学习scala笔记--6 scala map、tuple

 

scala  map  tuple

 

 

 

默认是map 是不可变map immutable.Map

 

scala> val a= Map("a"-> 1 ,"b"->2)

a: scala.collection.immutable.Map[String,Int] = Map(a -> 1, b -> 2)

 

 

 

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可以指定创建  可变map

 

scala> val b = scala.collection.mutable.Map("c"->3,"d"->4)

b: scala.collection.mutable.Map[String,Int] = Map(d -> 4, c -> 3)

 

 

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map的合并

 

scala> val b = scala.collection.mutable.Map("c"->3,"d"->4)

b: scala.collection.mutable.Map[String,Int] = Map(d -> 4, c -> 3)

 

scala> 

 

scala> val c  = scala.collection.mutable.Map("e"->3,"f"->4)

c: scala.collection.mutable.Map[String,Int] = Map(e -> 3, f -> 4)

 

scala> b ++= c

res0: b.type = Map(e -> 3, d -> 4, c -> 3, f -> 4)

 

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通过元组创建：

 

scala> val g = Map(("a",1),("b",2),("c",3))

g: scala.collection.immutable.Map[String,Int] = Map(a -> 1, b -> 2, c -> 3)

 

 

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获取 值

 

scala> val x = g("a")

x: Int = 1

 

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判断是否存在key

 

scala> g.contains("a")

res2: Boolean = true

 

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scala> g.getOrElse("d",0)

res3: Int = 0

 

 

 

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遍历  

 

scala> for((k,v)<-g) println(k+"->"+v)

a->1

b->2

c->3

 

 

遍历key

 

scala> for(k <- g.keySet) println(k)

a

b

c

 

遍历 value

 

scala> for(v <- g.values) println(v)

1

2

3

 

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SortedMap 自动安装key排序

 

scala> val sm = scala.collection.immutable.Map("d"->1,"a"->2,"e"->3)

sm: scala.collection.immutable.Map[String,Int] = Map(d -> 1, a -> 2, e -> 3)

 

scala> val sm = scala.collection.immutable.SortedMap("d"->1,"a"->2,"e"->3)

sm: scala.collection.immutable.SortedMap[String,Int] = Map(a -> 2, d -> 1, e -> 3)

 

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自动维护插入顺序的 LinkedHashMap

 

scala> val h = scala.collection.mutable.LinkedHashMap[String,Int]()

h: scala.collection.mutable.LinkedHashMap[String,Int] = Map()

 

scala> h +=("x"->1)

res14: h.type = Map(x -> 1)

 

scala> h +=("y"->2)

res15: h.type = Map(x -> 1, y -> 2)

 

scala> h +=("z"->3)

res16: h.type = Map(x -> 1, y -> 2, z -> 3)

 

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tuple  元组

 

 

scala> var t = ("name","tom")

t: (String, String) = (name,tom)

 

scala> var t2 = ("name","tom","age")

t2: (String, String, String) = (name,tom,age)

 

 

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zip操作

 

scala> val a = Array("a","b","c")

a: Array[String] = Array(a, b, c)

 

scala> val b = Array("1","2","3")

b: Array[String] = Array(1, 2, 3)

 

scala> var t = a.zip(b)

t: Array[(String, String)] = Array((a,1), (b,2), (c,3))

 

 

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访问 tuple 元素  t._1

 

scala> var t = ("name","tom")

t: (String, String) = (name,tom)

 

scala> t._1

res20: String = name

 

scala> t._2

res21: String = tom

 

 

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