將可變長度數組轉換爲scala中的元組

Question

我需要創建一個元組，但是此元組的大小需要根據變量值在運行時更改。我不能在scala中這樣做。所以我創建了一個數組：

val temp:Array[String] = new Array[String](x) 

如何將數組轉換爲元組。這可能嗎？我是一個斯卡拉新手。

Answer 1

爲了創建一個元組，你必須知道預期的大小。假設你有這個，那麼你可以做這樣的事情：

val temp = Array("1", "2") 
val tup = temp match { case Array(a,b) => (a,b) } 
// tup: (String, String) = (1,2) 

def expectsTuple(x: (String,String)) = x._1 + x._2 
expectsTuple(tup) 

而且允許您將元組傳遞給任何函數期望它。

---

如果你想更大膽的嘗試，你可以定義.toTuple方法：

implicit class Enriched_toTuple_Array[A](val seq: Array[A]) extends AnyVal { 
    def toTuple2 = seq match { case Array(a, b) => (a, b); case x => throw new AssertionError(s"Cannot convert array of length ${seq.size} into Tuple2: Array(${x.mkString(", ")})") } 
    def toTuple3 = seq match { case Array(a, b, c) => (a, b, c); case x => throw new AssertionError(s"Cannot convert array of length ${seq.size} into Tuple3: Array(${x.mkString(", ")})") } 
    def toTuple4 = seq match { case Array(a, b, c, d) => (a, b, c, d); case x => throw new AssertionError(s"Cannot convert array of length ${seq.size} into Tuple4: Array(${x.mkString(", ")})") } 
    def toTuple5 = seq match { case Array(a, b, c, d, e) => (a, b, c, d, e); case x => throw new AssertionError(s"Cannot convert array of length ${seq.size} into Tuple5: Array(${x.mkString(", ")})") } 
} 

這讓你做：

val tup = temp.toTuple2 
// tup: (String, String) = (1,2)