] extends IterableOnce[A] with Product with Serializable {
self =>
...
final def map[B](f: A => B): Option[B] =
if (isEmpty) None else Some(f(this.get))
...
}
If you view Option like a "collection" of either one element (Some(x)) or no elements (None), it might be easier to see the resemblance of how map transforms an Option versus, say, a List:
val f: Int => Int = _ + 1
List(42).map(f)
// res1: List[Int] = List(43)
List.empty[Int].map(f)
// res2: List[Int] = List()
Some(42).map(f)
// res3: Option[Int] = Some(43)
None.map(f)
// res4: Option[Int] = None

How map work on Options in Scala?

I have this two functions
def pattern(s: String): Option[Pattern] =
try {
Some(Pattern.compile(s))
} catch {
case e: PatternSyntaxException => None
}
and
def mkMatcher(pat: String): Option[String => Boolean] =
pattern(pat) map (p => (s: String) => p.matcher(s).matches)
Map is the higher-order function that applies a given function to each element of a list.
Now I am not getting that how map is working here as per above statement.
Map is the higher-order function that applies a given function to each element of a list.
This is an uncommonly restrictive definition of map.
At any rate, it works because it was defined by someone who did not hold to that.
For example, that someone wrote something akin to
sealed trait Option[+A] {
def map[B](f: A => B): Option[B] = this match {
case Some(value) => Some(f(value))
case None => None
}
}
as part of the standard library. This makes map applicable to Option[A]
It was defined because it makes sense to map many kinds of data structures not just lists.
Mapping is a transformation applied to the elements held by the data structure.
It applies a function to each element.
Option[A] can be thought of as a trivial sequence. It either has zero or one elements. To map it means to apply the function on its element if it has one.
Now it may not make much sense to use this facility all of the time, but there are cases where it is useful.
For example, it is one of a few distinct methods that, when present enable enable For Expressions to operate on a type. Option[A] can be used in for expressions which can be convenient.
For example
val option: Option[Int] = Some(2)
val squared: Option[Int] = for {
n <- option
if n % 2 == 0
} yield n * n
Interestingly, this implies that filter is also defined on Option[A].
If you just have a simple value it may well be clearer to use a less general construct.
Map is working the same way that it does with other collections types like List and Vector. It applies your function to the contents of the collection, potentially changing the type but keeping the collection type the same.
In many cases you can treat an Option just like a collection with either 0 or 1 elements. You can do a lot of the same operations on Option that you can on other collections.
You can modify the value
var opt = Option(1)
opt.map(_ + 3)
opt.map(_ * math.Pi)
opt.filter(_ == 1)
opt.collect({case i if i > 0 => i.toString })
opt.foreach(println)
and you can test the value
opt.contains(3)
opt.forall(_ > 0)
opt.exists(_ > 0)
opt.isEmpty
Using these methods you rarely need to use a match statement to unpick an Option.

Why does creating a map function whose parameter is of type `Nothing => U` appear to work?

I'm writing Scala code that uses an API where calls to the API can either succeed, fail, or return an exception. I'm trying to make an ApiCallResult monad to represent this, and I'm trying to make use of the Nothing type so that the failure and exception cases can be treated as a subtype of any ApiCallResult type, similar to None or Nil. What I have so far appears to work, but my use of Nothing in the map and flatMap functions has me confused. Here's a simplified example of what I have with just the map implementation:
sealed trait ApiCallResult[+T] {
def map[U]( f: T => U ): ApiCallResult[U]
}
case class ResponseException(exception: APICallExceptionReturn) extends ApiCallResult[Nothing] {
override def map[U]( f: Nothing => U ) = this
}
case object ResponseFailure extends ApiCallResult[Nothing] {
override def map[U]( f: Nothing => U ) = ResponseFailure
}
case class ResponseSuccess[T](payload: T) extends ApiCallResult[T] {
override def map[U]( f: T => U ) = ResponseSuccess( f(payload) )
}
val s: ApiCallResult[String] = ResponseSuccess("foo")
s.map( _.size ) // evaluates to ResponseSuccess(3)
val t: ApiCallResult[String] = ResponseFailure
t.map( _.size ) // evaluates to ResponseFailure
So it appears to work the way I intended with map operating on successful results but passing failures and exceptions along unchanged. However using Nothing as the type of an input parameter makes no sense to me since there is no instance of the Nothing type. The _.size function in the example has type String => Int, how can that be safely passed to something that expects Nothing => U? What's really going on here?
I also notice that the Scala standard library avoids this issue when implementing None by letting it inherit the map function from Option. This only furthers my sense that I'm somehow doing something horribly wrong.
Three things are aligning to make this happen, all having to do with covariance and contravariance in the face of a bottom type:
Nothing is the bottom type for all types, e.g. every type is its super.
The type signature of Function1[-T, +R], meaning it accepts any type which is a super of T and returns any type for which R is a super.
The type ApiCallResult[+R] means any type U for which R is a super of U is valid.
So any type is a super of Nothing means both any argument type is valid and the fact that you return something typed around Nothing is a valid return type.
I suggest that you don't need to distinguish failures and exceptions most of the time.
type ApiCallResult[+T] = Try[T]
case class ApiFailure() extends Throwable
val s: ApiCallResult[String] = Success("this is a string")
s.map(_.size)
val t: ApiCallResult[String] = Failure(new ApiFailure)
t.map(_.size)
To pick up the failure, use a match to select the result:
t match {
case Success(s) =>
case Failure(af: ApiFailure) =>
case Failure(x) =>
}

why scala Map does not implement unapply?

I wrote the following use case in scala:
val wordShortcut = Map("volume" -> "vol", "report" -> "rpt", ...)
object WordShortcutCase {
def unapply(key: String): Option[String] = wordShortcut.get(key)
}
val pluralR = "(.+)s".r
def encodeToken(token: String) = token match {
case WordShortcutCase(short) => short
case pluralR(singular) => singular
case _ => token
}
if scala Map would implement unapply, I wouldn't need the extra WordShortcutCase object (i could use case wordShortcut(short) => short instead`). This seems a common pattern to me.
And so the question is why scala Map does not implement the unapply method?
Map doesn't implement unapply because there is no sensible implementation that has the same characteristics as other collections.
In particular, you seem to want apply and unapply to do basically the same thing. But that's not how other collections work; they bind variables to contents and expect that the list is exhaustive (in the absence of a binding to "the rest"):
val xs = List("fish")
val ys = List("fish", "dish")
def iam(zs: List[String]) = zs match {
case List(x) => println(s"I am a $x")
case _ => println("Who am I??")
}
iam(xs) // Prints 'I am a fish'
iam(ys) // Prints 'Who am I??'
If Map were not a collection it would be free to implement unapply as another way to do an apply, more like regex does (though there, note that the key feature is being able to bind multiple variables to parts of the regex match). But since it is, having a regex-like unapply would be highly confusing because of the difference from other collections; and because maps are unordered and unapplySeq is ordered, having the same unapply as other collections would also be confusing. So it just doesn't have one.

Is there a way to ignore a non-matching case?

If I have a match expression, how can I make it automatically ignore a non-match without explicitly doing case _ =>? Is there a way to create a function that does something like this maybe?
You need a generic way to handle "ignoring". Options, among other classes, provide this (among other things). So you can:
val i = 7
Some(i) collect {
case 3 => "Yay!"
case 5 => "Boo!"
}
to get None (typed as an Option[String]). So basically, if you change x match to Some(x) collect you get the functionality you want. It is better to do this when one is comfortable with handling options.
Write a generic matcher:
object Match {
def default: PartialFunction[Any, Unit] = { case _ => }
def apply[T](x: T)(body: PartialFunction[T, Unit]) = (body orElse default)(x)
}
Example:
scala> 1 to 5 foreach (Match(_) {
| case 2 => println("two")
| case 3 => println("three")
| }
| )
two
three
You might be interested too in PartialFunction's companion object's methods cond and condOpt.
Any match block that can't handle all possible input values forms a Partial Function, which is absolutely possible to define and use in Scala.
PartialFunction is a subclass of Function, and adds the isDefinedAt method, which returns true if it has a defined match for the supplied value. Partial Functions are best used in places that test for definition, such as catch blocks or the collect method in the 2.8 collections API. Otherwise, you'll get an exception if you try to call it with a value that isn't defined as an input.