Reputation: 41306
Let's say I have a Spark Dataset
like this:
scala> import java.sql.Date
scala> case class Event(id: Int, date: Date, name: String)
scala> val ds = Seq(Event(1, Date.valueOf("2016-08-01"), "ev1"), Event(2, Date.valueOf("2018-08-02"), "ev2")).toDS
I want to create a new Dataset
with only the name and date fields. As far as I can see, I can either use ds.select()
with TypedColumn
or I can use ds.select()
with Column
and then convert the DataFrame
to Dataset
.
However, I can't get the former option working with the Date
type. For example:
scala> ds.select($"name".as[String], $"date".as[Date])
<console>:31: error: Unable to find encoder for type stored in a Dataset. Primitive types (Int, String, etc) and Product types (case classes) are supported by importing spark.implicits._ Support for serializing other types will be added in future releases.
ds.select($"name".as[String], $"date".as[Date])
^
The later option works:
scala> ds.select($"name", $"date").as[(String, Date)]
res2: org.apache.spark.sql.Dataset[(String, java.sql.Date)] = [name: string, date: date]
Is there a way to select Date
fields from Dataset
without going to DataFrame
and back?
Upvotes: 8
Views: 4280
Reputation: 32319
Been bashing my head against problems like these for the whole day. I think you can solve your problem with one line:
implicit val e: Encoder[(String, Date)] = org.apache.spark.sql.Encoders.kryo[(String,Date)]
At least that has been working for me.
In these cases, the problem is that for most Dataset
operations, Spark 2 requires an Encoder
that stores schema information (presumably for optimizations). The schema information takes the form of an implicit parameter (and a bunch of Dataset
operations have this sort of implicit parameter).
In this case, the OP found the correct schema for java.sql.Date
so the following works:
implicit val e = org.apache.spark.sql.Encoders.DATE
Upvotes: 5