kotlin集合包属性函数（待分组）(8)

partition根据函数返回true和false的元素分别产生集合 [原文来自：www.11jj.com]

val list = listOf(1, 2, 3, 4, 5, 6)assertEquals(    Pair(listOf(2, 4, 6), listOf(1, 3, 5)),    list.partition { it % 2 == 0 })

plus两个集合元素合并产生新集合

val list = listOf(1, 2, 3, 4, 5, 6)assertEquals(    listOf(1, 2, 3, 4, 5, 6, 7, 8),    list + listOf(7, 8))

zip两个集合相同索引的元素组成双值对，大小等于小集合

val list = listOf(1, 2, 3, 4, 5, 6)assertEquals(    listOf(Pair(1, 7), Pair(2, 8)),    list.zip(listOf(7, 8)))

unzip列表中双值对按位置分拆再合并成两个集合

val list = listOf(1, 2, 3, 4, 5, 6)assertEquals(    Pair(listOf(5, 6), listOf(7, 8)),    listOf(Pair(5, 7), Pair(6, 8)).unzip())

retainAll两个集合产生交集集合

val list = mutableListOf(1, 2, 3, 4, 5, 6)val list2 = listOf(3, 4, 5, 6, 7, 8, 9)assertEquals(listOf(3, 4, 5, 6), list.retainAll(list2))

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顺序操作符

reverse顺序反转

val unsortedList = listOf(3, 2, 7, 5)assertEquals(listOf(5, 7, 2, 3), unsortedList.reverse())

sort自然排序从小到大

val unsortedList = listOf(3, 2, 7, 5)assertEquals(listOf(2, 3, 5, 7), unsortedList.sort())

sortBy根据函数排序

val unsortedList = listOf(3, 2, 7, 5)assertEquals(listOf(3, 7, 2, 5), unsortedList.sortBy { it % 3 })

sortDescending降序排序从大到小

val unsortedList = listOf(3, 2, 7, 5)assertEquals(listOf(7, 5, 3, 2), unsortedList.sortDescending())

sortDescendingBy根据函数降序排序--------

val unsortedList = listOf(3, 2, 7, 5)assertEquals(listOf(2, 5, 7, 3), unsortedList.sortDescendingBy {it % 3 })