Ruckus
08/16/2017, 8:51 PMval filtered = (0..list.lastIndex).asSequence()
.filter { it in indexes }
.map(list::get)
.toList()
(I haven't tested it)karelpeeters
08/16/2017, 8:52 PMval result = set.map { list[it] }.toList()
also works.Ruckus
08/16/2017, 8:58 PMkarelpeeters
08/16/2017, 9:00 PMset.toList().sort().map { list[it] }
. Everything to avoid int ranges!