Succinct way to go from List[Any] to a specific case class

Question

I've got a List[Any] that looks like:

val a: List[Any] = List(1, "a", "b", 2.0)

I also have a case class that looks like:

case class Data(field1: Int, field2: String, field3: String, field4: Double)

I'd like to instantiate a case class Data with the values from a to look like:

val d = Data(1, "a", "b", 2.0)

I know I can iterate through a, but for a large list that seems cumbersome.

I don't see how iterating through `a` would get you anywhere. — jwvh, Oct 08 '18 at 02:10
Maybe a silly question, but are you trying to do CSV parsing, by chance? — erip, Oct 08 '18 at 02:24
@erip I'm trying to parse a Javascript array. {data: [1, "a", "b", 2.0]} — jstuartmill, Oct 08 '18 at 02:46
@jstuartmill Do you control the upstream source? Having a heterogeneous list is pretty abnormal even for JSON. — erip, Oct 08 '18 at 11:17
I wish I did. This is the first tie I've had to deal with this in all my years manipulating JSON @erip — jstuartmill, Oct 09 '18 at 03:23

prayagupa · Accepted Answer · 2018-10-08T02:24:46.980

6

If you are using shapeless,

libraryDependencies += "com.chuusai" %% "shapeless" % "2.3.3"

You can apply the tuple to your case class,

scala> import shapeless._
import shapeless._

scala> import syntax.std.traversable._
import syntax.std.traversable._

scala> case class Data(field1: Int, field2: String, field3: String, field4: Double)
defined class Data

scala> val a: List[Any] = List(1, "a", "b", 2.0)
a: List[Any] = List(1, a, b, 2.0)

scala> (Data.apply _) tupled a.toHList[Int::String::String::Double::HNil].get.tupled
res1: Data = Data(1,a,b,2.0)

typesafe way to apply tuple on data class,

scala> a.toHList[Int::String::String::Double::HNil].map(_.tupled).map(t => (Data.apply _) tupled t)
res2: Option[Data] = Some(Data(1,a,b,2.0))

edited Oct 08 '18 at 02:24

answered Oct 08 '18 at 02:17

prayagupa

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This answer is fine, but the unchecked `.get` should give you pause. – erip Oct 08 '18 at 02:21
I did `.get` because I'm confident there is value. But sure you can `.map` on it instead. Updated. Thank you. – prayagupa Oct 08 '18 at 02:23
I was a bit surprised to find that this works: `.map(Data.apply _ tupled _.tupled)` – jwvh Oct 08 '18 at 02:53

erip · Answer 2 · 2018-10-08T02:23:31.310

There is no way to do this in a type-safe way. Lists' lengths aren't known until runtime, so there's no way to ensure that a.length == Data#productArity (see? we're comparing a field of an instance to a field of a type).

You should think about how to not use a List to represent your data in the first place.

If you could represent your data as a tuple, you could use this answer as inspiration, but there be dragons: tuples are nearly as dubious as lists here. Try to instantiate your case class directly.

Succinct way to go from List[Any] to a specific case class

2 Answers2