大資料之scala（二） --- 對映，元組，簡單類，內部類，物件Object，Idea中安裝scala外掛，trait特質[介面]，包和包的匯入
阿新 • • 發佈：2018-11-09
一、對映<Map>
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1.建立一個不可變的對映Map<k,v> ==> Map(k -> v)
scala> val map = Map("tom" -> 1 , "tomasLee" -> 2 , "join" -> 3)
map: scala.collection.immutable.Map[String,Int] = Map(tom -> 1, tomasLee -> 2, join -> 3)
注意，這種方式建立的對映的元素是不可以改變的
2.訪問Map
scala> map("tom")
3.建立可變的對映
scala> val map = scala.collection.mutable.Map(100 -> "tom",200->"tomsLee",300 -> "join")
map: scala.collection.mutable.Map[Int,String] = Map(200 -> tomsLee, 100 -> tom, 300 -> join)
scala> map(100) = "222"
scala> map
res1: scala.collection.mutable.Map[Int,String] = Map(200 -> tomsLee, 100 -> 222, 300 -> join)
4.建立空的可變對映：如果想從一個空對映開始，你需要選擇一個對映實現hashmap和map的引數[Int,String]
scala> val map = new scala.collection.mutable.HashMap[Int,String]
map: scala.collection.mutable.HashMap[Int,String] = Map()
scala> map
res2: scala.collection.mutable.HashMap[Int,String] = Map()
scala> map(100)="111"
scala> map
res4: scala.collection.mutable.HashMap[Int,String] = Map(100 -> 111)
5.追加和移除元素
a.對於可變對映
scala> val map = scala.collection.mutable.HashMap[Int,String]()
map: scala.collection.mutable.HashMap[Int,String] = Map()
//新增元素
scala> map += (1->"100",2->"200")
res5: map.type = Map(2 -> 200, 1 -> 100)
//移除元素
scala> map -= 2
res9: map.type = Map(1 -> 100)
b.對於不可變對映
不可變對映是不能新增和移除元素的，但是可以變成另外一個對映
scala> val map = Map("tom" -> 1 , "tomasLee" -> 2 , "join" -> 3)
map: scala.collection.immutable.Map[String,Int] = Map(tom -> 1, tomasLee -> 2, join -> 3)
//immutable不可變集合新增元素是錯誤的
scala> map +=("xiao" -> 4)
<console>:12: error: value += is not a member of scala.collection.immutable.Map[String,Int]
map +=("xiao" -> 4)
//但是可以將map元素新增或者移除一個元素，之後重新賦值給一個新的不可變對映
scala> val map1 = map + ("xiao" -> 4)
map1: scala.collection.immutable.Map[String,Int] = Map(tom -> 1, tomasLee -> 2, join -> 3, xiao -> 4)
scala> map - "tom"
res15: scala.collection.immutable.Map[String,Int] = Map(tomasLee -> 2, join -> 3)
scala> map
res17: scala.collection.immutable.Map[String,Int] = Map(tom -> 1, tomasLee -> 2, join -> 3)
6.迭代
scala> for((k,v) <- map) println(k + ":" + v)
tom:1
tomasLee:2
join:3
//得到一個類似Set("tom", "tomasLee", "join")
scala> map.keys
res19: Iterable[String] = Set(tom, tomasLee, join)
scala> map.keySet
res20: scala.collection.immutable.Set[String] = Set(tom, tomasLee, join)
//遍歷values
scala> for(v <- map.values) println(v)
1
2
3
//遍歷keys
scala> for(v <- map.keys) println(v)
tom
tomasLee
join
//遍歷keySet
scala> for(v <- map.keySet) println(v)
tom
tomasLee
join
//遍歷values
scala> for(v <- map.keySet) println(map(v))
1
2
3
7.yield 迭代產生新的對映，比如kv調換
scala> map
res12: scala.collection.mutable.HashMap[Int,String] = Map(1 -> 100)
scala> for((k,v)<- map ) yield(v,k)
res11: scala.collection.mutable.HashMap[String,Int] = Map(100 -> 1)
二、元組tuple
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1.元組是不同型別的值的聚集 -- 對映是鍵值對的集合，鍵值對是元組的最簡單的形式
2.元組的值是通過將單個的值包含在圓括號中構成的
(1,3.14,"Fred")
3.定義一個元組Tuple,最多22個元素
scala> val t = (1,3.14,"Fred")
t: (Int, Double, String) = (1,3.14,Fred)
4.訪問元組中的元素
scala> t._1
res27: Int = 1
scala> t _2
res28: Double = 3.14
scala> t._3
res29: String = Fred
5.直接取出元組中的所有元素
scala> val(a,b,c) = t
a: Int = 1
b: Double = 3.14
c: String = Fred
6.陣列的zip(咬合/拉鍊)
scala> val h = Array(1,2,3)
h: Array[Int] = Array(1, 2, 3)
scala> val w = Array(4,5,6)
w: Array[Int] = Array(4, 5, 6)
scala> h.zip(w)
res31: Array[(Int, Int)] = Array((1,4), (2,5), (3,6))
三、簡單類
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1.定義類
class Person{
//定義變數,私有型別,必須初始化
//set/get也私有
private var id = 0 ;
//不宣告就是public，public的常量自帶get,沒有set
val age = 100 ;
//不宣告就是public，public的變數自帶get，set
var name = "tom" ;
//預設public
def incre(a:Int) = {id += a ;}
//如果定義函式時，沒有加(),呼叫就不能加()
def current() = id
}
2.使用類
scala> var p = new Person()
p: Person = Pers