Kotlin sort array by value in range

Question

Let's have a class Player(val position: Int, val time: Float) and we want to sort an array or list of players by position. If some of these players have the same position after first sorting, we want to sort them by time in groups. By group I mean set of players with the same position.

I know about

list.sortedWith(compareBy<Foo> { it.a }.thenByDescending { it.b }.thenBy { it.c })

But of course it does not solve this case.

Is there any smart way in Kotlin to achieve this simple task? We can sort it manually by checking positions and swapping items, but I wonder if Kotlin has something to say in this case.

You should make your question clearer: What kind of data structure do you have, do you want to have this one sorted always, i.e. whenever a new element is added it's supposed to be sorted in at the appropriate position or do you want to have a function which sorts on demand? What data structure do you need then? — s1m0nw1
– s1m0nw1, Commented Aug 27, 2017 at 14:54
yes, i wasn't specific about this. i wanted kotlin to showoff with case like this and converting array to map and then back to array is just not desired solution. i thought about some tricks like parallel sorting or even in place with some extra predicates, but you are right tho, we can sort it by converting list to map and then back again. i marked your answer correct. — Marcel Skotnicki
– Marcel Skotnicki, Commented Aug 28, 2017 at 9:01

s1m0nw1 · Accepted Answer · 2017-08-27 14:09:56Z

3

You could first sort by position and time and then group by time with standard Kotlin functionality.

Example

data class Player(val position: Int, val time: Float)

val p1 = Player(1, 10f)
val plys = arrayOf(p1, p1.copy(position = 3),
        p1.copy(time = 0f), p1.copy(time = 20f),
        p1.copy(position = 2), p1.copy(position = 2, time = 20f))

val groupBy = plys.sortedWith(compareBy(Player::position, Player::time))
                  .groupBy { it.position }

Description

sort the Array by the Player's position and time with sortedWith + compareBy
group it by the Player's position

Result

The result is a Map<Int,List<Player>, which in the example looks like this:

    {
     1=[Player(position=1, time=0.0), Player(position=1, time=10.0), Player(position=1, time=20.0)], 
     2=[Player(position=2, time=10.0), Player(position=2, time=20.0)],
     3=[Player(position=3, time=10.0)]
    }

edited Aug 27, 2017 at 14:09

answered Aug 27, 2017 at 10:27

s1m0nw1

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3 Comments

Marcel Skotnicki Over a year ago

thank you for your answer. I know about groupBy{}, but it results with a map instead of list as you wrote yourself. It changes structure of keeping players and in my case it is a deal breaker.

Les Over a year ago

maps are easily converted to lists using toList() ... this should be marked the correct answer, or the OP should clarify the desired result

Marcel Skotnicki Over a year ago

well it is not desired to sort an array through converting it to map and then back to array, but i wasn't specific about this, so i mark this answer correct

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