Kotlin: How to convert list to map with list?

Question

I have a list as below {("a", 1), ("b", 2), ("c", 3), ("a", 4)}

I want to convert it to a map of list as below {("a" (1, 4)), ("b", (2)), ("c", (3)))}

i.e. for a, we have a list of 1 and 4, since the key is the same.

The answer in How to convert List to Map in Kotlin? only show unique value (instead of duplicate one like mine).

I tried associateBy in Kotlin

    data class Combine(val alpha: String, val num: Int)
    val list = arrayListOf(Combine("a", 1), Combine("b", 2), Combine("c", 3), Combine("a", 4))
    val mapOfList = list.associateBy ( {it.alpha}, {it.num} )
    println(mapOfList)

But doesn't seems to work. How could I do it in Kotlin?

Answer 1

Code

fun main(args: Array<String>) {
    data class Combine(val alpha: String, val num: Int)
    val list = arrayListOf(Combine("a", 1), Combine("b", 2), Combine("c", 3), Combine("a", 4))
    val mapOfList = list.associateBy ( {it.alpha}, {it.num} )
    println(mapOfList)
    val changed = list
        .groupBy ({ it.alpha }, {it.num})
    println(changed)
}

Output

{a=4, b=2, c=3}
{a=[1, 4], b=[2], c=[3]}

How it works

• First it takes the list
• It groups the Combines by their alpha value to their num values

Answer 2

You may group the list by alpha first and then map the value to List<Int>:

data class Combine(val alpha: String, val num: Int)
val list = arrayListOf(Combine("a", 1), Combine("b", 2), Combine("c", 3), Combine("a", 4))
val mapOfList = list
        .groupBy { it.alpha }
        .mapValues { it.value.map { it.num } }
println(mapOfList)

Answer 3

Here's a slightly more concise version of Jacky Choi's solution.

It combines the grouping and the transforming into one call to groupBy().

val mapOfList = list
        .groupBy (
            keySelector = { it.name },
            valueTransform = { it.num },
        )