Kotlin: Retain, replace or remove each map entry in-place

Question

I have a mutable map (a LinkedHashMap to be specific) in Kotlin. For each map entry, I want to conditionally perform one of three actions:

• Retain the entry as-is
• Replace the entry's value
• Remove the entry

In old Java, I would probably use an Iterator over the map's entrySet() for this purpose (using Iterator.remove() or Entry.setValue() accordingly). However, I wondered if there is a more elegant, functional approach present in Kotlin to do the same (i.e. apply a lambda to each entry in-place, similar to using replaceAll(), but with the ability to also remove entries).

---

In Kotlin, the following combination of replaceAll() and retainAll() does the same thing:

val map = LinkedHashMap<String,Int>(mapOf("a" to 1, "b" to 2, "c" to 3))
map.replaceAll { key, value ->
    if(key.startsWith("a")) {
        value * 2
    } else {
        value
    }
}
map.entries.retainAll { entry ->
    entry.key.startsWith("a") || entry.key.startsWith("b")
}

However, this iterates the map twice and requires splitting up the predicate/transformation. I would prefer a compute()-style all-in-one solution, just for all entries.

Answer 1

In functional world, immutability is an essential principle. So in-place modifications are not convenient.

But if you prefer an in-place approach, possible options are retainAll or removeAll to filter items. For in-place replacement you need to iterate the collection. You can use forEach to do an in-place modification:

data class User(val name: String)

val users = mutableListOf( User("John"), User("Alice"), User("Bob") )
numbers.run{
    removeAll { it.name == "Alice" }
    forEach { it.name =  "Mr. ${it.name}"}
}