scala中的逻辑和if内部匹配案例

Question

我需要在scala中使用一堆match case，如果在match case里面怎么用for或者if，例如：

import scala.util.Random

val x = Random.nextInt(30)
val result = match x {
case 0 to 10 => bad
case 11 to 20 => average
case 20 to 30 => cool
}

第一个问题是我该怎么做而不是使用floor、ceil、round 或其他数学东西？

其次，对于字符串值，在 php 中我们可以轻松使用if (x == 'some')||(x == 'thing'){result}，但这在 scala 匹配情况下如何工作？

例如当 val x 是随机的 A 到 I 时：

val result = match x {
case A || B || C => bad
case D || E || F => average
case G || H || I => cool
}

谢谢！

Answer 1

这是一个更接近您的伪代码的替代方案：

val x:Int  

val result = x match {
  case x if (1 to 10).contains(x) => "bad"
  case x if (11 to 20).contains(x) => "average"
  case x if (21 to 30).contains(x) => "cool"
  case _ => "unknown"
}


val y:String 

val result = y match  {
    case "A" | "B" | "C" => "bad"
    case "D" | "E" | "F" => "average"
    case "G" | "H" | "I" => "cool"
    case _ => "unknown"
}

为了利用第一个示例中的范围，将字母分数视为字符会更方便：

val z:Char 

val result = z match {
  case x if ('A' to 'C').contains(x) => "bad"
  case x if ('D' to 'F').contains(x) => "average"
  case x if ('G' to 'I').contains(x) => "cool"
  case _ => "unknown"
}

Answer 2

使用 Guards 的模式匹配可以帮助您做到这一点

val x = Random.nextInt(30)

val result = x match {
 case x if x > 0 && x <= 10 => "bad"
 case x if x > 11 &&  x <= 20 => "average"
 case x if x > 20 && x <= 30 => "cool"
 case _ => "no idea"
}

对于字符串，您可以这样做

val str = "foo"

val result = str match {
 case "foo" => "found foo"
 case "bar" => "found bar"
 case _ => "found some thing"
}

你也可以使用|

val result = str match {
  case "foo" | "bar" => "match"
  case _ => "no match"
}

Scala REPL

scala> :paste
// Entering paste mode (ctrl-D to finish)

val x = Random.nextInt(30)

val result = x match {
 case x if x > 0 && x <= 10 => "bad"
 case x if x > 11 &&  x <= 20 => "average"
 case x if x > 20 && x <= 30 => "cool"
 case _ => "no idea"
}

// Exiting paste mode, now interpreting.

x: Int = 13
result: String = average