scala> val numbers = List(1, 2, 3, 4)
numbers: List[Int] = List(1, 2, 3, 4)
scala> Set(1, 1, 2)
res0: scala.collection.immutable.Set[Int] = Set(1, 2)
元组是在不使用类的前提下,将元素组合起来形成简单的逻辑集合。
scala> val hostPort = ("localhost", 80)
hostPort: (String, Int) = (localhost, 80)
scala> hostPort._1
res0: String = localhost
scala> hostPort._2
res1: Int = 80
Map(1 -> 2)
Map("foo" -> "bar")
Option 是一个表示有可能包含值的容器。 Option 本身是泛型的,并且有两个子类: Some[T] 或 None
scala> val numbers = Map("one" -> 1, "two" -> 2)
numbers: scala.collection.immutable.Map[java.lang.String,Int] = Map(one -> 1, two -> 2)
scala> numbers.get("two")
res0: Option[Int] = Some(2)
scala> numbers.get("three")
res1: Option[Int] = None
// We want to multiply the number by two, otherwise return 0.
val result = if (res1.isDefined) {
res1.get * 2
} else {
0
}
scala> numbers.map((i: Int) => i * 2)
res0: List[Int] = List(2, 4, 6, 8)
//或传入一个部分应用函数
scala> def timesTwo(i: Int): Int = i * 2
timesTwo: (i: Int)Int
scala> numbers.map(timesTwo _)
res0: List[Int] = List(2, 4, 6, 8)
scala> numbers.foreach((i: Int) => i * 2)
scala> numbers.filter((i: Int) => i % 2 == 0)
res0: List[Int] = List(2, 4)
scala> def isEven(i: Int): Boolean = i % 2 == 0
isEven: (i: Int)Boolean
scala> numbers.filter(isEven _)
res2: List[Int] = List(2, 4)
zip 将两个列表的内容聚合到一个对偶列表中
scala> List(1, 2, 3).zip(List("a", "b", "c"))
res0: List[(Int, String)] = List((1,a), (2,b), (3,c))
partition 将使用给定的谓词函数分割列表。
scala> val numbers = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> numbers.partition(_ % 2 == 0)
res0: (List[Int], List[Int]) = (List(2, 4, 6, 8, 10),List(1, 3, 5, 7, 9))
scala> numbers.find((i: Int) => i > 5)
res0: Option[Int] = Some(6)
dropWhile 将删除元素直到找到第一个匹配谓词函数的元素。例如,如果我们在 numbers 列表上使用 dropWhile 奇数的函数, 1 将被丢弃(但 3 不会被丢弃,因为他被 2 “保护”了)。
scala> numbers.drop(5)
res0: List[Int] = List(6, 7, 8, 9, 10)
scala> numbers.dropWhile(_ % 2 != 0)
res0: List[Int] = List(2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> numbers.foldLeft(0)((m: Int, n: Int) => m + n)
res0: Int = 55
scala> numbers.foldLeft(0) { (m: Int, n: Int) => println("m: " + m + " n: " + n); m + n }
m: 0 n: 1
m: 1 n: 2
m: 3 n: 3
m: 6 n: 4
m: 10 n: 5
m: 15 n: 6
m: 21 n: 7
m: 28 n: 8
m: 36 n: 9
m: 45 n: 10
res0: Int = 55
flatten 将嵌套结构扁平化为一个层次的集合。
scala> List(List(1, 2), List(3, 4)).flatten
res0: List[Int] = List(1, 2, 3, 4)
flatMap 是一种常用的组合子,结合映射 [mapping] 和扁平化 [flattening]。flatMap 需要一个处理嵌套列表的函数,然后将结果串连起来
scala> val nestedNumbers = List(List(1, 2), List(3, 4))
nestedNumbers: List[List[Int]] = List(List(1, 2), List(3, 4))
scala> nestedNumbers.flatMap(x => x.map(_ * 2))
res0: List[Int] = List(2, 4, 6, 8)