Scala学习笔记04_Map与Tuple

in #cn • 7 years ago

创建Map

// 创建一个不可变的Map
scala> val ages = Map("Leo" -> 30, "Jen" -> 25, "Jack" -> 23)

scala> val ages = Map("Leo" -> 30, "Jen" -> 25, "Jack" -> 23)
ages: scala.collection.immutable.Map[String,Int] = Map(Leo -> 30, Jen -> 25, Jack -> 23)

scala> ages("Leo") = 31
<console>:15: error: value update is not a member of scala.collection.immutable.Map[String,Int]
       ages("Leo") = 31
       ^
// 创建一个可变的Map
scala> val ages = scala.collection.mutable.Map("Leo" -> 30, "Jen" -> 25, "Jack" -> 23)
ages: scala.collection.mutable.Map[String,Int] = Map(Jen -> 25, Jack -> 23, Leo -> 30)

scala> ages("Leo") = 31

scala> ages("Leo")
res43: Int = 31

scala> ages
res44: scala.collection.mutable.Map[String,Int] = Map(Jen -> 25, Jack -> 23, Leo -> 31)
// 使用另外一种方式定义Map元素
scala> val ages = Map(("Leo",30), ("Jen",25), ("Jack",23))
ages: scala.collection.immutable.Map[String,Int] = Map(Leo -> 30, Jen -> 25, Jack -> 23)
// 创建一个空的HashMap
scala> val ages = new scala.collection.mutable.HashMap[String,Int]
ages: scala.collection.mutable.HashMap[String,Int] = Map()

scala> val ages = new scala.collection.mutable.HashMap[String,Int]()
ages: scala.collection.mutable.HashMap[String,Int] = Map()

scala> val ages = new scala.collection.mutable.Map[String,Int]()
<console>:13: error: trait Map is abstract; cannot be instantiated
       val ages = new scala.collection.mutable.Map[String,Int]()
                  ^

访问Map的元素

// 获取指定key对应的value，如果key不存在，会报错
scala> val ages = scala.collection.mutable.Map("Leo" -> 30, "Jen" -> 25, "Jack" -> 23)
ages: scala.collection.mutable.Map[String,Int] = Map(Jen -> 25, Jack -> 23, Leo -> 30)

scala> val leoAge = ages("Leo")
leoAge: Int = 30

scala> val leoAge = ages("leo")
java.util.NoSuchElementException: key not found: leo
  at scala.collection.MapLike$class.default(MapLike.scala:228)
  at scala.collection.AbstractMap.default(Map.scala:59)
  at scala.collection.mutable.HashMap.apply(HashMap.scala:65)
  ... 32 elided
// 使用contains函数检查key是否存在
scala> val leoAge = if(ages.contains("leo")) ages("leo") else 0
leoAge: Int = 0
// getOrElse函数
scala> val leoAge = ages.getOrElse("leo",0)
leoAge: Int = 0

修改Map的元素

// 更新Map的元素
scala> ages
res45: scala.collection.mutable.Map[String,Int] = Map(Jen -> 25, Jack -> 23, Leo -> 30)

scala> ages("Leo") = 31

scala> ages
res47: scala.collection.mutable.Map[String,Int] = Map(Jen -> 25, Jack -> 23, Leo -> 31)
// 增加多个元素
scala> ages += ("Mike" -> 35, "Tom" -> 50)
res48: ages.type = Map(Jen -> 25, Mike -> 35, Tom -> 50, Jack -> 23, Leo -> 31)
// 移除元素
scala> ages -= "Mike"
res49: ages.type = Map(Jen -> 25, Tom -> 50, Jack -> 23, Leo -> 31)
// 创建不可变的Map
scala> val ages = Map("Leo" -> 30, "Jack" -> 40, "Jen" -> 25)
ages: scala.collection.immutable.Map[String,Int] = Map(Leo -> 30, Jack -> 40, Jen -> 25)
// 更新不可变的Map
scala> val ages2 = ages + ("Mike" -> 36, "Tom" -> 60)
ages2: scala.collection.immutable.Map[String,Int] = Map(Mike -> 36, Tom -> 60, Leo -> 30, Jack -> 40, Jen -> 25)
// 原来Map已有的元素也可更新
scala> val ages2 = ages + ("Leo" -> 31)
ages2: scala.collection.immutable.Map[String,Int] = Map(Leo -> 31, Jack -> 40, Jen -> 25)
// 移除不可变Map的元素
scala> val ages3 = ages - "Jack"
ages3: scala.collection.immutable.Map[String,Int] = Map(Leo -> 30, Jen -> 25)

遍历Map

// 遍历Map的entrySet
scala> ages
res50: scala.collection.immutable.Map[String,Int] = Map(Leo -> 30, Jack -> 40, Jen -> 25)

scala> for((key,value) <- ages) println(key + ": " + value)
Leo: 30
Jack: 40
Jen: 25
// 遍历Map的key
scala> for(key <- ages.keySet) println(key)
Leo
Jack
Jen
// 遍历Map的Value
scala> for(value <- ages.values) println(value)
30
40
25
// 生成新Map，反转key和value
scala> for((key, value) <- ages) yield (value, key)
res54: scala.collection.immutable.Map[Int,String] = Map(30 -> Leo, 40 -> Jack, 25 -> Jen)

SortedMap和LinkedHashMap

// SortedMap可以自动对Map的key的排序
scala> val ages = scala.collection.immutable.SortedMap("Leo" -> 30, "Jack" -> 20, "Jen" ->25)
ages: scala.collection.immutable.SortedMap[String,Int] = Map(Jack -> 20, Jen -> 25, Leo -> 30)
// LinkedHashMap可以记住插入entry的顺序
scala> val ages = new scala.collection.mutable.LinkedHashMap[String, Int]
ages: scala.collection.mutable.LinkedHashMap[String,Int] = Map()

scala> ages("leo") = 30

scala> ages("jack") = 20

scala> ages("jen") = 25

scala> ages
res3: scala.collection.mutable.LinkedHashMap[String,Int] = Map(leo -> 30, jack -> 20, jen -> 25)

Map的元素类型Tuple

// 简单Tuple
scala> val t = ("leo", 30)
t: (String, Int) = (leo,30)
// 访问Tuple
scala> t._1
res4: String = leo

scala> t._2
res5: Int = 30
// zip操作
scala> val names = Array("leo", "jack", "jen")
names: Array[String] = Array(leo, jack, jen)

scala> val ages = Array(30, 20, 25)
ages: Array[Int] = Array(30, 20, 25)

scala> val nameAges = names.zip(ages)
nameAges: Array[(String, Int)] = Array((leo,30), (jack,20), (jen,25))

scala> for((name, age) <- nameAges) println(name + ": " + age)
leo: 30
jack: 20
jen: 25

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