How to convert list to mutable map with pre processing in Scala?

Question

I have a list (size of the list is variable):

val ids = List(7, 8, 9)

and would like to get the following map:

val result= Map("foo:7:bar" -> "val1",
                "foo:8:bar" -> "val1",
                "foo:9:bar" -> "val1")

everything in the map is hard-coded except ids and value is the same for everyone, but map has to be mutable, I'd like to update one of its values later:

result("foo:8:bar") = "val2"
val result= Map("foo:7:bar" -> "val1",
                "foo:8:bar" -> "val2",
                "foo:9:bar" -> "val1")

score 12 · Accepted Answer · edited May 23 '17 at 11:45

12

You can do that like this: First map over the list to produce a list of tuples, then call toMap on the result which will make an immutable Map out of the tuples:

val m = ids.map(id => ("foo:" + id + ":bar", "val1")).toMap

Then convert the immutable Map to a mutable Map, for example like explained here:

val mm = collection.mutable.Map(m.toSeq: _*)

edit - the intermediate immutable Map is not necessary as the commenters noticed, you can also do this:

val mm = collection.mutable.Map(ids.map(id => ("foo:" + id + ":bar", "val1")): _*)

edited May 23 '17 at 11:45

Community

1
1

answered Oct 13 '15 at 07:47

Jesper

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It seems unnecessary to first call `toMap` and then `toSeq`. – Peter Neyens Oct 13 '15 at 07:49
7

This could be simplified somewhat to `collection.mutable.Map(ids map { ix => s"foo:$ix:bar" -> "val1" } : _*)` – Shadowlands Oct 13 '15 at 07:50
@Shadowlands you may want to convert your comment to an answer, it’s better than the other existing answers. – Abhijit Sarkar Apr 24 '19 at 07:02

score 1 · Answer 2 · answered Oct 13 '15 at 08:05

Try this:

import scala.collection.mutable
val ids = List(7, 8, 9)
val m = mutable.Map[String, String]()
ids.foreach { id => m.update(s"foo:$id:bar", "val1") }

scala> m
Map(foo:9:bar -> val1, foo:7:bar -> val1, foo:8:bar -> val1)

You, don't need to create any intermediate objects, that map does.

score 1 · Answer 3 · answered Apr 24 '19 at 07:42

1

Why not use foldLeft?

ids.foldLeft(mutable.Map.empty[String, String]) { (m, i) => 
  m += (s"foo:$i:bar" -> "val1")
}

answered Apr 24 '19 at 07:42

Abhijit Sarkar

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Hackaholic · Answer 4 · 2015-10-13T11:40:05.420

Try Like this:

scala> import scala.collection.mutable
scala> val result = mutable.Map[String, String]()
result: scala.collection.mutable.Map[String,String] = Map()

scala> val ids = List(7, 8, 9)
ids: List[Int] = List(7, 8, 9)

scala> for(x <- ids){
     | result("foo:%d:bar".format(x)) = "val1"
     | }

scala> result
res3: scala.collection.mutable.Map[String,String] = Map(foo:9:bar -> val1, foo:7:bar -> val1, foo:8:bar -> val1)

How to convert list to mutable map with pre processing in Scala?

4 Answers4