前提知识
kotlin的集合分为可变和不可变，可变的集合带有mutable形容词。
| 创建方式 | 事例 | 是否可变 | 说明 |
|---|---|---|---|
| arrayListOf() mutableListOf() | val array = arrayListOf(1,4,7) | 可变 | 必须制定元素类型 |
| listOf() | val array = listOf(1,4,7) | 不可变 | 必须制定元素类型，必须指定初始化数据类型 |
| arrayMap/SetOf() mutableMap/SetOf() | val array = arrayMapOf(Pair("k","v")) val set = arrayMapOf(1,2,3) | 可变 | 必须使用Pair包装 ,Set集合会去重 |
| map/setOf() | val array = map/arraySetOf(Pair("k","v")) | 不可变 | 必须制定元素类型，必须指定初始化数据类型，Set集合会去重 |
转换操作符
list转array，set转list，list转MutableList
val sourceList = mutableListOf(1, 2, 3)
val readOnlyCopyList = sourceList.toList() 转为不可变集合val sourceList = mutableListOf(1, 2, 3)
val copySet = sourceList.toMutableSet() 转为可变Set
映射操作符
- map
val numbers = setOf(1, 2, 3)
println(numbers.map { it * 3 })
println(numbers.mapIndexed { idx, value -> value * idx })
打印结果
[3, 6, 9]
[0, 2, 6]
- flatMap
val containers = listOf(listOf("one", "two", "three"),listOf("four", "five", "six"),listOf("seven", "eight")
)
println(containers.flatMap { it.values })
//打印 [one, two, three, four, five, six, seven, eight]
过滤操作符
- filter
val numbers = listOf("one", "two", "three", "four")
val longerThan3 = numbers.filter { it.length > 3 }
//打印的结果是 three four
- take
val numbers = listOf("1", "2", "3", "4","5","6")
println(numbers.take(3).toList())
//打印的结果是 1 2 3val numbers = listOf("one", "two", "three", "four", "five", "six")
println(numbers.take(3)) 保留前三个
println(numbers.takeLast(3)) 保留最后三个
//打印
[one, two, three]
[four, five, six]
- drop
val numbers = listOf("one", "two", "three", "four", "five", "six")
println(numbers.drop(1)) //删除第一个
println(numbers.dropLast(5)) //删除最后5个
//打印
[two, three, four, five, six]
[one]
- predicates
val numbers = listOf("one", "two", "three", "four")
println(numbers.any { it.endsWith("e") }) //只要有一个以e结尾
println(numbers.none { it.endsWith("a") }) //没有一个以a结尾
println(numbers.all { it.endsWith("e") }) //每一个都是以e结尾
打印结果
true
true
false
- slice
val numbers = listOf("one", "two", "three", "four", "five", "six")
println(numbers.slice(1..3)) //位置以1到3进行打算
println(numbers.slice(0..4 step 2)) //0到4 每次价格2个 0,2,2+2
println(numbers.slice(setOf(3, 5, 0)))
打印结果
[two, three, four]
[one, three, five]
[four, six, one]
- find
val numbers = listOf(1, 2, 3, 4)
println(numbers.find { it % 2 == 0 }) 找到能被2 整除的数据
println(numbers.findLast { it % 2 == 0 }) 找到最后一个能被2整除的数
打印
2
4