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Is there a way to serialize a single None field to "null" ?
For example:
// When None, I'd like to serialize only f2 to `null`
case class Example(f1: Option[Int], f2: Option[Int])
val printer = Printer.noSpaces.copy(dropNullValues = true)
Example(None, None).asJson.pretty(printer) === """{"f2":null}"""
You can do this pretty straightforwardly by mapping a filter over the output of an encoder (which can be derived, defined with Encoder.forProductN, etc.):
import io.circe.{ Json, ObjectEncoder }
import io.circe.generic.semiauto.deriveEncoder
case class Example(f1: Option[Int], f2: Option[Int])
val keepSomeNulls: ((String, Json)) => Boolean = {
case ("f1", v) => !v.isNull
case (_, _) => true
}
implicit val encodeExample: ObjectEncoder[Example] =
deriveEncoder[Example].mapJsonObject(_.filter(keepSomeNulls))
And then:
scala> import io.circe.syntax._
import io.circe.syntax._
scala> Example(Some(1), Some(2)).asJson.noSpaces
res0: String = {"f1":1,"f2":2}
scala> Example(Some(1), None).asJson.noSpaces
res1: String = {"f1":1,"f2":null}
scala> Example(None, Some(2)).asJson.noSpaces
res2: String = {"f2":2}
scala> Example(None, None).asJson.noSpaces
res3: String = {"f2":null}
Note that configuring a printer to drop null values will still remove "f2": null here. This is part of the reason I think in general it's best to make the preservation of null values solely the responsibility of the printer, but in a case like this, the presence or absence of null-valued fields is clearly semantically meaningful, so you kind of have to mix up the levels.
I am trying to create a Macro to give me a list of val for a specific case class.
object CaseClass {
def valList[T]: List[String] = macro implValList[T]
def implValList[T](c: whitebox.Context): c.Expr[List[String]] = {
import c.universe._
val listApply = Select(reify(List).tree, TermName("apply"))
val vals = weakTypeOf[T].decls.collect {
case m: TermSymbol if m.isVal => q"${m.name}"
}
c.Expr[List[String]](Apply(listApply, vals.toList))
}
}
So given
case class AClass(
val a: String,
val b: Int
)
I want a list of CaseClass.valList[AClass] = List("a", "b")
Not an expert on macros, so take it with a grain of salt. But I tested it with Intellij.
First, to use weakTypeOf you need to take a WeakTypeTag as an implicit in your macro impl like this:
def implValList[T](c: whitebox.Context)(implicit wt: c.WeakTypeTag[T]) ...
Second, to create literals, you use this construct instead of your quasiquote, (which, I believe, actually does nothing):
Literal(Constant(m.name.toString))
Last, I recommend using this guard instead of isVal:
m.isCaseAccessor && m.isGetter
Which is properly checking for case class parameter and also being a getter (case class parameters are duplicated, one as isGetter, other one as isParam). The reason for this being that isVal names for case classes surprisingly produce a name ending in whitespace.
The final implementation that works for me is as follows:
object CaseClass {
def valList[T]: List[String] = macro implValList[T]
def implValList[T](c: whitebox.Context)(implicit wt: c.WeakTypeTag[T]): c.Expr[List[String]] = {
import c.universe._
val listApply = Select(reify(List).tree, TermName("apply"))
val vals = weakTypeOf[T].decls.collect {
case m: TermSymbol if m.isCaseAccessor && m.isGetter => Literal(Constant(m.name.toString))
}
c.Expr[List[String]](Apply(listApply, vals.toList))
}
}
As an alternative (because macros are somewhat of a pain to set up - you cannot use macro in the same subproject that defines it), and you don't need it very often, you might be able to get away with a shapeless one-liner:
import shapeless._
import shapeless.ops.record.Keys
case class Foo(a: Int, b: String)
Keys[the.`LabelledGeneric[Foo]`.Repr].apply().toList.map(_.name) // List("a", "b")
Let's say I have this example case class
case class Test(key1: Int, key2: String, key3: String)
And I have a map
myMap = Map("k1" -> 1, "k2" -> "val2", "k3" -> "val3")
I need to convert this map to my case class in several places of the code, something like this:
myMap.asInstanceOf[Test]
What would be the easiest way of doing that? Can I somehow use implicit for this?
Two ways of doing this elegantly. The first is to use an unapply, the second to use an implicit class (2.10+) with a type class to do the conversion for you.
1) The unapply is the simplest and most straight forward way to write such a conversion. It does not do any "magic" and can readily be found if using an IDE. Do note, doing this sort of thing can clutter your companion object and cause your code to sprout dependencies in places you might not want:
object MyClass{
def unapply(values: Map[String,String]) = try{
Some(MyClass(values("key").toInteger, values("next").toFloat))
} catch{
case NonFatal(ex) => None
}
}
Which could be used like this:
val MyClass(myInstance) = myMap
be careful, as it would throw an exception if not matched completely.
2) Doing an implicit class with a type class creates more boilerplate for you but also allows a lot of room to expand the same pattern to apply to other case classes:
implicit class Map2Class(values: Map[String,String]){
def convert[A](implicit mapper: MapConvert[A]) = mapper conv (values)
}
trait MapConvert[A]{
def conv(values: Map[String,String]): A
}
and as an example you'd do something like this:
object MyObject{
implicit val new MapConvert[MyObject]{
def conv(values: Map[String, String]) = MyObject(values("key").toInt, values("foo").toFloat)
}
}
which could then be used just as you had described above:
val myInstance = myMap.convert[MyObject]
throwing an exception if no conversion could be made. Using this pattern converting between a Map[String, String] to any object would require just another implicit (and that implicit to be in scope.)
Here is an alternative non-boilerplate method that uses Scala reflection (Scala 2.10 and above) and doesn't require a separately compiled module:
import org.specs2.mutable.Specification
import scala.reflect._
import scala.reflect.runtime.universe._
case class Test(t: String, ot: Option[String])
package object ccFromMap {
def fromMap[T: TypeTag: ClassTag](m: Map[String,_]) = {
val rm = runtimeMirror(classTag[T].runtimeClass.getClassLoader)
val classTest = typeOf[T].typeSymbol.asClass
val classMirror = rm.reflectClass(classTest)
val constructor = typeOf[T].decl(termNames.CONSTRUCTOR).asMethod
val constructorMirror = classMirror.reflectConstructor(constructor)
val constructorArgs = constructor.paramLists.flatten.map( (param: Symbol) => {
val paramName = param.name.toString
if(param.typeSignature <:< typeOf[Option[Any]])
m.get(paramName)
else
m.get(paramName).getOrElse(throw new IllegalArgumentException("Map is missing required parameter named " + paramName))
})
constructorMirror(constructorArgs:_*).asInstanceOf[T]
}
}
class CaseClassFromMapSpec extends Specification {
"case class" should {
"be constructable from a Map" in {
import ccFromMap._
fromMap[Test](Map("t" -> "test", "ot" -> "test2")) === Test("test", Some("test2"))
fromMap[Test](Map("t" -> "test")) === Test("test", None)
}
}
}
Jonathan Chow implements a Scala macro (designed for Scala 2.11) that generalizes this behavior and eliminates the boilerplate.
http://blog.echo.sh/post/65955606729/exploring-scala-macros-map-to-case-class-conversion
import scala.reflect.macros.Context
trait Mappable[T] {
def toMap(t: T): Map[String, Any]
def fromMap(map: Map[String, Any]): T
}
object Mappable {
implicit def materializeMappable[T]: Mappable[T] = macro materializeMappableImpl[T]
def materializeMappableImpl[T: c.WeakTypeTag](c: Context): c.Expr[Mappable[T]] = {
import c.universe._
val tpe = weakTypeOf[T]
val companion = tpe.typeSymbol.companionSymbol
val fields = tpe.declarations.collectFirst {
case m: MethodSymbol if m.isPrimaryConstructor ⇒ m
}.get.paramss.head
val (toMapParams, fromMapParams) = fields.map { field ⇒
val name = field.name
val decoded = name.decoded
val returnType = tpe.declaration(name).typeSignature
(q"$decoded → t.$name", q"map($decoded).asInstanceOf[$returnType]")
}.unzip
c.Expr[Mappable[T]] { q"""
new Mappable[$tpe] {
def toMap(t: $tpe): Map[String, Any] = Map(..$toMapParams)
def fromMap(map: Map[String, Any]): $tpe = $companion(..$fromMapParams)
}
""" }
}
}
This works well for me，if you use jackson for scala：
def from[T](map: Map[String, Any])(implicit m: Manifest[T]): T = {
val mapper = new ObjectMapper() with ScalaObjectMapper
mapper.convertValue(map)
}
Reference from：Convert a Map<String, String> to a POJO
I don't love this code, but I suppose this is possible if you can get the map values into a tuple and then use the tupled constructor for your case class. That would look something like this:
val myMap = Map("k1" -> 1, "k2" -> "val2", "k3" -> "val3")
val params = Some(myMap.map(_._2).toList).flatMap{
case List(a:Int,b:String,c:String) => Some((a,b,c))
case other => None
}
val myCaseClass = params.map(Test.tupled(_))
println(myCaseClass)
You have to be careful to make sure the list of values is exactly 3 elements and that they are the correct types. If not, you end up with a None instead. Like I said, not great, but it shows that it is possible.
commons.mapper.Mappers.mapToBean[CaseClassBean](map)
Details: https://github.com/hank-whu/common4s
Here's an update to Jonathon's answer for Scala 3 (which no longer has TypeTag). Be aware that this won't work for case classes nested inside of other classes. But for top-level case classes it seems to work fine.
import scala.reflect.ClassTag
object Reflect:
def fromMap[T <: Product : ClassTag](m: Map[String, ?]): T =
val classTag = implicitly[ClassTag[T]]
val constructor = classTag.runtimeClass.getDeclaredConstructors.head
val constructorArgs = constructor.getParameters()
.map { param =>
val paramName = param.getName
if (param.getType == classOf[Option[_]])
m.get(paramName)
else
m.get(paramName)
.getOrElse(throw new IllegalArgumentException(s"Missing required parameter: $paramName"))
}
constructor.newInstance(constructorArgs: _*).asInstanceOf[T]
And a test for the above:
case class Foo(a: String, b: Int, c: Option[String] = None)
case class Bar(a: String, b: Int, c: Option[Foo])
class ReflectSuite extends munit.FunSuite:
test("fromMap") {
val m = Map("a" -> "hello", "b" -> 42, "c" -> "world")
val foo = Reflect.fromMap[Foo](m)
assertEquals(foo, Foo("hello", 42, Some("world")))
val n = Map("a" -> "hello", "b" -> 43)
val foo2 = Reflect.fromMap[Foo](n)
assertEquals(foo2, Foo("hello", 43))
val o = Map("a" -> "yo", "b" -> 44, "c" -> foo)
val bar = Reflect.fromMap[Bar](o)
assertEquals(bar, Bar("yo", 44, Some(foo)))
}
test("fromMap should fail when required parameter is missing") {
val m = Map("a" -> "hello", "c" -> "world")
intercept[java.lang.IllegalArgumentException] {
Reflect.fromMap[Foo](m)
}
}
I need a Map where I put different types of values (Double, String, Int,...) in it, key can be String.
Is there a way to do this, so that I get the correct type with map.apply(k) like
val map: Map[String, SomeType] = Map()
val d: Double = map.apply("double")
val str: String = map.apply("string")
I already tried it with a generic type
class Container[T](element: T) {
def get: T = element
}
val d: Container[Double] = new Container(4.0)
val str: Container[String] = new Container("string")
val m: Map[String, Container] = Map("double" -> d, "string" -> str)
but it's not possible since Container takes an parameter. Is there any solution to this?
This is not straightforward.
The type of the value depends on the key. So the key has to carry the information about what type its value is. This is a common pattern. It is used for example in SBT (see for example SettingsKey[T]) and Shapeless Records (Example). However, in SBT the keys are a huge, complex class hierarchy of its own, and the HList in shapeless is pretty complex and also does more than you want.
So here is a small example of how you could implement this. The key knows the type, and the only way to create a Record or to get a value out of a Record is the key. We use a Map[Key, Any] internally as storage, but the casts are hidden and guaranteed to succeed. There is an operator to create records from keys, and an operator to merge records. I chose the operators so you can concatenate Records without having to use brackets.
sealed trait Record {
def apply[T](key:Key[T]) : T
def get[T](key:Key[T]) : Option[T]
def ++ (that:Record) : Record
}
private class RecordImpl(private val inner:Map[Key[_], Any]) extends Record {
def apply[T](key:Key[T]) : T = inner.apply(key).asInstanceOf[T]
def get[T](key:Key[T]) : Option[T] = inner.get(key).asInstanceOf[Option[T]]
def ++ (that:Record) = that match {
case that:RecordImpl => new RecordImpl(this.inner ++ that.inner)
}
}
final class Key[T] {
def ~>(value:T) : Record = new RecordImpl(Map(this -> value))
}
object Key {
def apply[T] = new Key[T]
}
Here is how you would use this. First define some keys:
val a = Key[Int]
val b = Key[String]
val c = Key[Float]
Then use them to create a record
val record = a ~> 1 ++ b ~> "abc" ++ c ~> 1.0f
When accessing the record using the keys, you will get a value of the right type back
scala> record(a)
res0: Int = 1
scala> record(b)
res1: String = abc
scala> record(c)
res2: Float = 1.0
I find this sort of data structure very useful. Sometimes you need more flexibility than a case class provides, but you don't want to resort to something completely type-unsafe like a Map[String,Any]. This is a good middle ground.
Edit: another option would be to have a map that uses a (name, type) pair as the real key internally. You have to provide both the name and the type when getting a value. If you choose the wrong type there is no entry. However this has a big potential for errors, like when you put in a byte and try to get out an int. So I think this is not a good idea.
import reflect.runtime.universe.TypeTag
class TypedMap[K](val inner:Map[(K, TypeTag[_]), Any]) extends AnyVal {
def updated[V](key:K, value:V)(implicit tag:TypeTag[V]) = new TypedMap[K](inner + ((key, tag) -> value))
def apply[V](key:K)(implicit tag:TypeTag[V]) = inner.apply((key, tag)).asInstanceOf[V]
def get[V](key:K)(implicit tag:TypeTag[V]) = inner.get((key, tag)).asInstanceOf[Option[V]]
}
object TypedMap {
def empty[K] = new TypedMap[K](Map.empty)
}
Usage:
scala> val x = TypedMap.empty[String].updated("a", 1).updated("b", "a string")
x: TypedMap[String] = TypedMap#30e1a76d
scala> x.apply[Int]("a")
res0: Int = 1
scala> x.apply[String]("b")
res1: String = a string
// this is what happens when you try to get something out with the wrong type.
scala> x.apply[Int]("b")
java.util.NoSuchElementException: key not found: (b,Int)
This is now very straightforward in shapeless,
scala> import shapeless._ ; import syntax.singleton._ ; import record._
import shapeless._
import syntax.singleton._
import record._
scala> val map = ("double" ->> 4.0) :: ("string" ->> "foo") :: HNil
map: ... <complex type elided> ... = 4.0 :: foo :: HNil
scala> map("double")
res0: Double with shapeless.record.KeyTag[String("double")] = 4.0
scala> map("string")
res1: String with shapeless.record.KeyTag[String("string")] = foo
scala> map("double")+1.0
res2: Double = 5.0
scala> val map2 = map.updateWith("double")(_+1.0)
map2: ... <complex type elided> ... = 5.0 :: foo :: HNil
scala> map2("double")
res3: Double = 5.0
This is with shapeless 2.0.0-SNAPSHOT as of the date of this answer.
I finally found my own solution, which worked best in my case:
case class Container[+T](element: T) {
def get[T]: T = {
element.asInstanceOf[T]
}
}
val map: Map[String, Container[Any]] = Map("a" -> Container[Double](4.0), "b" -> Container[String]("test"))
val double: Double = map.apply("a").get[Double]
val string: String = map.apply("b").get[String]
(a) Scala containers don't track type information for what's placed inside them, and
(b) the return "type" for an apply/get method with a simple String parameter/key is going to be static for a given instance of the object the method is to be applied to.
This feels very much like a design decision that needs to be rethought.
I don't think there's a way to get bare map.apply() to do what you'd want. As the other answers suggest, some sort of container class will be necessary. Here's an example that restricts the values to be only certain types (String, Double, Int, in this case):
sealed trait MapVal
case class StringMapVal(value: String) extends MapVal
case class DoubleMapVal(value: Double) extends MapVal
case class IntMapVal(value: Int) extends MapVal
val myMap: Map[String, MapVal] =
Map("key1" -> StringMapVal("value1"),
"key2" -> DoubleMapVal(3.14),
"key3" -> IntMapVal(42))
myMap.keys.foreach { k =>
val message =
myMap(k) match { // map.apply() in your example code
case StringMapVal(x) => "string: %s".format(x)
case DoubleMapVal(x) => "double: %.2f".format(x)
case IntMapVal(x) => "int: %d".format(x)
}
println(message)
}
The main benefit of the sealted trait is compile-time checking for non-exhaustive matches in pattern matching.
I also like this approach because it's relatively simple by Scala standards. You can go off into the weeds for something more robust, but in my opinion you're into diminishing returns pretty quickly.
If you want to do this you'd have to specify the type of Container to be Any, because Any is a supertype of both Double and String.
val d: Container[Any] = new Container(4.0)
val str: Container[Any] = new Container("string")
val m: Map[String, Container[Any]] = Map("double" -> d, "string" -> str)
Or to make things easier, you can change the definition of Container so that it's no longer type invariant:
class Container[+T](element: T) {
def get: T = element
override def toString = s"Container($element)"
}
val d: Container[Double] = new Container(4.0)
val str: Container[String] = new Container("string")
val m: Map[String, Container[Any]] = Map("double" -> d, "string" -> str)
There is a way but it's complicated. See Unboxed union types in Scala. Essentially you'll have to type the Map to some type Int |v| Double to be able to hold both Int and Double. You'll also pay a high price in compile times.
Is there a way in Scala to convert a List[Int] to java.util.List[java.lang.Integer]?
I'm interfacing with Java (Thrift).
JavaConversions supports List --> java.util.List, and implicits exist between Int --> java.lang.Integer, but from what I can tell I would still need an extra pass to manually do the conversion:
val y = List(1)
val z: java.util.List[Integer] = asList(y) map { (x: Int) => x : java.lang.Integer }
Apparently you need both conversions. However, you can group them in a single implicit conversion:
implicit def toIntegerList( lst: List[Int] ) =
seqAsJavaList( lst.map( i => i:java.lang.Integer ) )
Example:
scala> def sizeOf( lst: java.util.List[java.lang.Integer] ) = lst.size
scala> sizeOf( List(1,2,3) )
res5: Int = 3
Because the underlying representation of Int is Integer you can cast directly to java.util.List[java.lang.Integer]. It will save you an O(n) operation and some implicit stuff.
import collection.JavaConversions._
class A {
def l() = asList(List(1,2)).asInstanceOf[java.util.List[java.lang.Integer]]
}
Then you can use from Java like this:
A a = new A();
java.util.List<Integer> l = a.l();
Note that on 2.9.0 ,I get a deprecation warning on asList (use seqAsJavaList instead)
Did you try:
val javalist = collection.JavaConversions.asJavaList (y)
I'm not sure, whether you need a conversion Int=>Integer or Int=>int here. Can you try it out?
Update:
The times, they are a changing. Today you'll get a deprecated warning for that code. Use instead:
import scala.collection.JavaConverters._
val y = List (1)
> y: List[Int] = List(1)
val javalist = (y).asJava
> javalist: java.util.List[Int] = [1]
This doesn't have an implicit at the outmost layer, but I like this generic approach and have implemented it for a couple of types of collections (List, Map).
import java.util.{List => JList}
import scala.collection.JavaConverters._
def scalaList2JavaList[A, B](scalaList: List[A])
(implicit a2bConversion: A => B): JList[B] =
(scalaList map a2bConversion).asJava
Since an implicit conversion from Int to Integer is part of standard lib, usage in this case would just look like this:
scalaList2JavaList[Int, Integer](someScalaList)
In the other direction!
(since I have these available anyway as they were my original implementations...)
import java.util.{List => JList}
import scala.collection.JavaConversions._
def javaList2ScalaList[A, B](javaList: JList[A])
(implicit a2bConversion: A => B): List[B] =
javaList.toList map a2bConversion
Usage:
javaList2ScalaList[Integer, Int](someJavaList)
This can then be re-used for all lists so long as an implicit conversion of the contained type is in scope.
(And in case you're curious, here is my implementation for map...)
def javaMap2ScalaMap[A, B, C, D](javaMap: util.Map[A, B])(implicit a2cConversion: A => C, b2dConversion: B => D): Map[C, D] =
javaMap.toMap map { case (a, b) => (a2cConversion(a), b2dConversion(b)) }
Starting Scala 2.13, the standard library includes scala.jdk.CollectionConverters which provides Scala to Java implicit collection conversions.
Which we can combine with java.lang.Integer::valueOf to convert Scala's Int to Java's Integer:
import scala.jdk.CollectionConverters._
List(1, 2, 3).map(Integer.valueOf).asJava
// java.util.List[Integer] = [1, 2, 3]
I was trying to pass a Map[String, Double] to a Java method. But the problem was JavaConversions converts the Map to a java Map, but leaves the scala Double as is, instead of converting it to java.lang.Double. After a few hours of seaching I found [Alvaro Carrasco's answer])https://stackoverflow.com/a/40683561/1612432), it is as simple as doing:
val scalaMap = // Some Map[String, Double]
val javaMap = scalaMap.mapValues(Double.box)
After this, javaMap is a Map[String, java.lang.Double]. Then you can pass this to a java function that expects a Map<String, Double> and thanks to implicit conversions the Scala Map will be converted to java.util.Map
In your case would be the same, but with Int.box:
val y = List(1)
val javay = y.map(Int.box)