// 初始化一个空 map val scores01 = new HashMap[String, Int] // 从指定的值初始化 Map(方式一) val scores02 = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // 从指定的值初始化 Map(方式二) val scores03 = Map(("hadoop", 10), ("spark", 20), ("storm", 30))
采用上面方式获得的都是不可变 Map(immutable map),想要获得可变 Map(mutable map),则须要使用:java
val scores04 = scala.collection.mutable.Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30)
object ScalaApp extends App { val scores = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // 1.获取指定 key 对应的值 println(scores("hadoop")) // 2. 若是对应的值不存在则使用默认值 println(scores.getOrElse("hadoop01", 100)) }
可变 Map 容许进行新增、修改、删除等操做。git
object ScalaApp extends App { val scores = scala.collection.mutable.Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // 1.若是 key 存在则更新 scores("hadoop") = 100 // 2.若是 key 不存在则新增 scores("flink") = 40 // 3.能够经过 += 来进行多个更新或新增操做 scores += ("spark" -> 200, "hive" -> 50) // 4.能够经过 -= 来移除某个键和值 scores -= "storm" for (elem <- scores) {println(elem)} } // 输出内容以下 (spark,200) (hadoop,100) (flink,40) (hive,50)
不可变 Map 不容许进行新增、修改、删除等操做,可是容许由不可变 Map 产生新的 Map。github
object ScalaApp extends App { val scores = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) val newScores = scores + ("spark" -> 200, "hive" -> 50) for (elem <- scores) {println(elem)} } // 输出内容以下 (hadoop,10) (spark,200) (storm,30) (hive,50)
object ScalaApp extends App { val scores = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // 1. 遍历键 for (key <- scores.keys) { println(key) } // 2. 遍历值 for (value <- scores.values) { println(value) } // 3. 遍历键值对 for ((key, value) <- scores) { println(key + ":" + value) } }
能够使用 yield
关键字从现有 Map 产生新的 Map。编程
object ScalaApp extends App { val scores = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // 1.将 scores 中全部的值扩大 10 倍 val newScore = for ((key, value) <- scores) yield (key, value * 10) for (elem <- newScore) { println(elem) } // 2.将键和值互相调换 val reversalScore: Map[Int, String] = for ((key, value) <- scores) yield (value, key) for (elem <- reversalScore) { println(elem) } } // 输出 (hadoop,100) (spark,200) (storm,300) (10,hadoop) (20,spark) (30,storm)
在使用 Map 时候,若是不指定,默认使用的是 HashMap,若是想要使用 TreeMap
或者 LinkedHashMap
,则须要显式的指定。数组
object ScalaApp extends App { // 1.使用 TreeMap,按照键的字典序进行排序 val scores01 = scala.collection.mutable.TreeMap("B" -> 20, "A" -> 10, "C" -> 30) for (elem <- scores01) {println(elem)} // 2.使用 LinkedHashMap,按照键值对的插入顺序进行排序 val scores02 = scala.collection.mutable.LinkedHashMap("B" -> 20, "A" -> 10, "C" -> 30) for (elem <- scores02) {println(elem)} } // 输出 (A,10) (B,20) (C,30) (B,20) (A,10) (C,30)
object ScalaApp extends App { val scores = scala.collection.mutable.TreeMap("B" -> 20, "A" -> 10, "C" -> 30) // 1. 获取长度 println(scores.size) // 2. 判断是否为空 println(scores.isEmpty) // 3. 判断是否包含特定的 key println(scores.contains("A")) }
import java.util import scala.collection.{JavaConverters, mutable} object ScalaApp extends App { val scores = Map("hadoop" -> 10, "spark" -> 20, "storm" -> 30) // scala map 转 java map val javaMap: util.Map[String, Int] = JavaConverters.mapAsJavaMap(scores) // java map 转 scala map val scalaMap: mutable.Map[String, Int] = JavaConverters.mapAsScalaMap(javaMap) for (elem <- scalaMap) {println(elem)} }
元组与数组相似,可是数组中全部的元素必须是同一种类型,而元组则能够包含不一样类型的元素。oop
scala> val tuple=(1,3.24f,"scala") tuple: (Int, Float, String) = (1,3.24,scala)
能够经过模式匹配来获取元组中的值并赋予对应的变量:大数据
scala> val (a,b,c)=tuple a: Int = 1 b: Float = 3.24 c: String = scala
若是某些位置不须要赋值,则能够使用下划线代替:spa
scala> val (a,_,_)=tuple a: Int = 1
object ScalaApp extends App { val array01 = Array("hadoop", "spark", "storm") val array02 = Array(10, 20, 30) // 1.zip 方法获得的是多个 tuple 组成的数组 val tuples: Array[(String, Int)] = array01.zip(array02) // 2.也能够在 zip 后调用 toMap 方法转换为 Map val map: Map[String, Int] = array01.zip(array02).toMap for (elem <- tuples) { println(elem) } for (elem <- map) {println(elem)} } // 输出 (hadoop,10) (spark,20) (storm,30) (hadoop,10) (spark,20) (storm,30)
更多大数据系列文章能够参见 GitHub 开源项目: 大数据入门指南scala