CODEWITHSUNDEEP
kotlinlambdagroup-by
user1359660
user1359660

Reputation: 69

Group pairs by a second value, and map it where the key is the second element and a value is a list of first elements associated with the second

As I mentioned in the title. I would like to group List of Pairs by the second value. I came up with a code below that works fine, but I'm wondering if there is a possibility to do it more elegant using lambdas.

val data = mapOf(
        Pair("a1", listOf("b1", "b2", "b3", "b4", "b5")),
        Pair("a2", listOf("b1", "b7", "b8", "b9", "b10")),
        Pair("a3", listOf("b6", "b7", "b8", "b9", "b10")),
        Pair("a4", listOf("b6", "b7", "b8", "b9", "b11")),
        Pair("a5", listOf("b6", "b2", "b12", "b9", "b13"))
)

val map = mutableMapOf<String, MutableList<String>>()

data.forEach { (k, v) ->
    v.forEach {
        if (map.containsKey(it)) {
            map[it]?.add(k)
        } else {
            map[it] = mutableListOf(k)
        }
    }
}

Upvotes: 1

Views: 851

Answers (1)

s1m0nw1
s1m0nw1

Reputation: 81997

According to your given example, the expected result is

{b1=[a1, a2], b2=[a1, a5], b3=[a1], b4=[a1], b5=[a1], b7=[a2, a3, a4], b8=[a2, a3, a4], b9=[a2, a3, a4, a5], b10=[a2, a3], b6=[a3, a4, a5], b11=[a4], b12=[a5], b13=[a5]}

You could expand the given data structure in a way that every element from your second element is being associated to the first element. But be aware that this results in additional space. Grouping becomes easier then:

data.entries
    .flatMap { (k, v) -> v.map { k to it } }
    .groupBy(keySelector = Pair<*, *>::second, valueTransform = Pair<*, *>::first) //group by list element, only add key to group

The iterative approach can be simplified to:

val map = mutableMapOf<String, List<String>>().withDefault { listOf() }

for ((k, v) in data) {
    for (e in v) map[e] = map.getValue(e) + k
}

Upvotes: 1

Related Questions