Scala入门之List使用详解

//字符串类型List
scala> val fruit=List("Apple","Banana","Orange")
fruit: List[String] = List(Apple, Banana, Orange)

//前一个语句与下面语句等同
scala> val fruit=List.apply("Apple","Banana","Orange")
fruit: List[String] = List(Apple, Banana, Orange)

//数值类型List
scala> val nums=List(1,2,3,4,5)
nums: List[Int] = List(1, 2, 3, 4, 5)

//多重List，List的子元素为List
scala> val list = List(List(1, 2, 3), List("adfa", "asdfa", "asdf"))
list: List[List[Any]] = List(List(1, 2, 3), List(adfa, asdfa, asdf))

//遍历List
scala> for(i <- list; from=i; j<-from)println(j)
1
2
3
adfa
asdfa
asdf

scala>nums(3)=4
<console>:10: error: value update is not a member of List[Int]
       nums(3)=4
       ^

scala>var listStr:List[Object] = List("This", "Is", "Covariant", "Example")
listStr:List[Object] = List(This, Is, Covariant, Example)

//空的List,其类行为Nothing,Nothing在Scala的继承层次中的最底层
//,即Nothing是任何Scala其它类型如String,Object等的子类
scala> var listStr = List()
listStr:List[Nothing] = List()

scala>var listStr:List[String] = List()
listStr:List[String] = List()

//1、常用::及Nil进行列表构建
scala> val nums = 1 :: (2:: (3:: (4 :: Nil)))
nums: List[Int] = List(1, 2, 3, 4)


//由于::操作符的优先级是从右向左的，因此上一条语句等同于下面这条语句
scala> val nums = 1::2::3::4::Nil
nums:List[Int] = List(1, 2, 3, 4)
至于::操作符的使用将在下面介绍

//判断是否为空
scala> nums.isEmpty
res5: Boolean = false

//取第一个元素
scala> nums.head
res6: Int = 1

//取列表第二个元素
scala>nums.tail.head
res7: Int = 2

//取第三个元素
scala>nums.tail.tail.head
res8: Int = 3

//插入操作
//在第二个位置插入一个元素
scala>nums.head::(3::nums.tail)
res11: List[Int] = List(1, 3, 2, 3, 4)

scala> nums.head::(nums.tail.head::(4::nums.tail.tail))
res12: List[Int] = List(1, 2, 4, 3, 4)

//插入排序算法实现
def isort(xs: List[Int]):List[Int] = {
  if(xs.isEmpty) Nil
  else insert(xs.head, issort(xs.tail))
}

def insert(x:Int, xs:List[Int]):List[Int] = {
  if(xs.isEmpty || x <= xs.head) x::xs
  else xs.head :: insert(x, xs.tail)
}

//连接操作
scala>List(1, 2, 3):::List(4, 5, 6)
res13: List[Int] = List(1, 2, 3, 4, 5, 6)

//去除最后一个元素外的元素，返回的是列表
scala> nums.init
res13: List[Int] = List(1, 2, 3)

//取出列表最后一个元素
scala>nums.last
res14: Int = 4

//列表元素倒置
scala> nums.reverse
res15: List[Int] = List(4, 3, 2, 1)

//一些好玩的方法调用
scala> nums.reverse.reverse == nums


//丢弃前面n个元素
scala>nums drop 3
res16: List[Int] = List(4)

//获取前面n个元素
scala>nums take 1
res17: List[Int] = List[1]

//将列表进行分割
scala> nums.splitAt(2)
res18: (List[Int], List[Int]) = (List(1, 2),List(3, 4))

//前一个操作与下列语句等同
scala> (nums.take(2),nums.drop(2))
res19: (List[Int], List[Int]) = (List(1, 2),List(3, 4))

//Zip操作
scala> val nums=List(1,2,3,4)
nums: List[Int] = List(1, 2, 3, 4)

scala> val chars=List('1','2','3','4')
chars: List[Char] = List(1, 2, 3, 4)

//返回的是List类型的元组(Tuple），返回的元素个数与最小的List集合的元素个数一样
scala> nums zip chars
res20: List[(Int, Char)] = List((1,1), (2,2), (3,3), (4,4))

//List toString方法
scala> nums.toString
res21: String = List(1, 2, 3, 4)

//List mkString方法
scala> nums.mkString
res22: String = 1234

//转换成数组
scala> nums.toArray
res23: Array[Int] = Array(1, 2, 3, 4)

//apply方法
scala> List.apply(1, 2, 3)
res24: List[Int] = List(1, 2, 3)

//range方法，构建某一值范围内的List
scala> List.range(2, 6)
res25: List[Int] = List(2, 3, 4, 5)

//步长为2
scala> List.range(2, 6,2)
res26: List[Int] = List(2, 4)

//步长为-1
scala> List.range(2, 6,-1)
res27: List[Int] = List()

scala> List.range(6,2 ,-1)
res28: List[Int] = List(6, 5, 4, 3)

//构建相同元素的List
scala> List.make(5, "hey")
res29: List[String] = List(hey, hey, hey, hey, hey)

//unzip方法
scala> List.unzip(res20)
res30: (List[Int], List[Char]) = (List(1, 2, 3, 4),List(1, 2, 3, 4))

//list.flatten，将列表平滑成第一个无素
scala> val xss =
   | List(List('a', 'b'), List('c'), List('d', 'e'))
xss: List[List[Char]] = List(List(a, b), List(c), List(d, e))
scala> xss.flatten
res31: List[Char] = List(a, b, c, d, e)

//列表连接
scala> List.concat(List('a', 'b'), List('c'))
res32: List[Char] = List(a
, b, c)

scala> val twoThree = List(2, 3)
scala> val oneTwoThree = 1 :: twoThree
scala> oneTwoThree
oneTwoThree: List[Int] = List(1, 2, 3)

scala> val one = List('A', 'B')
val one = List('A', 'B')
scala> val two = List('C', 'D')

scala> one:::two
res1: List[Char] = List(A, B, C, D)