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scala的多种集合的使用(6)之映射Map的操作方法

發布時間：2023/12/18 编程问答 38 豆豆

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1.創建映射

1)創建不可變的映射

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

2)創建可變的映射，要么用導入的方式將其帶入作用域中，要么就在創建實例時指定scala.collection.mutable.Map類的完整路徑。

scala> var status = collection.mutable.Map((1,"a"),(2,"b")) status: scala.collection.mutable.Map[Int,String] = Map(2 -> b, 1 -> a)

3)在創建的時候創建一空可變的Map，之后在添加元素。

scala> var status = collection.mutable.Map[Int,String]() status: scala.collection.mutable.Map[Int,String] = Map()scala> status += ((1,"a")) res38: scala.collection.mutable.Map[Int,String] = Map(1 -> a)scala> status += ((3,"c"),(2,"b")) res39: scala.collection.mutable.Map[Int,String] = Map(2 -> b, 1 -> a, 3 -> c)

2.Map使用的場景

1)想要返回元素按照鍵有序的映射，請使用SortedMap。

scala> import scala.collection.SortedMap import scala.collection.SortedMapscala> val grades = SortedMap(("kim",90),("al",86),("mes",88),("ema",78),("han",93)) grades: scala.collection.SortedMap[String,Int] = Map(al -> 86, ema -> 78, han ->93, kim -> 90, mes -> 88)scala> val grades = SortedMap((1,90),(3,86),(2,88),(5,78),(4,93)) grades: scala.collection.SortedMap[Int,Int] = Map(1 -> 90, 2 -> 88, 3 -> 86, 4 -> 93, 5 -> 78)

2)按插入時的順序返回元素，只有可變的LinkedHashMap。

scala> import scala.collection.mutable.LinkedHashMap import scala.collection.mutable.LinkedHashMapscala> var status = LinkedHashMap((5,"apple")) status: scala.collection.mutable.LinkedHashMap[Int,String] = Map(5 -> apple)scala> status += ((3,"orange")) res40: scala.collection.mutable.LinkedHashMap[Int,String] = Map(5 -> apple, 3 -> orange)scala> status += ((6,"banana")) res41: scala.collection.mutable.LinkedHashMap[Int,String] = Map(5 -> apple, 3 ->orange, 6 -> banana)

3)按插入時的相反順序返回元素，可以是可變的或者不可變的ListMap。

scala> import scala.collection.mutable.ListMap import scala.collection.mutable.ListMapscala> var status = ListMap((1,"a")) status: scala.collection.mutable.ListMap[Int,String] = Map(1 -> a)scala> status += ((1,"a")) res43: scala.collection.mutable.ListMap[Int,String] = Map(1 -> a)scala> status += ((2,"b")) res44: scala.collection.mutable.ListMap[Int,String] = Map(2 -> b, 1 -> a)scala> status += ((3,"c")) res45: scala.collection.mutable.ListMap[Int,String] = Map(3 -> c, 1 -> a, 2 -> b)

3.可變映射的添加、更新和刪除元素

1)通過給鍵指定值的方式為可變映射添加元素。

scala> var status = scala.collection.mutable.Map[String,String]() status: scala.collection.mutable.Map[String,String] = Map()scala> status("a1") = "a1a"scala> status res47: scala.collection.mutable.Map[String,String] = Map(a1 -> a1a)

2)通過+=方法添加一個或者多個元素。

scala> var status = scala.collection.mutable.Map[String,String]() status: scala.collection.mutable.Map[String,String] = Map()scala> status += (("a1","a1a")) res50: status.type = Map(a1 -> a1a)scala> status res51: scala.collection.mutable.Map[String,String] = Map(a1 -> a1a)scala> status += (("a1","a1a"),("a2","a2a")) res52: status.type = Map(a1 -> a1a, a2 -> a2a)scala> status res53: scala.collection.mutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a)

3)用++=從另一個集合添加多個元素。

4)用-=的方法通過指定元素的鍵從映射中刪除一個或者多個元素。

scala> status ++= List(("a1","a1a"),("a2","a2a")) res56: status.type = Map(a1 -> a1a, a2 -> a2a)scala> status -= "a1" res57: status.type = Map(a2 -> a2a)scala> status res58: scala.collection.mutable.Map[String,String] = Map(a2 -> a2a)scala> status -= ("a1","a2") res60: status.type = Map()scala> status res61: scala.collection.mutable.Map[String,String] = Map()

5)用--=刪除集合里的指定的元素。

scala> var status = scala.collection.mutable.Map[String,String]() status: scala.collection.mutable.Map[String,String] = Map()scala> status ++= List(("a1","a1a"),("a2","a2a")) res67: status.type = Map(a1 -> a1a, a2 -> a2a)scala> status --= List("a1","a2") res68: status.type = Map()scala> status res69: scala.collection.mutable.Map[String,String] = Map()

6)通過賦值值給元素的鍵更新元素。

scala> var status = scala.collection.mutable.Map[String,String]() status: scala.collection.mutable.Map[String,String] = Map()scala> status ++= List(("a1","a1a"),("a2","a2a")) res72: status.type = Map(a1 -> a1a, a2 -> a2a)scala> status("a1") = "hello world"scala> status res74: scala.collection.mutable.Map[String,String] = Map(a1 -> hello world, a2 -> a2a)

4.不可變映射的添加、更新和刪除元素

1)用+個方法添加一個或者多個元素，在這個過程中將結果賦給一個新的變量。

scala> val a = Map(("a1","a1a")) a: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a)scala> val b = a + (("a2","a2a")) b: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a)scala> val c = b + (("a3","a31"),("a4","a4a")) c: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a31, a4 -> a4a)

2)更新一個不可變映射的鍵值對，需要用+方法對鍵/值重新賦值，新值替換舊值。

scala> val a = Map(("a1","a1a")) a: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a)scala> val b = a + (("a2","a2a")) b: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a)scala> val c = b + (("a1","hello world")) c: scala.collection.immutable.Map[String,String] = Map(a1 -> hello world, a2 -> a2a)

3)使用-方法刪除一個或者多個元素。

scala> val a = Map(("a1","a1a")) a: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a)scala> val b = a + (("a2","a2a"),("a3","a3a"),("a4","a4a")) b: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a3a, a4 -> a4a)scala> val c = b - "a1" - "a2" c: scala.collection.immutable.Map[String,String] = Map(a3 -> a3a, a4 -> a4a)scala> val d = c - "a4" d: scala.collection.immutable.Map[String,String] = Map(a3 -> a3a)

當一個不可變的變量聲明為var時，它仍然是一個不可變的映射，不能給映射中的元素重新賦值。

5.映射值的訪問

1)訪問保存在映射中單獨的值，如果鍵不存在，會拋出異常。為了避免這個問題，可以在創建映射時使用withDefaultValue的方法。該方法會創建一個默認值，如果鍵沒有找到，映射會返回這個值。

scala> val status = Map((1,"a"),(2,"b"),(3,"c")).withDefaultValue("Not Found") status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)scala> status(4) res5: String = Not Foundscala> status(3) res6: String = c

2)尋找鍵時可以使用getOrElse方法，當指定的鍵找不到時，會返回默認值。

scala> val status = Map((1,"a"),(2,"b"),(3,"c")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)scala> val s = status.getOrElse(6,"Not such value") s: String = Not such value

3)可以使用get方法返回Option對象。

scala> val status = Map((1,"a"),(2,"b"),(3,"c")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c)scala> val s = status.get(5) s: Option[String] = Nonescala> val s = status.get(2) s: Option[String] = Some(b)

6.映射的遍歷

1)for循環遍歷所有的映射元素。

scala> val status = Map((1,"a"),(2,"b"),(3,"c"),(4,"d")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4 -> d)scala> for((k,v) <- status) println(s"key: $k, value: $v") key: 1, value: a key: 2, value: b key: 3, value: c key: 4, value: d

2)匹配表達式配合foreach方法。

scala> val status = Map((1,"a"),(2,"b"),(3,"c"),(4,"d")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4 -> d)scala> status.foreach{| case(k,v) => println(s"key: $k,value: $v")| } key: 1,value: a key: 2,value: b key: 3,value: c key: 4,value: d

3)使用Tuple語法訪問鍵/值字段。

scala> val status = Map((1,"a"),(2,"b"),(3,"c"),(4,"d")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4 -> d)scala> status.foreach(x => println(s"key: ${x._1},value: ${x._2}")) key: 1,value: a key: 2,value: b key: 3,value: c key: 4,value: d

4)如果想要映射中所有的鍵，keys方法返回Iterable。

scala> val status = Map((1,"a"),(2,"b"),(3,"c"),(4,"d")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4 -> d)scala> status.keys.foreach((key) => println(key)) 1 2 3 4

5)如果想要映射中所有的value的值，用values方法可以遍歷映射中所有的值。

scala> val status = Map((1,"a"),(2,"b"),(3,"c"),(4,"d")) status: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4 -> d)scala> status.values.foreach((value) => println(value)) a b c d

7.操作映射的值

1)如果要遍歷映射并對每個值進行操作，mapValues是個不錯的選擇。它可以在每個映射值上執行一個函數，然后返回修改后的映射。

scala> var x = collection.mutable.Map((1,"a"),(2,"b")) x: scala.collection.mutable.Map[Int,String] = Map(2 -> b, 1 -> a)scala> val y = x.mapValues(_.toUpperCase) y: scala.collection.Map[Int,String] = Map(2 -> B, 1 -> A)

2)transform方法可以同時使用鍵/值實現一個變換方法。

scala> val map = Map((1,10),(2,20),(3,30)) map: scala.collection.immutable.Map[Int,Int] = Map(1 -> 10, 2 -> 20, 3 -> 30)scala> val newMap = map.transform((k,v) => k + v) newMap: scala.collection.immutable.Map[Int,Int] = Map(1 -> 11, 2 -> 22, 3 -> 33)

8.從映射中獲取所有的鍵和值

1)用keySet的方法將以集合的方式獲取所有的鍵。

scala> val status = Map(("a1","a1a"),("a2","a2a"),("a3","a3a")) status: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a3a)scala> status.keySet res14: scala.collection.immutable.Set[String] = Set(a1, a2, a3)

2)用keys方法獲得一個Iterable。

scala> status.keys res16: Iterable[String] = Set(a1, a2, a3)

3)用keysIterator方法獲取所有作為迭代器的鍵。

scala> status.keysIterator res17: Iterator[String] = non-empty iterator

4)使用values方法可以獲取映射中所有的值，并將結果轉化為Iterable對象。

scala> status.values res18: Iterable[String] = MapLike(a1a, a2a, a3a)

5)使用valuesIterator方法，返回Iterator對象。

scala> status.valuesIterator res19: Iterator[String] = non-empty iterator

keysIterator和valuesIterator方法都會從映射數據返回一個迭代器。這些方法不會創建一個新的集合，僅僅是提供遍歷已存在的迭代器。

9.反轉鍵值

可以用for推導反轉映射的鍵值，然后將結果賦給一個新的變量。但是映射中，key是不重復的，value是可以重復的，但是在反轉時可能會丟掉數據。

scala> val status = Map(("a1","a1a"),("a2","a2a"),("a3","a3a")) status: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a3a)scala> val newMap = for((k,v) <- status) yield (v,k) newMap: scala.collection.immutable.Map[String,String] = Map(a1a -> a1, a2a -> a2, a3a -> a3)

10.測試映射中鍵/值的存在

1)用contains方法測試映射中是否包含鍵。

scala> val status = Map(("a1","a1a"),("a2","a2a"),("a3","a3a")) status: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a3a)scala> if (status.contains("a1")) println("found a1") else println("not found") found a1scala> if (status.contains("a5")) println("found a5") else println("not found") not found

2)使用valuesIterator方法搜索值，結合exists和contains。

scala> val status = Map(("a1","a1a"),("a2","a2a"),("a3","a3a")) status: scala.collection.immutable.Map[String,String] = Map(a1 -> a1a, a2 -> a2a, a3 -> a3a)scala> status.valuesIterator.exists(_.contains("a2a")) res22: Boolean = truescala> status.valuesIterator.exists(_.contains("a2a2")) res23: Boolean = false

11.根據鍵或者值對映射排序

1)可以用sortBy方法對值(_2)/鍵(_1)從低到高進對映射進行排序。

scala> val grade = Map((1,98),(2,89),(3,88),(4,93),(5,95)) grade: scala.collection.immutable.Map[Int,Int] = Map(5 -> 95, 1 -> 98, 2 -> 89, 3 -> 88, 4 -> 93)scala> import scala.collection.immutable.ListMap import scala.collection.immutable.ListMapscala> ListMap(grade.toSeq.sortBy(_._2):_*) res25: scala.collection.immutable.ListMap[Int,Int] = Map(3 -> 88, 2 -> 89, 4 -> 93, 5 -> 95, 1 -> 98)

2)可以用sortWith方法對值(_2)/鍵(_1)按照升序或者降序對值進行排序。

//按照值升序: scala> ListMap(grade.toSeq.sortWith(_._2 < _._2):_*) res28: scala.collection.immutable.ListMap[Int,Int] = Map(3 -> 88, 2 -> 89, 4 ->93, 5 -> 95, 1 -> 98) //按照值降序: scala> ListMap(grade.toSeq.sortWith(_._2 > _._2):_*) res29: scala.collection.immutable.ListMap[Int,Int] = Map(1 -> 98, 5 -> 95, 4 ->93, 2 -> 89, 3 -> 88)

關于_*:
它的作用是將數據轉換，然后將其作為多個參數傳給ListMap。

12.映射中鍵值的最值

1)對鍵進行排序。

scala> val grade = Map((1,98),(2,89),(3,88),(4,93),(5,95)) grade: scala.collection.immutable.Map[Int,Int] = Map(5 -> 95, 1 -> 98, 2 -3 -> 88, 4 -> 93)scala> grade.max res31: (Int, Int) = (5,95)scala> grade.keysIterator.max res32: Int = 5scala> grade.keysIterator.reduceLeft((x,y) => if(x > y) x else y) res33: Int = 5

2)對值進行排序?。

scala> val grade = Map((1,98),(2,89),(3,88),(4,93),(5,95)) grade: scala.collection.immutable.Map[Int,Int] = Map(5 -> 95, 1 -> 98, 2 -> 89,3 -> 88, 4 -> 93)scala> grade.valuesIterator.max res40: Int = 98scala> grade.valuesIterator.reduceLeft((x,y) => if(x > y) x else y) res41: Int = 98scala> grade.valuesIterator.reduceLeft(_ max _) res42: Int = 98

轉載于:https://www.cnblogs.com/zhaohadoopone/p/9534242.html

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