Kotlin sort list of list based on parameter in inner list

Question

I have a list of list that I want to sort based on a date value in inner list. Here's example of my list:

ClassA(id: String, classb : List<ClassB>) and ClassB is ClassB(id: String, date: Date) so I have list of ClassA which also contain list of ClassB and I want to order/sort List based on descending order of dates in inner classB

do you want a list of ClassB ordered by date with its corresponding ClassA or do you want a list of ClassA where only the ClassB are ordered by date for each ClassA... or in other words: what should happen if there are two ClassB-dates of any ClassA-object that are both before and after other ClassA-object-contained ClassB-dates? — Roland
– Roland, Commented Sep 24, 2019 at 10:10

Roland · Accepted Answer · 2019-09-24 10:38:38Z

1

Assuming the following:

val int = AtomicInteger(0)
data class A(val bs: List<B>, val id: Int = int.incrementAndGet())
data class B(val date: LocalDate, val id: Int = int.incrementAndGet())

and a list (listOfA) as follows:

A(bs=[B(date=2019-09-24, id=1), B(date=2019-09-25, id=2), B(date=2019-09-23, id=3)], id=4)
A(bs=[B(date=2019-09-21, id=5), B(date=2019-09-22, id=6), B(date=2019-09-23, id=7)], id=8)
A(bs=[B(date=2019-09-19, id=9), B(date=2019-09-23, id=10), B(date=2019-09-29, id=11)], id=12)

only ordering the class B elements (actually I create a new list with completely new A-objects... if you do not want that, this can still serve as a starting point):

val listOfAWithOrderedB = listOfA.map {
  it.copy(bs = it.bs.sortedByDescending(B::date))
}

which leads to:

A(bs=[B(date=2019-09-25, id=2), B(date=2019-09-24, id=1), B(date=2019-09-23, id=3)], id=4)
A(bs=[B(date=2019-09-23, id=7), B(date=2019-09-22, id=6), B(date=2019-09-21, id=5)], id=8)
A(bs=[B(date=2019-09-29, id=11), B(date=2019-09-23, id=10), B(date=2019-09-19, id=9)], id=12)

ordering by all B-dates and keeping a reference to the actual A:

val bSortedByDateAndTheirA = listOfA.flatMap { anA ->
  anA.bs.map {
    it to anA
  }
}
    .sortedByDescending { (b) -> b.date }

which leads to a List<Pair<B, A>> (again... a possible starting point) as follows:

(B(date=2019-09-29, id=11), A(bs=[B(date=2019-09-19, id=9), B(date=2019-09-23, id=10), B(date=2019-09-29, id=11)], id=12))
(B(date=2019-09-25, id=2), A(bs=[B(date=2019-09-24, id=1), B(date=2019-09-25, id=2), B(date=2019-09-23, id=3)], id=4))
(B(date=2019-09-24, id=1), A(bs=[B(date=2019-09-24, id=1), B(date=2019-09-25, id=2), B(date=2019-09-23, id=3)], id=4))
(B(date=2019-09-23, id=3), A(bs=[B(date=2019-09-24, id=1), B(date=2019-09-25, id=2), B(date=2019-09-23, id=3)], id=4))
(B(date=2019-09-23, id=7), A(bs=[B(date=2019-09-21, id=5), B(date=2019-09-22, id=6), B(date=2019-09-23, id=7)], id=8))
(B(date=2019-09-23, id=10), A(bs=[B(date=2019-09-19, id=9), B(date=2019-09-23, id=10), B(date=2019-09-29, id=11)], id=12))
(B(date=2019-09-22, id=6), A(bs=[B(date=2019-09-21, id=5), B(date=2019-09-22, id=6), B(date=2019-09-23, id=7)], id=8))
(B(date=2019-09-21, id=5), A(bs=[B(date=2019-09-21, id=5), B(date=2019-09-22, id=6), B(date=2019-09-23, id=7)], id=8))
(B(date=2019-09-19, id=9), A(bs=[B(date=2019-09-19, id=9), B(date=2019-09-23, id=10), B(date=2019-09-29, id=11)], id=12))

answered Sep 24, 2019 at 10:38

Roland

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5 Comments

Roland Over a year ago

if you want to have a mixture of both... sorting list A corresponding to its B-dates, then you need to tell what should happen if there are several dates within one A that could be before/after any other A'.

saintjab Over a year ago

I will need mixture of both (sorting list A corresponding to its B-dates). My dates are actually DateTime and there's a guarantee that two dates in classB in two classA won't be the same, at least there will be a time difference. Thanks for the above, will try it

Roland Over a year ago

so do you have cases, such as in my example, where A(id=4) contains B-dates that are actually both, before and after some B-dates of A(id=12)... how would you want to order that?

saintjab Over a year ago

sorry for late reply. I would show the outer entries with higher size first. This that possible? Also I see your solution involves Pairs<A, B> output, how can I get the output in original format i.e instead of List<Pair<A, B>> but List<List<B>>

Roland Over a year ago

hmmm... maybe I wasn't clear enough or I don't understand what you really want... in my second approach I basically have each corresponding A in the pair... however each A still has its complete list of all possible Bs... so if you say List<List<B>> what do you really expect? if you only need the B-dates to be sorted, you can also just do listOfAs.flatMap { bs }.sortedByDescending { it.date }

Black mamba · Accepted Answer · 2019-09-24 09:47:31Z

0

you can use Comparator to use this... here the code sample that you can use

Comparator object

companion object : Comparator<MyDate> {

override fun compare(a: MyDate, b: MyDate): Int = when {
    a.year != b.year -> a.year - b.year
    a.month != b.month -> a.month - b.month
    else -> a.day - b.day
}}

date model class

data class MyDate (val year: Int, val month: Int, val day: Int) {}

and this is how you can perform sorting

fun main(args: Array<String>) {
val myDates = listOfdays
myDates.sortedWith(CompareObjects).forEach { 

    add sorted list
}}

answered Sep 24, 2019 at 9:47

Black mamba

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1 Comment

Roland Over a year ago

instead of using your own comparator, you may also be interested in compareBy... that way you could also write something like listOfB.sortedWith(compareBy { it.date })...

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