scala的多種集合的使用(2)之集合常用方法

一、常用的集合方法

1.可遍歷集合的常用方法

下表列出了Traverable在所有集合常用的方法。接下來的符號：

• c代表一個集合

• f代表一個函數

• p代表一個謂詞

• n代表一個數字

• op代表一個簡單的操作(通常是一個簡單的函數)

 

2.可變的集合方法

下表裏的是可變集合常用的操作符：

3.不可變的集合方法

注意不可變集合不可被修改，所以在第一行的每個表達式的結果必須賦給一個新變量。

操作符	描述
c1 ++ c2	把c2集合的元素附加到c1裏創造一個新的集合

 

 

 

scala> val c1 = Vector(1,2,3,4,5)
c1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

scala> val c2 = Vector(6,7,8)
c2: scala.collection.immutable.Vector[Int] = Vector(6, 7, 8)

scala> c1 ++ c2
res35: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5, 6, 7, 8)

---

 

操作符	描述
c :+ e	返回把元素e附加到集合c的新集合

 

 

 

scala> val v1 = Vector(1,2,3,4,5)
c1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

scala> val e = 10
e: Int = 10

scala> v1 :+ e
res38: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5, 10)

---

 

操作符	描述
e +: c2	返回一個把元素e前插到集合c的新集合

 

 

 

scala> val v1 = Vector(1,2,3,4,5)
c1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

scala> val e = 10
e: Int = 10

scala> e +: v1
res39: scala.collection.immutable.Vector[Int] = Vector(10, 1, 2, 3, 4, 5)

---

 

操作符	描述
e :: list	返回一個把e前插到名爲list列表的集合(::只在List集合上用)

 

 

 

scala> val list = List(7,8,9)
list: List[Int] = List(7, 8, 9)

scala> val e = 10
e: Int = 10

scala> e :: list
res75: List[Int] = List(10, 7, 8, 9)

---

 

操作符	描述
c drop n	把c2集合的元素附加到c1裏創造一個新的集合

 

 

 

scala> val v1 = Vector(1,2,3,4,5)
v1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

scala> val n = 3
n: Int = 3

scala> v1 drop n
res1: scala.collection.immutable.Vector[Int] = Vector(4, 5)

---

 

操作符	描述
c1 take n	把集合的前兩個元素列出來

 

 

 

scala> val c1 = Vector(1,2,3,4,5)
c1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

scala> c1 take 2
res82: scala.collection.immutable.Vector[Int] = Vector(1, 2)

 4.可變和不可變的map的常用方法

• m代表一個不可變map

• mm代表一個可變的map

• k代表一個鍵

• v代表一個值

• p代表一個謂詞

• c代表一個集合

如果是var修飾，引用可變，支持讀寫。
如果是val修飾，引用不可變，支持讀寫。

操作符	描述
m - k	返回刪除key(以及相應的value的值)後的map

 

 

 

scala> val v1 = Vector(1,2,3,4,5)
v1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

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

scala> map - 1
res5: scala.collection.immutable.Map[Int,String] = Map(2 -> b, 3 -> c)

---

 

操作符	描述
m - (k1,k2,k3)	返回刪除k1,k2,k3後的map

 

 

 

scala> val map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d",5 -> "e",6 -> "f")
map: scala.collection.immutable.Map[Int,String] = Map(5 -> e, 1 -> a, 6 -> f, 2-> b, 3 -> c, 4 -> d)

scala> map - (1,3,5)
res6: scala.collection.immutable.Map[Int,String] = Map(6 -> f, 2 -> b, 4 -> d)

---

 

操作符	描述
m -- c	返回刪除key後的map，c是指集合

 

 

 

scala> val map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d",5 -> "e",6 -> "f")
map: scala.collection.immutable.Map[Int,String] = Map(5 -> e, 1 -> a, 6 -> f, 2-> b, 3 -> c, 4 -> d)

scala> val c = Vector(1,3,5)
c: scala.collection.immutable.Vector[Int] = Vector(1, 3, 5)

scala> map -- c
res7: scala.collection.immutable.Map[Int,String] = Map(6 -> f, 2 -> b, 4 -> d)

---

 

操作符	描述
m -- List(k1,k2)	返回刪除key後的map

 

 

 

1
scala> val map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d",5 -> "e",6 -> "f")
map: scala.collection.immutable.Map[Int,String] = Map(5 -> e, 1 -> a, 6 -> f, 2-> b, 3 -> c, 4 -> d)

scala> val list = List(1,3,5)
list: List[Int] = List(1, 3, 5)

scala> map -- list
res8: scala.collection.immutable.Map[Int,String] = Map(6 -> f, 2 -> b, 4 -> d)
2
scala> val map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d",5 -> "e",6 -> "f")
map: scala.collection.immutable.Map[Int,String] = Map(5 -> e, 1 -> a, 6 -> f, 2
-> b, 3 -> c, 4 -> d)

scala> map -- List(1,2,3)
res9: scala.collection.immutable.Map[Int,String] = Map(5 -> e, 6 -> f, 4 -> d)

---

 

操作符	描述
m(k)	返回k的value

 

 

 

scala> var map = Map(1 -> 1,2 -> 2,3 -> 3,4 -> 4)
map: scala.collection.immutable.Map[Int,Int] = Map(1 -> 1, 2 -> 2, 3 -> 3, 4 ->4)

scala> map(2)
res34: Int = 2

---

 

操作符	描述
m contains k	如果map包含k則返回true

 

 

 

scala> var map = Map(1 -> 1,2 -> 2,3 -> 3,4 -> 4)
map: scala.collection.immutable.Map[Int,Int] = Map(1 -> 1, 2 -> 2, 3 -> 3, 4 ->4)

scala> map contains 3
res36: Boolean = true

---

 

操作符	描述
m get k	如果k存在返回Some[A]作爲key的值，否則None

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map get 2
res44: Option[String] = Some(b)

scala> map get 5
res45: Option[String] = None

---

 

操作符	描述
m getOrElse(k,d)	如果k找到的話，返回k的值，否則返回默認值d

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map getOrElse(2,"hello")
res46: String = b

scala> map getOrElse(5,"hello")
res47: String = hello

---

 

操作符	描述
m isDefinedAt k	如果map包含k則返回true

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map isDefinedAt 2
res48: Boolean = true

scala> map isDefinedAt 5
res49: Boolean = false

---

 

操作符	描述
m keys	把map中的keys作爲Iterable返回

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map keys
warning: there was one feature warning; re-run with -feature for details
res50: Iterable[Int] = Set(1, 2, 3, 4)

---

 

操作符	描述
m keySet	把map中的keys作爲Set返回

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map keySet
warning: there was one feature warning; re-run with -feature for details
res54: scala.collection.immutable.Set[Int] = Set(1, 2, 3, 4)

---

 

操作符	描述
m values	把map中的value作爲Iterable返回

 

 

 

scala> var map = Map(1 -> "a",2 -> "b",3 -> "c",4 -> "d")
map: scala.collection.immutable.Map[Int,String] = Map(1 -> a, 2 -> b, 3 -> c, 4-> d)

scala> map values
warning: there was one feature warning; re-run with -feature for details
res56: Iterable[String] = MapLike(a, b, c, d)

二、在創建集合時聲明一個類型

想要創建一個混合型的集合，scala不會自動分配想要的類型。

創建一個手動的集合如下：

scala> trait Animal
defined trait Animal

scala> trait FurryAnimal extends Animal
defined trait FurryAnimal

scala> case class Dog(name: String) extends Animal
defined class Dog

scala> case class Cat(name: String) extends Animal
defined class Cat

scala> val x = Array(Dog("Fido"),Cat("Felix"))
x: Array[Product with Serializable with Animal] = Array(Dog(Fido), Cat(Felix))

如上所示，scala給product指定了Serializable和Animal類型。如果想要一個Array[Animal],手動需要指定類型：

scala> val x = Array[Animal](Dog("Fido"),Cat("Felix"))
x: Array[Animal] = Array(Dog(Fido), Cat(Felix))

三、可變變量集合與不可變量集合

但把一個不可變的Vector賦給一個可變的var時，似乎可以給它添加元素。

scala> var sisters = Vector("Melinda")
sisters: scala.collection.immutable.Vector[String] = Vector(Melinda)

scala> sisters = sisters :+ "Melissa"
sisters: scala.collection.immutable.Vector[String] = Vector(Melinda, Melissa)

scala> sisters = sisters :+ "Marisa"
sisters: scala.collection.immutable.Vector[String] = Vector(Melinda, Melissa, Ma
risa)

scala> sisters.foreach(println)
Melinda
Melissa
Marisa

儘管你看起來改變了一個不可改變的集合，但實際上是每次用:+方法時sisters變量會指向一個新的集合。sisters變量是可變的，所以每次執行時相當於重新賦值一個新的集合。

總結：

• 一個可變的變量var可以被賦予一個新的數據。
• 一個不可變的變量val就如java中的final變量，不可能被重新賦值。
• 在可變集合的元素中可以被改變（ArrayBuffer）。
• 在不可被改變的集合元素中不可被改變（Vector）。