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scala的多种集合的使用(7)之集Set的操作方法

1.给集添加元素

1)用+=、++=和add给可变集添加元素。

scala> var set = scala.collection.mutable.Set[Int]()set: scala.collection.mutable.Set[Int] = Set()scala> set += 1res48: scala.collection.mutable.Set[Int] = Set(1)scala> set += (2,3)res49: scala.collection.mutable.Set[Int] = Set(1, 2, 3)scala> set ++= Vector(4,5)res50: scala.collection.mutable.Set[Int] = Set(1, 5, 2, 3, 4)scala> set.add(6)res51: Boolean = truescala> setres52: scala.collection.mutable.Set[Int] = Set(1, 5, 2, 6, 3, 4)scala> set.add(5)res53: Boolean = false

2)使用+和++方法,通过向前一个集添加元素生成一个新的集。

scala> val set = Set(1,3,5,2,7)set: scala.collection.immutable.Set[Int] = Set(5, 1, 2, 7, 3)scala> val set1 = set + (8,9)set1: scala.collection.immutable.Set[Int] = Set(5, 1, 9, 2, 7, 3, 8)scala> val set2 = set1 ++ List(10,11)set2: scala.collection.immutable.Set[Int] = Set(5, 10, 1, 9, 2, 7, 3, 11, 8)

2.从集中删除元素

1)处理可变集时,用-=和--=从集中删除元素。

scala> var set = scala.collection.mutable.Set(1,2,3,4,5)set: scala.collection.mutable.Set[Int] = Set(1, 5, 2, 3, 4)scala> set -= 1res57: scala.collection.mutable.Set[Int] = Set(5, 2, 3, 4)scala> set -= (2,3)res58: scala.collection.mutable.Set[Int] = Set(5, 4)scala> set --= Array(4,5)res59: scala.collection.mutable.Set[Int] = Set()

2)处理可变集,retain和clear删除集中元素。

scala> var set = scala.collection.mutable.Set(1,2,3,4,5)set: scala.collection.mutable.Set[Int] = Set(1, 5, 2, 3, 4)scala> set.retain(_ > 2);println(set)Set(5, 3, 4)scala> set.clear;println(set)Set()

3)处理可变集时,remove的返回值可以提示集是否有元素被删除。

scala> var set = scala.collection.mutable.Set(1,2,3,4,5)set: scala.collection.mutable.Set[Int] = Set(1, 5, 2, 3, 4)scala> set.remove(2);println(set)Set(1, 5, 3, 4)scala> set.remove(3)res64: Boolean = truescala> setres65: scala.collection.mutable.Set[Int] = Set(1, 5, 4)

4)处理不可变集时,可以使用-和--操作符删除元素,同时将结果重新赋给一个新的变量。

scala> val s1 = Set(1,2,3,4,5)s1: scala.collection.immutable.Set[Int] = Set(5, 1, 2, 3, 4)scala> val s2 = s1 -1s2: scala.collection.immutable.Set[Int] = Set(5, 2, 3, 4)scala> val s3 = s2 - (2,3)s3: scala.collection.immutable.Set[Int] = Set(5, 4)scala> val s4 = s3 -- Array(4,5)s4: scala.collection.immutable.Set[Int] = Set()

3.使用可排序集

1)SortedSet返回元素时有序的。

scala> val s = scala.collection.SortedSet(3,1,2,5,6,9)s: scala.collection.SortedSet[Int] = TreeSet(1, 2, 3, 5, 6, 9)scala> val s = scala.collection.SortedSet("c","g","a","b")s: scala.collection.SortedSet[String] = TreeSet(a, b, c, g)

2)LinkedHashSet按照插入顺序保存元素的。

scala> val s = scala.collection.mutable.LinkedHashSet(10,8,3,5,7)s: scala.collection.mutable.LinkedHashSet[Int] = Set(10, 8, 3, 5, 7)

  

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