2010-05-04 62 views

回答

10

我第一次嘗試是這樣的:

scala> val country2capitalList = List("England" -> "London", "Germany" -> "Berlin") 
country2capitalList: List[(java.lang.String, java.lang.String)] = List((England,London), (Germany,Berlin)) 

scala> val country2capitalMap = country2capital.groupBy(e => e._1).map(e => (e._1, e._2(0)._2)) 
country2capitalMap: scala.collection.Map[java.lang.String,java.lang.String] = Map(England -> London, Germany -> Berlin) 

但這裏是最好的解決辦法:

scala> val betterConversion = Map(country2capitalList:_*) 
betterConversion: scala.collection.immutable.Map[java.lang.String,java.lang.String] = Map(England -> London, Germany -> Berlin) 

:_*需要給編譯器的提示使用該列表作爲可變參數的參數。否則,它會給你:

scala> Map(country2capitalList) 
<console>:6: error: type mismatch; 
found : List[(java.lang.String, java.lang.String)] 
required: (?, ?) 
     Map(country2capitalList) 
     ^

從斯卡拉2.8上,你可以使用toMap

scala> val someList = List((1, "one"), (2, "two")) 
someList: List[(Int, java.lang.String)] = List((1,one), (2,two)) 

scala> someList.toMap 
res0: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two)) 
14

它不能簡單:

Map(listOf2Tuples: _*) 

使用apply方法Map同伴對象。

+0

僅當'listOf2Tuples'顯式而不是變量/ val時纔有效 – 2010-05-04 08:18:32

+0

修復了它的問題。 – 2010-05-04 08:56:44

+0

+1,這個IMO是做這件事的最好方法。 :) – missingfaktor 2010-05-04 09:03:02

7

斯卡拉2.8:

scala> import scala.collection.breakOut 
import scala.collection.breakOut 

scala> val ls = List("a","bb","ccc") 
ls: List[java.lang.String] = List(a, bb, ccc) 

scala> val map: Map[String,Int] = ls.map{ s => (s,s.length) }(breakOut) 
map: Map[String,Int] = Map((a,1), (bb,2), (ccc,3)) 

scala> val map2: Map[String,Int] = ls.map{ s => (s,s.length) }.toMap 
map2: Map[String,Int] = Map((a,1), (bb,2), (ccc,3)) 

scala> 
9

在2.8,你可以使用toMap方法:

scala> val someList = List((1, "one"), (2, "two")) 
someList: List[(Int, java.lang.String)] = List((1,one), (2,two)) 

scala> someList.toMap 
res0: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two)) 

這將適用於任何成對的集合。請注意,文檔有此說關於它的重複的策略:

重複鍵將 後來鍵被覆蓋:如果這是一個無序 集合,關鍵是在 結果圖中是不確定的。

相關問題