convert play getIntList to scala List[Int] simply - scala

I have a Play app where I'd like to read a list of integers. The Configuration object returns Option[java.util.List[java.lang.Integer]] but I'd like to return the scala type Option[List[Int]].
I have used JavaConversions to return the scala List mylist.map(_.toList).
But I'm unsure about converting the underlying element type from java.lang.Integer to Int in a neat way and without using any additional libraries.
Any ideas?

Just map the list:
import scala.collection.JavaConverters._
mylist.map(_.asScala.toList.map(_.intValue))
(it's generally recommended to prefer JavaConverters, which add toJava and toScala methods, to JavaConversions).

Related

Does Scala's JavaConverters produce incomplete result?

Following code converts Scala List into java.util.List (Tested in Scala 2.11)
import scala.collection.JavaConverters._
val a = List(1, 2, 3)
val b = a.asJava
However, the conversion result seems incomplete. Because some methods in java.util.List do not work.
scala> b.remove(2)
java.lang.UnsupportedOperationException
at java.util.AbstractList.remove(AbstractList.java:161)
... 29 elided
My workaround is as follows:
val c = new java.util.ArrayList(a.asJava)
This works but seems redundant in API-design perspective.
Is this the correct way of using asJava method?
Why does Scala's JavaConverters produce incomplete result?
Because some methods in java.util.List do not work.
These methods are explicitly optional, because java.util.List covers both mutable and immutable lists, and some of implementations in Java standard library don't support them either:
Removes the first occurrence of the specified element from this list, if it is present (optional operation)...
Throws: ... UnsupportedOperationException - if the remove operation is not supported by this list
Same for other Java collection interfaces. So the result does completely satisfy the interface.
It appears that what you're getting from the asJava is an immutable list, as you've started with an immutable Scala list. Try the following
val a = List(1,2,3).to[scala.collection.mutable.ListBuffer].asJava

Scala: Invoking Standard Methods on Wrapper Maps After Java to Scala Conversion

I have two java maps which map a String to a set of Strings.
I want to convert the java maps to Scala and "add" map1 to map2 such that if they both have the same keys the value in the resultant map will be the union of the 2 value sets. I found a nice solution for the map addition:
map1 ++ map2.map { case (k,v) => k -> (v ++ map1.getOrElse(k,v))}
The issue happens when I convert the java maps to Scala via the 'asScala' call. When we do this conversation we get:
After the conversion to Scala, I am no longer able to run the solution above on these wrapper collections. The ++ operation on the map is still defined. However the SetWrapper class does not define a ++ operation. As far as I can tell the SetWrapper does not define any operations at all! Here is the full definition from the docs:
SetWrapper seems to extend an abstract class and does not define any of the functionality.
How can I get around this issue? Is there another conversation step to a real Set object?
JMapWrapper wraps a Java Map in a Scala Map; SetWrapper wraps a Scala Set in a Java Set. So it has Java's methods for Set, not Scala's.
You don't provide types, but it appears you start with a java.util.Map[SomeKey, java.util.Set[SomeType]]. Calling asScala on it will only convert the outer Map, not its values. You can fix this by using javaMap.asScala.mapValues(_.asScala) (with the usual caveat that it will call asScala each time a value is accessed; use .map instead if this isn't acceptable).

Convert a Seq[String] to a case class in a typesafe way

I have written a parser which transforms a String to a Seq[String] following some rules. This will be used in a library.
I am trying to transform this Seq[String] to a case class. The case class would be provided by the user (so there is no way to guess what it will be).
I have thought to shapeless library because it seems to implement the good features and it seems mature, but I have no idea to how to proceed.
I have found this question with an interesting answer but I don't find how to transform it for my needs. Indeed, in the answer there is only one type to parse (String), and the library iterates inside the String itself. It probably requires a deep change in the way things are done, and I have no clue how.
Moreover, if possible, I want to make this process as easy as possible for the user of my library. So, if possible, unlike the answer in link above, the HList type would be guess from the case class itself (however according to my search, it seems the compiler needs this information).
I am a bit new to the type system and all these beautiful things, if anyone is able to give me an advice on how to do, I would be very happy!
Kind Regards
--- EDIT ---
As ziggystar requested, here is some possible of the needed signature:
//Let's say we are just parsing a CSV.
#onUserSide
case class UserClass(i:Int, j:Int, s:String)
val list = Seq("1,2,toto", "3,4,titi")
// User transforms his case class to a function with something like:
val f = UserClass.curried
// The function created in 1/ is injected in the parser
val parser = new Parser(f)
// The Strings to convert to case classes are provided as an argument to the parse() method.
val finalResult:Seq[UserClass] = parser.parse(list)
// The transfomation is done in two steps inside the parse() method:
// 1/ first we have: val list = Seq("1,2,toto", "3,4,titi")
// 2/ then we have a call to internalParserImplementedSomewhereElse(list)
// val parseResult is now equal to Seq(Seq("1", "2", "toto"), Seq("3","4", "titi"))
// 3/ finally Shapeless do its magick trick and we have Seq(UserClass(1,2,"toto"), UserClass(3,4,"titi))
#insideTheLibrary
class Parser[A](function:A) {
//The internal parser takes each String provided through argument of the method and transforms each String to a Seq[String]. So the Seq[String] provided is changed to Seq[Seq[String]].
private def internalParserImplementedSomewhereElse(l:Seq[String]): Seq[Seq[String]] = {
...
}
/*
* Class A and B are both related to the case class provided by the user:
* - A is the type of the case class as a function,
* - B is the type of the original case class (can be guessed from type A).
*/
private def convert2CaseClass[B](list:Seq[String]): B {
//do something with Shapeless
//I don't know what to put inside ???
}
def parse(l:Seq[String]){
val parseResult:Seq[Seq[String]] = internalParserImplementedSomewhereElse(l:Seq[String])
val finalResult = result.map(convert2CaseClass)
finalResult // it is a Seq[CaseClassProvidedByUser]
}
}
Inside the library some implicit would be available to convert the String to the correct type as they are guessed by Shapeless (similar to the answered proposed in the link above). Like string.toInt, string.ToDouble, and so on...
May be there are other way to design it. It's just what I have in mind after playing with Shapeless few hours.
This uses a very simple library called product-collecions
import com.github.marklister.collections.io._
case class UserClass(i:Int, j:Int, s:String)
val csv = Seq("1,2,toto", "3,4,titi").mkString("\n")
csv: String =
1,2,toto
3,4,titi
CsvParser(UserClass).parse(new java.io.StringReader(csv))
res28: Seq[UserClass] = List(UserClass(1,2,toto), UserClass(3,4,titi))
And to serialize the other way:
scala> res28.csvIterator.toList
res30: List[String] = List(1,2,"toto", 3,4,"titi")
product-collections is orientated towards csv and a java.io.Reader, hence the shims above.
This answer will not tell you how to do exactly what you want, but it will solve your problem. I think you're overcomplicating things.
What is it you want to do? It appears to me that you're simply looking for a way to serialize and deserialize your case classes - i.e. convert your Scala objects to a generic string format and the generic string format back to Scala objects. Your serialization step presently is something you seem to already have defined, and you're asking about how to do the deserialization.
There are a few serialization/deserialization options available for Scala. You do not have to hack away with Shapeless or Scalaz to do it yourself. Try to take a look at these solutions:
Java serialization/deserialization. The regular serialization/deserialization facilities provided by the Java environment. Requires explicit casting and gives you no control over the serialization format, but it's built in and doesn't require much work to implement.
JSON serialization: there are many libraries that provide JSON generation and parsing for Java. Take a look at play-json, spray-json and Argonaut, for example.
The Scala Pickling library is a more general library for serialization/deserialization. Out of the box it comes with some binary and some JSON format, but you can create your own formats.
Out of these solutions, at least play-json and Scala Pickling use macros to generate serializers and deserializers for you at compile time. That means that they should both be typesafe and performant.

how to read properties file in scala

I am new to Scala programming and I wanted to read a properties file in Scala.
I can't find any APIs to read a property file in Scala.
Please let me know if there are any API for this or other way to read properties files in Scala.
Beside form Java API, there is a library by Typesafe called config with a good API for working with configuration files of different types.
You will have to do it in similar way you would with with Scala Map to java.util.Map. java.util.Properties extends java.util.HashTable whiche extends java.util.Dictionary.
scala.collection.JavaConverters has functions to convert to and fro from Dictionary to Scala mutable.Map:
val x = new Properties
//load from .properties file here.
import scala.collection.JavaConverters._
scala> x.asScala
res4: scala.collection.mutable.Map[String,String] = Map()
You can then use Map above. To get and retrieve. But if you wish to convert it back to Properties type (to store back etc), you might have to type cast it manually then.
You can just use the Java API.
Consider something along the lines
def getPropertyX: Option[String] = Source.fromFile(fileName)
.getLines()
.find(_.startsWith("propertyX="))
.map(_.replace("propertyX=", ""))

Create an immutable list from a java.lang.Iterator

I'm using a library (JXPath) to query a graph of beans in order to extract matching elements. However, JXPath returns groups of matching elements as an instance of java.lang.Iterator and I'd rather like to convert it into an immutable scala list. Is there any simpler way of doing than iterating over the iterator and creating a new immutable list at each iteration step ?
You might want to rethink the need for a List, although it feels very familiar when coming from Java, and List is the default implementation of an immutable Seq, it often isn't the best choice of collection.
The operations that list is optimal for are those already available via an iterator (basically taking consecutive head elements and prepending elements). If an iterator doesn't already give you what you need, then I can pretty much guarantee that a List won't be your best choice - a vector would be more appropriate.
Having got that out the way... The recommended technique to convert between Java and Scala collections (since Scala 2.8.1) is via scala.collection.JavaConverters. This gives you more control than JavaConversions and avoids some possible implicit conflicts.
You won't have a direct implicit conversion this way. Instead, you get asScala and asJava methods pimped onto collections, allowing you to perform the conversions explicitly.
To convert a Java iterator to a Scala iterator:
javaIterator.asScala
To convert a Java iterator to a Scala List (via the scala iterator):
javaIterator.asScala.toList
You may also want to consider converting toSeq instead of toList. In the case of iterators, this'll return a Stream - allowing you to retain the lazy behaviour of iterators within the richer Seq interface.
EDIT:
There's no toVector method, but (as Daniel pointed out) there's a toIndexedSeq method that will return a Vector as the default IndexedSeq subclass (just as List is the default Seq).
javaIterator.asScala.toIndexedSeq
EDIT: You should probably look at Kevin Wright's answer, which provides a better solution available since Scala 2.8.1, with less implicit magic.
You can import the implicit conversions from scala.collection.JavaConversions and then create a new Scala collection seamlessly, e.g. like this:
import collection.JavaConversions._
println(List() ++ javaIterator)
Your Java iterator is converted to a Scala iterator by JavaConversions.asScalaIterator. A Scala iterator with elements of type A implements TraversableOnce[A], which is the argument type needed to concatenate collections with ++.
If you need another collection type, just change List() to whatever you need (e.g., IndexedSeq() or collection.mutable.Seq(), etc.).