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scala的多种集合的使用(6)之映射Map的操作方法

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1.创建映射

1)创建不可变的映射 

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)创建可变的映射，要么用导入的方式将其带入作用域中，要么就在创建实例时指定scala.collection.mutable.Map类的完整路径。 

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

3)在创建的时候创建一空可变的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使用的场景

1)想要返回元素按照键有序的映射，请使用SortedMap。 

scala> import scala.collection.SortedMap
import scala.collection.SortedMap

scala> 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)按插入时的顺序返回元素，只有可变的LinkedHashMap。 

scala> import scala.collection.mutable.LinkedHashMap
import scala.collection.mutable.LinkedHashMap

scala> 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)按插入时的相反顺序返回元素，可以是可变的或者不可变的ListMap。 

scala> import scala.collection.mutable.ListMap
import scala.collection.mutable.ListMap

scala> 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.可变映射的添加、更新和删除元素

1)通过给键指定值的方式为可变映射添加元素。 

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)通过+=方法添加一个或者多个元素。 

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++=从另一个集合添加多个元素。 

scala> var status = scala.collection.mutable.Map[String,String]()
status: scala.collection.mutable.Map[String,String] = Map()

scala> status ++= List(("a1","a1a"),("a2","a2a"))
res55: status.type = Map(a1 -> a1a, a2 -> a2a)

4)用-=的方法通过指定元素的键从映射中删除一个或者多个元素。 

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--=删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)通过赋值值给元素的键更新元素。 

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.不可变映射的添加、更新和删除元素

1)用+个方法添加一个或者多个元素，在这个过程中将结果赋给一个新的变量。 

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)更新一个不可变映射的键值对，需要用+方法对键/值重新赋值，新值替换旧值。 

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-方法删除一个或者多个元素。 

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)

当一个不可变的变量声明为var时，它仍然是一个不可变的映射，不能给映射中的元素重新赋值。

5.映射值的访问

1)访问保存在映射中单独的值，如果键不存在，会抛出异常。为了避免这个问题，可以在创建映射时使用withDefaultValue的方法。该方法会创建一个默认值，如果键没有找到，映射会返回这个值。 

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 Found

scala> status(3)
res6: String = c

2)寻找键时可以使用getOrElse方法，当指定的键找不到时，会返回默认值。 

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对象。 

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] = None

scala> val s = status.get(2)
s: Option[String] = Some(b)

6.映射的遍历

1)for循环遍历所有的映射元素。 

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)匹配表达式配合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语法访问键/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键，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方法可以遍历映射中所有的值。 

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)如果要遍历映射并对每个值进行操作，mapValues是个不错的选择。它可以在每个映射值上执行一个函数，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方法可以同时使用键/值实现一个变换方法。 

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.从映射中获取所有的键和值

1)用keySet的方法将以集合的方式获取所有的键。 

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方法获得一个Iterable。 

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

3)用keysIterator方法获取所有作为迭代器的键。 

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

4)使用values方法可以获取映射中所有的值，并将结果转化为Iterable对象。 

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

5)使用valuesIterator方法，返回Iterator对象。 

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

keysIterator和valuesIterator方法都会从映射数据返回一个迭代器。这些方法不会创建一个新的集合，仅仅是提供遍历已存在的迭代器。

9.反转键值

可以用for推导反转映射的键值，然后将结果赋给一个新的变量。但是映射中，key是不重复的，value是可以重复的，但是在反转时可能会丢掉数据。 

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.测试映射中键/值的存在

1)用contains方法测试映射中是否包含键。 

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 a1

scala> if (status.contains("a5")) println("found a5") else println("not found")
not found

2)使用valuesIterator方法搜索值，结合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 = true

scala> status.valuesIterator.exists(_.contains("a2a2"))
res23: Boolean = false

11.根据键或者值对映射排序

1)可以用sortBy方法对值(_2)/键(_1)从低到高进对映射进行排序。 

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.ListMap

scala> 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方法对值(_2)/键(_1)按照升序或者降序对值进行排序。 

//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)

关于_*:
它的作用是将数据转换，然后将其作为多个参数传给ListMap。

12.映射中键值的最值

1)对键进行排序。 

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 = 5

scala> grade.keysIterator.reduceLeft((x,y) => if(x > y) x else y)
res33: Int = 5

2)对值进行排序 。 

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 = 98

scala> grade.valuesIterator.reduceLeft((x,y) => if(x > y) x else y)
res41: Int = 98

scala> grade.valuesIterator.reduceLeft(_ max _)
res42: Int = 98

 

原文：https://www.cnblogs.com/zhaohadoopone/p/9534242.html

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