Scala过滤一个序列以映射符合和不符合过滤条件的元素

Question

我有一个字符串序列：

val results = Seq("", "one", "two", "three")

使用两个过滤器，我得到：

val emptyStrings: Seq[String] = results.filter( s => s.isEmpty )
val notEmptyStrings: Seq[String] = results.filterNot( s => s.isEmpty )

是否可以使用一个过滤器 a Map[Boolean, Seq[String]]：

• 键 = isEmpty 真/假
• values = 空或非空字符串

Answer 1

$ scala
Welcome to Scala 2.13.0 (OpenJDK 64-Bit Server VM, Java 1.8.0_262).
Type in expressions for evaluation. Or try :help.

scala> val results = Seq("", "one", "two", "three")
results: Seq[String] = List("", one, two, three)

scala> results.groupBy(_.isEmpty)
res0: scala.collection.immutable.Map[Boolean,Seq[String]] = HashMap(false -> List(one, two, three), true -> List(""))

更好的选择是使用partitiondue 它返回一个具有 2 个值的元组而不是一个映射：

scala> results.partition(_.isEmpty)
res1: (Seq[String], Seq[String]) = (List(""),List(one, two, three))

甚至更好地将结果与命名值进行模式匹配：

scala> val (emptyStrings, nonEmptyStrings) = results.partition(_.isEmpty)
emptyStrings: Seq[String] = List("")
nonEmptyStrings: Seq[String] = List(one, two, three)