如何在scala中實現部分縮減？

Question

這裏是場景：只在某些特定情況下減少元素。

例如， val sen = List("Jone", "had", "a", "great", "time", ".")，如果元素以「e」結尾，則將其與下一個元素連接起來。對於sen，結果可能是("Jone had", "a", "great", "time .")。

是否有任何優雅的方式來實現這些，而不是一個循環？

Answer 1

您可以簡單地使用foldLeft爲：

val result = sen.foldLeft(List[String]()) { 
    (rs, s) => 
     if (rs.nonEmpty && rs.head.endsWith("e")) 
     (rs.head + " " + s) :: rs.tail 
     else 
     s :: rs 
    }.reverse 

只需更換rs.head.endsWith("e")用你自己的功能。

Answer 2

作爲@kaktusito張貼的foldLeft是最好的方法。以下是它的工作原理....大部分。

def partialReduce(list : List[String]): List[String] = { 
    @scala.annotation.tailrec 
    def internalRecurse(previousWord: String)(reducedList: List[String], finalList: List[String]): List[String] = 
    reducedList match { 
     case Nil => finalList :+ previousWord //No more words, just add the previous to the list and return 
     case head :: rest => { 
     //if ends with e, run through the rest with the new word being the old + the head 
     if(previousWord endsWith "e") internalRecurse(previousWord + " " + head)(rest, finalList) 
     //if doesn't end with e, append the previous word to the final list 
     //and run the rest with the head as the next to check 
     else internalRecurse(head)(rest, finalList :+ previousWord) 
     } 
    } 

    //Call the internal method only if there is at least one item in the list 
    list match { 
    case Nil => Nil 
    case head :: rest => internalRecurse(head)(rest, List()) 
    } 
} 

此外，我要發佈我的模式匹配的版本，因爲我喜歡看更好：

val result = sen.foldLeft(List[String]()) { 
    (curList, s) => { 
     curList match { 
     case Nil => List(s) 
     case head :: rest => { 
      if(head endsWith "e") head + " " + s :: rest 
      else s :: curList 
     } 
     } 
    } 
    }.reverse