CODEWITHSUNDEEP
kotlin
Jonas
Jonas

Reputation: 287

How to sort in kotlin with condition

I have pairs of list and i want to sort by Pair.second if Pair.first is not null

my list

val list = listOf<Pair<Int?, Int>>(
    Pair(1, 199),
    Pair(null, 180),
    Pair(10, 210),
    Pair(null, 178)
)

Desired result

 Pair(1, 199),
 Pair(10, 210),
 Pair(null, 178),
 Pair(null, 180)

Upvotes: 1

Views: 2444

Answers (3)

lukas.j
lukas.j

Reputation: 7163

You could split the list into two lists, one containing the pairs with the null value, the other containing the others, and then sort each group by the second value:

val list = listOf(
  Pair(1, 199),
  Pair(null, 180),
  Pair(10, 210),
  Pair(null, 178)
)

val result = list
  .groupBy { it.first == null }
  .flatMap { (_, subList) -> subList.sortedBy { it.second } }

result.forEach(::println)

Output:

(1, 199)
(10, 210)
(null, 178)
(null, 180)

Upvotes: 1

Pawel
Pawel

Reputation: 17248

I suggest checking out this question: How to sort based on/compare multiple values in Kotlin?

You can combine multiple sort criteria, first push nulls to the end then compare the fields:

val result = list.sortedWith(
    compareBy<Pair<Int?, Int>> { it.first == null } // sort nulls explicitly or they'll end up before non-null values
        .thenBy{ it.first }
        .thenBy{ it.second }
)

Upvotes: 3

cactustictacs
cactustictacs

Reputation: 19524

You can just use an elvis operator to default to using the second item if the first is null

list.sortedBy { it.first ?: it.second }

Upvotes: 1

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