如何在scala中切片元組

Question

我正在嘗試切片一個元組，刪除最后兩個項目。 我嘗試使用list drop / take方法，但我無法成功獲得元組。

這是我嘗試的方法：

scala> val myTuple = (1, 2, 4, 5, 0, 5)
myTuple: (Int, Int, Int, Int, Int, Int) = (1,2,4,5,0,5)

scala> val myList = myTuple.productIterator.toList
myList: List[Any] = List(1, 2, 4, 5, 0, 5)

scala> val mySubList = myList.dropRight(2)
mySubList: List[Any] = List(1, 2, 4, 5)

scala> val mySubTuple = ???

我在這里看到，在scala中，元組的列表不是（但是？）。

是否有其他方法可以獲得該子組（不處理myTuple._1，myTuple._2 ...）？

Answer 1

這是無形的可以通用的方式做的事情，涉及轉換為HList 。

首先 - 變得沒型 。 然后運行scala並打開依賴方法類型 （默認情況下在2.10中打開 ）：

C:\Scala\sdk\scala-2.9.2\bin>scala -Ydependent-method-types
Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_04).
Type in expressions to have them evaluated.
Type :help for more information.

在classpath中添加shapeless：

scala> :cp C:\Users\cmarsha\Downloads\shapeless_2.9.2-1.2.2.jar
Added 'C:\Users\cmarsha\Downloads\shapeless_2.9.2-1.2.2.jar'.  Your new classpath is:
"C:\tibco\tibrv\8.2\lib\tibrvnative.jar;C:\Users\cmarsha\Downloads\shapeless_2.9.2-1.2.2.jar"

現在讓我們玩吧！

scala> (1, 2.3, 'a, 'b', "c", true)
res0: (Int, Double, Symbol, Char, java.lang.String, Boolean) = (1,2.3,'a,b,c,true)

我們必須進口無形

scala> import shapeless._; import Tuples._; import Nat._
import shapeless._
import Tuples._
import Nat._

我們將元組轉換為HList

scala> res0.hlisted
res2: shapeless.::[Int,shapeless.::[Double,shapeless.::[Symbol,shapeless.::[Char,shapeless.::[java.lang.String,shapeless.::[Boolean,shapeless.HNil]]]]]] = 1 :: 2.3 :: 'a :: b :: c :: true :: HNil

然后我們取第4個（注意_4是一個類型參數， 而不是方法參數 ）

scala> res2.take[_4]
res4: shapeless.::[Int,shapeless.::[Double,shapeless.::[Symbol,shapeless.::[Char, shapeless.HNil]]]] = 1 :: 2.3 :: 'a :: b :: HNil

現在轉換回元組

scala> res4.tupled
res5: (Int, Double, Symbol, Char) = (1,2.3,'a,b)

我們可以縮短這個：

val (a, b, c, d) = sixtuple.hlisted.take[_4].tupled 
//a, b, c and d would all have the correct inferred type

這當然推廣到N元組的前M元素

Answer 2

scala> val myTuple = (1, 2, 4, 5, 0, 5)
myTuple: (Int, Int, Int, Int, Int, Int) = (1,2,4,5,0,5)

scala> myTuple match {
     |   case (a, b, c, d, _, _) => (a, b, c, d)
     | }
res0: (Int, Int, Int, Int) = (1,2,4,5)

Answer 3

怎么樣：

scala> val myTuple = (1,2,4,5,0,5)
myTuple: (Int, Int, Int, Int, Int, Int) = (1,2,4,5,0,5)

scala> val (left,right):Tuple2[List[Int],List[Int]] = myTuple.productIterator.toList.splitAt(myTuple.productArity - 2)
left: List[Int] = List(1, 2, 4, 5)
right: List[Int] = List(0, 5)

scala> val mytuple2 = (right(0),right(1))
mytuple2: (Int, Int) = (0,5)