Please take a look at stringToTimestamp() in
./sql/catalyst/src/main/scala/org/apache/spark/sql/catalyst/util/DateTimeUtils.scala
Representing timestamp with long should work.
Cheers
On Fri, Jul 31, 2015 at 2:50 PM, Joanne Contact <joannenetw...@gmail.com>
wrote:
> Hi Guys,
>
> I have struggled for a while on this seeming simple thing:
>
> I have a sequence of timestamps and want to create a dataframe with 1
> column.
>
> Seq[java.sql.Timestamp]
>
> //import collection.breakOut
>
> var seqTimestamp = scala.collection.Seq(listTs:_*)
>
> seqTimestamp: Seq[java.sql.Timestamp] = List(2015-07-22 16:52:00.0,
> 2015-07-22 16:53:00.0, ....., )
>
> I tried a lot of ways to create a dataframe and below is another failed
> way:
>
> import sqlContext.implicits._
> var rddTs = sc.parallelize(seqTimestamp)
> rddTs.toDF("minInterval")
>
> <console>:108: error: value toDF is not a member of
> org.apache.spark.rdd.RDD[java.sql.Timestamp] rddTs.toDF("minInterval")
>
> So, any guru could please tell me how to do this????
>
> I am not familiar with Scala or Spark. I wonder if learning Scala will
> help this at all? It just sounds a lot of time of trial/error and
> googling.
>
> docs like
>
> https://spark.apache.org/docs/1.3.0/api/java/org/apache/spark/sql/DataFrame.html
>
> https://spark.apache.org/docs/1.3.0/api/java/org/apache/spark/sql/SQLContext.html#createDataFrame(scala.collection.Seq
> ,
> scala.reflect.api.TypeTags.TypeTag)
> does not help.
>
> Btw, I am using Spark 1.4.
>
> Thanks in advance,
>
> J
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: user-unsubscr...@spark.apache.org
> For additional commands, e-mail: user-h...@spark.apache.org
>
>
