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Editorial Team
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Asked: 2026-06-12T16:49:16+00:00

case class DataItem(name: String, timestamp: Long, value: String) val dataitems = List(DataItem(SpindleSpeed, 1223334444, 20.3333),

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case class DataItem(name: String, timestamp: Long, value: String)
val dataitems = List(DataItem(SpindleSpeed, 1223334444, 20.3333),
                     DataItem(SpindleSpeed, 1223334450, 21.3333),
                     DataItem(SpindleSpeed, 1223334460, 19.3333),
                     DataItem(Load, 1223334444, 70.0023),
                     DataItem(Load, 1223334446, 72.0023),
                     DataItem(Pressure, 1223334444, 20.3333))

I have a list somewhat like this, I need to filter out the dataitems which has the lowest timestamp. There can be more than one dateitems having the same timestamp, in that case I need all those dataitems.

In the above case, I expect the filtered list to be,

List(DataItem(SpindleSpeed, 1223334444, 20.3333),
     DataItem(Load, 1223334444, 70.0023),
     DataItem(Pressure, 1223334444, 20.3333))

What is the functional way of doing it? I tried sorting the list and returning the head. But that returns only a single dataitem which doesn’t seem to right.

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1 Answer

  1. Editorial Team
    scala> val SpindleSpeed = "S"
SpindleSpeed: java.lang.String = S

scala> val Pressure = "P"
Pressure: java.lang.String = P

scala> val Load = "L"
Load: java.lang.String = L

scala> case class DataItem(name: String, timestamp: Long, value: Double)
defined class DataItem

scala> val dataitems = List(DataItem(SpindleSpeed, 1223334444, 20.3333),
     |                      DataItem(SpindleSpeed, 1223334450, 21.3333),
     |                      DataItem(SpindleSpeed, 1223334460, 19.3333),
     |                      DataItem(Load, 1223334444, 70.0023),
     |                      DataItem(Load, 1223334446, 72.0023),
     |                      DataItem(Pressure, 1223334444, 20.3333))
dataitems: List[DataItem] = List(DataItem(S,1223334444,20.3333), DataItem(S,1223334450,21.3333), DataItem(S,1223334460,19.3333), DataItem(L,1223334444,70.0023), DataItem(L,1223334446,72.0023), DataItem(P,1223334444,20.3333))

scala> dataitems.groupBy(_.timestamp).minBy(_._1)._2
res2: List[DataItem] = List(DataItem(S,1223334444,20.3333), DataItem(L,1223334444,70.0023), DataItem(P,1223334444,20.3333))

scala>
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