1、数组
- Array可以定义不可变数组，这里的不可变指的是长度不可变
scala> var arr = Array("233", "33") arr: Array[String] = Array(233, 33) scala> arr(1) = "22" scala> println(arr(1)) 22 scala> arr(2) java.lang.ArrayIndexOutOfBoundsException: 2 ... 48 elided
可以指定数组的具体类型，也可以不指定，Scala会自行判定
scala> val arrStr = Array("scala","spark") arrStr: Array[String] = Array(scala, spark) scala> val arrInt = new Array[Int](10) arrInt: Array[Int] = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) scala> val arrStr = Array("scala",1) arrStr: Array[Any] = Array(scala, 1)
遍历数组
scala> var arr = Array("string","spark") arr: Array[String] = Array(string, spark) scala> for(i <- 0 until arr.length)print(arr(i) + " ") string spark scala> for(ex <- arr)println(ex) string spark
转换为ArrayBuffer
scala> var arr = Array("233","23333") arr: Array[String] = Array(233, 23333) scala> var arr1 = arr.toBuffer arr1: scala.collection.mutable.Buffer[String] = ArrayBuffer(233, 23333) scala> arr1 += "222" res1: scala.collection.mutable.Buffer[String] = ArrayBuffer(233, 23333, 222)
- ArrayBuffer
ArrayBuffer是可变数组，使用前需要显示导入包
scala> import scala.collection.mutable.ArrayBuffer import scala.collection.mutable.ArrayBuffer scala> var arrBufInt = ArrayBuffer[Int]() arrBufInt: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer()
+=在尾端添加元素
scala> arrBufInt+= 1
res3: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1)
在尾端添加多个元素
scala> arrBufInt += (2,3,4,5)
res4: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5)
++=操作符可以追加任何集合
scala> var arrBufInt2 = ArrayBuffer[Int]() //创建一个新的ArrayBuffer arrBufInt2: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer() scala> arrBufInt2 += (3,4,5) //从尾端添加3，4，5 res5: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(3, 4, 5) scala> arrBufInt ++= arrBufInt2 //从尾端添加一个ArrayBuffer res6: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 3, 4, 5) scala> for(i <- 0 until arrBufInt.length)print(arrBufInt(i) + " ") 1 2 3 4 5 3 4 5
ArrayBuffer的一些方法
scala> arrBufInt.trimEnd(2) //移除最后两个元素 scala> for(i <- 0 until arrBufInt.length)print(arrBufInt(i) + " ") 1 2 3 4 5 3 4 5 6 scala> arrBufInt.remove(2) //移除索引号为2的元素 res13: Int = 3 scala> for(i <- 0 until arrBufInt.length)print(arrBufInt(i) + " ") 1 2 4 5 3 4 5 6 scala> arrBufInt.remove(2,4) //移除从索引号为2开始的4个元素 scala> for(i <- 0 until arrBufInt.length)print(arrBufInt(i) + " ") 1 2 5 6 scala> var arry = arrBufInt.toArray //转换为Array arry: Array[Int] = Array(1, 2, 5, 6)
2、Map
Scala映射就是键值对的集合Map,默认情况下，Scala使用不可变的映射。如果想使用可变集，必须明确地导入scala.collection.mutable.Map
- 构建不可变映射
val map = Map("name" -> "steve","age" -> "12","sex" -> "man") println(map("name")) //获取key为name的值 println("map中是否存在name：" + map.contains("name")) //判断map中是否有name作为key的键值对 println(map.getOrElse("age", "24")) //若map中有age作为key的键值对那么返回对应的value,若不存在就返回24 println(map.getOrElse("ages", "24")) //遍历map map.keys.foreach{i => print("key = " + i) println(" value = " + map(i))}
输出：
steve map中是否存在name：true 12 24 key = age value = 12 key = name value = steve key = sex value = man
- 构造可变映射
val map = scala.collection.mutable.Map("name" -> "steve","age" -> "12") map += ("sex" -> "boy")//添加键值对 println("map中是否存在sex：" + map.contains("sex")) map -= ("sex" )//删除键值对 println("map中是否存在sex：" + map.contains("sex")) //遍历map for((k, v) <- map){println(k + " " + v)} //遍历ma的键值 for(key <- map.keySet){println(key)}
输出：
map中是否存在sex：true map中是否存在sex：false age 12 name steve age name
3、Tuple元组
Tuple元组就是由两个小括号包住可以存放任意类型的数据集合
val t = ("1",2,3.0) t: (String, Int, Double) = (1,2,3.0) val t = new Tuple3(1, 3.14, "Fred") t: (Int, Double, String) = (1,3.14,Fred)
元组的实际类型取决于它的元素的类型，要注意的是元组的长度是不可变的，目前 Scala 支持的元组最大长度为 22。对于更大长度你可以使用集合，或者扩展元组
scala> t._1 res5: Int = 1 scala> t._2 res6: Double = 3.14 scala> t._3 res7: String = Fred
这样就能访问元组的元素
scala> t.productIterator.foreach{ i =>println("Value = " + i )}
Value = 1
Value = 3.14
Value = Fred
使用 Tuple.productIterator() 方法来迭代输出元组的所有元素
4、列表List
列表List是不可变的，其中元素不能被改变，列表List是一个链表
- 定义
val list1:List[String] = List("apple","oranges","pears") //定义String类型的List val list2:List[Int] = List(1,2,3,4) //定义Int类型的List val list3:List[Nothing] = List() //定义空List val list4 = List("apple","oranges","pears") //不指定类型，Scala将自行判断定义空List val list5:List[List[Int]] = List(List(1,2,3),List(4,5,6),List(7,8,9)) //定义二维List val list6 = "apple"::("orange"::("pears"::Nil))//使用Nil定义String类型的List val list7 = 1::(2::(3::(4::(5::Nil)))) //使用Nil定义Int类型的List val list8 = Nil //使用Nil定义空List val list9 = (1::(2::(3::(4::Nil))))::(1::(2::(3::(4::Nil))))::(1::(2::(3::(4::Nil))))::Nil //使用Nil定义二维List
- 常用方法
val list1:List[String] = List("apple","oranges","pears") //定义String类型的List val list2:List[Nothing] = List() //定义空List val list3:List[List[Int]] = List(List(1,2,3),List(4,5,6),List(7,8,9)) //定义二维List println("list1 head " + list1.head) //返回列表的第一个元素 println( "list1 tail " + list1.tail ) //返回列表除了第一个元素的其他元素 println("list3 head " + list3.head) println( "list3 tail " + list3.tail ) println( "list1是否为空 " + list1.isEmpty ) println( "list2是否为空 " + list2.isEmpty )
输出：
list1 head apple list1 tail List(oranges, pears) list3 head List(1, 2, 3) list3 tail List(List(4, 5, 6), List(7, 8, 9)) list1是否为空 false list2是否为空 true
- 连接列表
可以使用 ::: 运算符或 List.:::() 方法或 List.concat() 方法来连接两个或多个列表
val list1:List[String] = List("apple","oranges","pears") //定义String类型的List val list2:List[String] = List("apple1","oranges1","pears1") //定义String类型的List val list3:List[List[Int]] = List(List(1,2,3),List(4,5,6),List(7,8,9)) //定义二维List val list4:List[List[Int]] = List(List(11,21,31),List(41,51,61),List(71,81,91)) //定义二维List val list5 = list1:::list2 val list6 = list3:::list4 val list7 = list1.:::(list2) val list8 = list3.:::(list4) val list9 = List.concat(list1,list2) val list10 = List.concat(list3,list4) println("list1:::list2 " + list5) println("list3:::list4 " + list6) println("list1.:::(list2) " + list7) println("list3.:::(list4) " + list8) println("List.concat(list1,list2) " + list9) println("List.concat(list3,list4) " + list10)
输出
list1:::list2 List(apple, oranges, pears, apple1, oranges1, pears1) list3:::list4 List(List(1, 2, 3), List(4, 5, 6), List(7, 8, 9), List(11, 21, 31), List(41, 51, 61), List(71, 81, 91)) list1.:::(list2) List(apple1, oranges1, pears1, apple, oranges, pears) list3.:::(list4) List(List(11, 21, 31), List(41, 51, 61), List(71, 81, 91), List(1, 2, 3), List(4, 5, 6), List(7, 8, 9)) List.concat(list1,list2) List(apple, oranges, pears, apple1, oranges1, pears1) List.concat(list3,list4) List(List(1, 2, 3), List(4, 5, 6), List(7, 8, 9), List(11, 21, 31), List(41, 51, 61), List(71, 81, 91))
5、Set
Set是不包含重复元素的集合
- 定义不可变Set集合
var s:Set[Nothing] = Set() //定义空的Set var s1:Set[Int] = Set(1,2,3,4) //定义Int类型的Set var s2 = Set(1,2,3,4) //定义Int类型的Set
- 定义可变的Set集合需要引入scala.collection.mutable.Set
var s:Set[Int] = Set(1,2,3,4) //定义Int类型的Set println("s: " + s) s.add(5) println("s: " + s) s.remove(4) println("s: " + s) s += 6 println("s: " + s) s -= 2 println("s: " + s)
输出是：
s: Set(1, 2, 3, 4) s: Set(1, 5, 2, 3, 4) s: Set(1, 5, 2, 3) s: Set(1, 5, 2, 6, 3) s: Set(1, 5, 6, 3)
- 常用方法
var s:Set[Int] = Set(1,2,3,4) //定义Int类型的Set println("s.head " + s.head) println("s.tail " + s.tail) println("s.isEmpty " + s.isEmpty)
输出：
s.head 1 s.tail Set(2, 3, 4) s.isEmpty false
- 连接集合
可以使用 ++ 运算符或 Set.++() 方法来连接两个集合。如果元素有重复的就会移除重复的元素
var s1:Set[Int] = Set(1,2,3,4) var s2:Set[Int] = Set(4,5,6,1) var s3 = s1 ++ s2 var s4 = s1.++(s2) println("s1 ++ s2 " + s3) println("s1.++(s2) " +s4)
输出：
s1 ++ s2 Set(1, 5, 2, 6, 3, 4)
s1.++(s2) Set(1, 5, 2, 6, 3, 4)