c += x:把x元素添加到集合c中spa
scala> val a = collection.mutable.ArrayBuffer(1,2,3) a: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3) scala> a += 4 res15: a.type = ArrayBuffer(1, 2, 3, 4)
c += (x,y,z):把x,y,z添加到集合c中scala
scala> val a = collection.mutable.ArrayBuffer(1,2,3) a: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3) scala> a += (4,5,6) res16: a.type = ArrayBuffer(1, 2, 3, 4, 5, 6)
c1 ++= c2:把集合c2的元素所有添加到集合c1中code
scala> val c1 = mutable.ArrayBuffer(1,2,3) c1: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3) scala> val c2 = mutable.ArrayBuffer(4,5,6) c2: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(4, 5, 6) scala> c1 ++= c2 res20: c1.type = ArrayBuffer(1, 2, 3, 4, 5, 6)
c -= x:在集合c中刪除元素xorm
scala> val c = mutable.ArrayBuffer(1,2,3) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3) scala> c -= 3 res21: c.type = ArrayBuffer(1, 2) scala> c -= 7 res22: c.type = ArrayBuffer(1, 2)
c -= (x,y,z):在集合c中刪除元素x,y,zrem
scala> val c = mutable.ArrayBuffer(1,2,3,4,5) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> c -= (3,4) res23: c.type = ArrayBuffer(1, 2, 5)
c1 --= c2:刪除集合c1中全部在c2中存在的元素it
scala> val c1 = mutable.ArrayBuffer(1,2,3,4,5) c1: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> val c2 = mutable.ArrayBuffer(2,4) c2: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(2, 4) scala> c1 --= c2 res24: c1.type = ArrayBuffer(1, 3, 5)
c(n) = x:把集合c中第n個元素賦值爲xio
scala> val c = mutable.ArrayBuffer(1,2,3,4,5) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> c(2) = 8 scala> c res26: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 8, 4, 5)
c clear:刪除集合c中的全部元素table
scala> val c = mutable.ArrayBuffer(1,2,3,4,5) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> c.clear scala> c res28: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer()
c remove(n):在集合c中刪除第n個位置的元素,並返回該元素class
scala> val c = mutable.ArrayBuffer(1,2,3,4,5) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> c.remove(3) res29: Int = 4 scala> c res30: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 5)
c remove(n,len):在集合c的第n個元素開始,刪除len個元素容器
scala> val c = mutable.ArrayBuffer(1,2,3,4,5) c: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> c.remove(1,2) scala> c res32: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 4, 5)
下面咱們來展現不可變容器的通用操做方法。不可變容器內容是不可變的,因此制定一個新的變量來接受這些方法的返回值。
c1 ++ c2:返回集合c1和集合c2的並集
scala> val v1 = Vector(1,2,3) v1: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3) scala> val v2 = Vector(4,5,6) v2: scala.collection.immutable.Vector[Int] = Vector(4, 5, 6) scala> val v = v1 ++ v2 v: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5, 6)
c :+ e:返回一個新的集合,內容爲集合c添加元素e造成的新集合
scala> Vector(1,2,3) :+ 4 res34: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4)
e +: c:返回一個新的集合,內容爲元素e追加到集合c前面造成的新集合
scala> 4 +: Vector(3,2,1) res35: scala.collection.immutable.Vector[Int] = Vector(4, 3, 2, 1)
e :: List:返回新的List,內容爲元素e追加到List集合前造成的新List
scala> 4 :: List(3,2,1) res37: List[Int] = List(4, 3, 2, 1)