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我正在尝试切片一个元组，删除最后两项。我尝试使用列表删除/获取方法，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 = ???

我在这里看到元组列表在scala中（还没有？）是可能的。

还有其他方法可以获得该子元组（不处理 myTuple._1、myTuple._2...）？

4

3 回答 3

40

这是shapeless可以以通用方式做的事情，包括转换为HList.

首先 -变得无形。然后在打开依赖方法类型的情况下运行 scala（默认情况下在 2.10 中打开）：

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一个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

这当然可以推广到-tuple的第一个M元素N

于 2012-06-19T11:12:40.240 回答
5

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)
于 2012-06-26T08:38:46.430 回答
5 
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)
于 2012-06-19T09:54:07.473 回答