如何在scala中切片元组

Question

我正在尝试对元组进行切片，删除最后两项。我尝试使用列表 drop/take 方法，但无法成功取回元组。

这是我尝试过的方法：

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 = ???

我看到 here 元组列表在 scala 中（还没有？）是可能的。

是否有其他方法可以获取该子元组（无需处理 myTuple._1、myTuple._2...）？

Answer 1

这是shapeless 可以以通用 方式完成的事情，包括转换为HList。

首先 - get shapeless。然后打开依赖方法类型运行 scala（在 2.10 中默认打开）：

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.

将 shapeless 添加到类路径中：

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"

现在让我们玩吧！

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

我们必须导入无形

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

我们把我们的元组变成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

然后我们取前4个（注意_4是类型参数，不是方法参数）

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

现在转换回元组

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

我们可以缩短这个：

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

这当然推广到N-tuple 的第一个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

怎么样：

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) 左：List[Int] = List(1, 2, 4, 5) 右：List[Int] = List(0, 5) scala> val mytuple2 = (right(0),right(1)) mytuple2: (Int, Int) = (0,5)