Scala flatten a map with list as key and string as value

Question

I have a peculiar case where I want to declare simple configuration like so

val config = List((("a", "b", "c"), ("first")), 
               (("d", "e"), ("second")),
               (("f"), ("third")))

which at run time, I would like to have a map, which maps like

"a" -> "first"
"b" -> "first"
"c" -> "first"
"d" -> "second"
"e" -> "second"
"f" -> "third"

Using toMap, I was able to convert the config to a Map

scala> config.toMap
res42: scala.collection.immutable.Map[java.io.Serializable,String] = Map((a,b,c) -> first, (d,e) -> second, f -> third)

But I am not able to figure out how to flatten the list of keys into keys so I get the final desirable form. How do I solve this?

That is not a simple configuration because you have tuples of different lengths inside each tuple (Note that the key type is java.io.Serializable). If you use Lists rather than tuples in the inner data it will be much easier. — Tim
– Tim, Commented Apr 27, 2020 at 7:07
You are right. Declaring the key as a list rather than tuples combined with the answer posted below, I was able to get my desired result. Thanks! — Sundeep
– Sundeep, Commented Apr 27, 2020 at 7:15

Tim · Accepted Answer · 2020-04-27 07:17:37Z

2

If you structure your config using List the code is very simple:

val config = List(
  (List("a", "b", "c"), ("first")),
  (List("d", "e"), ("second")),
  (List("f"), ("third")))

config.flatMap{ case (k, v) => k.map(_ -> v) }.toMap

answered Apr 27, 2020 at 7:17

Tim

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Comments

Sangeeta · Accepted Answer · 2020-04-27 09:51:10Z

1

You can try the solution below:

val config = List(
      (("a", "b", "c"), ("first")),
      (("d", "e"), ("second")),
      (("f"), ("third")))

val result = config.map {
  case (k,v) =>
    (
      k.toString().replace(")", "")
        .replace("(", "")
        .split(","), v)
  }

val res = result.map {
  case (key,value) => key.map{ data =>
    (data,value)
  }.toList
}.flatten.toMap

In case you change the config structure to something like below, solution is much more simpler:

val config1 = List (
  (List("a", "b", "c"), "first"),
  (List("d", "e"), "second"),
  (List("f"), "third")
)


config1.flatMap{
  case (k,v) => k.map{data => (data,v)}
}.toMap

edited Apr 27, 2020 at 9:51

answered Apr 27, 2020 at 7:07

Sangeeta

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1 Comment

Haemin Yoo Over a year ago

In your second example, here: case (k,v) => k.map{data => (data,v)}.toMap, toMap doesn't actually change anything. You can safely remove that and save some cpu cycles.

Haemin Yoo · Accepted Answer · 2020-04-27 08:47:36Z

1

I think the above answers are good practical answers. If you're in a situation where you have no control over the input and you're stuck with Tuples instead of Lists, I'd do it this way:

val result: Map[String, String] = config.flatMap { 
  case (s: String, v) => List(s -> v)
  case (ks: Product, v) => ks.productIterator.collect { case s: String => s -> v }
  case _ => Nil //Prevent throwing
}.toMap

This will throw away anything that's not a String in the keys.

answered Apr 27, 2020 at 8:47

Haemin Yoo

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Comments

QuickSilver · Accepted Answer · 2020-04-27 09:03:43Z

0

by using in built spark sql functions

val config = List((Array("a", "b", "c"), ("first")),
      (Array("d", "e"), ("second")),
      (Array("f"), ("third"))).toDF(List("col1","col2") : _*)

config.withColumn("exploded",functions.explode_outer($"col1")).drop("col1").show()

answered Apr 27, 2020 at 9:03

QuickSilver

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