在 scala 列表中搜索与属性匹配的内容

Question

执行此操作的惯用 scala 方法是什么？我有一个列表，如果我找到符合某些条件的内容，则希望返回“Y”，否则返回“N”。我有一个“有效”的解决方案，但我不喜欢它......

def someMethod( someList: List[Something]) : String = {

someList.foreach( a =>
  if (a.blah.equals("W") || a.bar.equals("Y") ) {
    return "Y"
  }
 )
  "N"


}

Answer 1

简单：

if (someList.exists{ a=> a.blah == "W" || a.bar == "Y"}) 
   "Y"
else 
   "N"

Answer 2

def condition(i: Something) = i.blah.equals("W") || i.bar.equals("Y")

somelist.find(condition(_)).map(ignore => "Y").getOrElse("N")

或者简单地

if (somelist exists condition) "Y" else "N"

Answer 3

假设 Something 有一个 blah 成员。

def someMethod(someList: List[Something]): String = 
  if (someList forall { a => a.blah == "W" || a.blah == "Y" }) "Y" else "N"

Answer 4

val l = List(1,2,3,4)

(l.find(x => x == 0 || x == 1)) match { case Some(x) => "Y"; case _ => "N"}
res: java.lang.String = Y

Answer 5

假设 Something 可以模式匹配，例如，当它是一个像...这样的案例类时

case class Something(blah:String, bar:String) 

，您可以写...

 def someMethod(someList: List[Something]) = someList.collect{
   case Something("W",_) | Something(_,"Y") => "Y"}.headOption.getOrElse("N")