斯卡拉 - 扁平列表索引製作Tuplets

我正在學習Scala。我想創建一個3人組元組。斯卡拉 - 扁平列表索引製作Tuplets

我的目錄列表

List((B,2), (H,1), (t,1), (S,1)) 
List((H,1), (t,1), (B,2), (D,1))

期望的結果：扁平化索引（的ListIndex，X，Y）列表...

(1,B,2), (1,H,1), (1,t,1), (1,S,1)) 
(2,H,1), (2,t,1), (2,B,2), (2,D,1))

來源

2014-03-03 user3001937

生育治療有時會形成連環。 –

val input = List(List((B,2), (H,1), (t,1), (S,1)), List((H,1), (t,1), (B,2), (D,1))) 
var output = input.zipWithIndex.flatMap({ case (l, i) => l.map(p => (i + 1, p._1, p._2)) })

來源

2014-03-03 21:08:59 Lee

試試這個：

val list = List(List(("B",2), ("H",1), ("t",1), ("S",1)), List(("H",1), ("t",1), ("B",2), ("D",1))) 

val result = list.foldLeft(List[(Int,String,Int)]())((res,sub) => res ++ sub.map(elem => (list.indexOf(sub) + 1,elem._1,elem._2)))

返回result爲：

List((1,B,2), (1,H,1), (1,t,1), (1,S,1), (2,H,1), (2,t,1), (2,B,2), (2,D,1))

雖然應該有一個更簡單的方法。

來源

2014-03-03 21:27:37 Peter

scala> val lst = List(
     List(('B',2), ('H',1), ('t',1), ('S',1)), 
     List(('H',1), ('t',1), ('B',2), ('D',1)) 
     ) 
lst: List[List[(Char, Int)]] = List(List((B,2), (H,1), (t,1), (S,1)), List((H,1), (t,1), (B,2), (D,1))) 

scala> lst.flatten.map{ x => (1, x) } 
      .groupBy{ y => y._2 } 
      .map{ z => (z._2.length, z._1._1, z._1._2) } 

res24: scala.collection.immutable.Iterable[(Int, Char, Int)] = 
     List((1,D,1), (2,t,1), (2,H,1), (2,B,2), (1,S,1))

來源

2014-03-03 21:47:53 tuxdna

斯卡拉 - 扁平列表索引製作Tuplets

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