Question

假设我有一个元组列表：

val xs: List[(Seq[String], Option[String])] = List(
  (Seq("Scala", "Python", "Javascript"), Some("Java")),
  (Seq("Wine", "Beer"), Some("Beer")),
  (Seq("Dog", "Cat", "Man"), None)
  )

和一个函数，如果它在字符串序列中，则返回字符串的索引：

def getIndex(s: Seq[String], e: Option[String]): Option[Int] = 
  if (e.isEmpty) None
  else Some(s.indexOf(e.get))

现在我尝试使用xs映射getIndex并仅返回我找到有效索引的那些。一种方法如下：

xs.map{case (s, e) => {
  val ii = getIndex(s, e) // returns an Option
  ii match {  // unpack the option
    case Some(idx) => (e, idx)
    case None => (e, -1) // give None entries a placeholder with -1
  }
}}.filter(_._2 != -1) // filter out invalid entries

这种方法对我来说非常冗长和笨拙。 flatMap在这里不起作用，因为我返回一个元组而不仅仅是索引。这样做的惯用方法是什么？

Answer 1

for comprehension是达到此目的的一种方法：

scala> val xs: List[(Seq[String], Option[String])] = List(
  (Seq("Scala", "Python", "Javascript"), Some("Java")),
  (Seq("Wine", "Beer"), Some("Beer")),
  (Seq("Dog", "Cat", "Man"), None)
)
xs: List[(Seq[String], Option[String])] = List((List(Scala, Python, Javascript),Some(Java)), (List(Wine, Beer),Some(Beer)), (List(Dog, Cat, Man),None))

scala> def getIndex(seq: Seq[String], e: Option[String]): Option[Int] =
  e.map(seq.indexOf(_)).filter(_ != -1) // notice we're doing the filter here
getIndex: getIndex[](val seq: Seq[String],val e: Option[String]) => Option[Int]

scala> for {
  (seq, string) <- xs
  index <- getIndex(seq, string)
  s <- string
} yield (s, index)
res0: List[(String, Int)] = List((Beer,1))

Answer 2

有很多方法可以做到这一点。其中之一是：

val result = xs.flatMap { tuple =>
  val (seq, string) = tuple
  string.map(s => (s, seq.indexOf(s))).filter(_._2 >= 0)
}

Answer 3

也许这看起来更像惯用语：

#!/usr/bin/osascript

set output to "Title,Notes,Completed,Completion Date,List,Creation Date,Due Date,Modification Date,Remind Me Date,Priority\n"

tell application "Reminders"
    set rs to reminders in application "Reminders"
    repeat with r in rs
        set output to output & "\"" & name of r & "\",\"" & body of r & "\",\"" & completed of r & "\",\"" & completion date of r & "\",\"" & name of container of r & "\",\"" & creation date of r & "\",\"" & due date of r & "\",\"" & modification date of r & "\",\"" & remind me date of r & "\",\"" & priority of r & "\"\n"
    end repeat
end tell
return output

Answer 4

我们可以使用collect方法合并map和filter：

xs.collect { case (s, e) if e.isDefined => (e, s.indexOf(e.get)) }
    .filter { case (e, i) => i > 0 }

Answer 5

map和getOrElse可能会让事情变得更加清晰：

// use map you will get Some(-1) if the element doesn't exist or None if the element is None
xs.map{case (s, e) => (e, e.map(s.indexOf(_)))}.

// check if the index is positive and use getOrElse to return false if it's None
   filter{case (e, idx) => idx.map(_ >= 0).getOrElse(false)}

// res16: List[(Option[String], Option[Int])] = List((Some(Beer),Some(1)))

或者：

xs.map{ case (s, e) => (e, e.map(s.indexOf).getOrElse(-1)) }.filter(_._2 != -1)
// res17: List[(Option[String], Int)] = List((Some(Beer),1)

在scala

5 个答案: