Eran Medan created SPARK-6513:
---------------------------------
Summary: Regression Adding zipWithUniqueId (and other missing RDD
APIs) to RDDApi.scala
Key: SPARK-6513
URL: https://issues.apache.org/jira/browse/SPARK-6513
Project: Spark
Issue Type: Bug
Components: SQL
Affects Versions: 1.3.0
Environment: Windows 7 64bit, Scala 2.11.6, JDK 1.7.0_21 (though I
don't think it's relevant)
Reporter: Eran Medan
I'm sure this has an Issue somewhere but I can't find it.
I see this as a regression bug, since it compiled in 1.2.1 but stopped in 1.3
without any earlier deprecation warnings, but I am sure the authors are well
aware, so please change it to an enhancement request if you disagree this is a
regression. It's such an obvious and blunt regression that I doubt it was done
without a lot of thought and I'm sure there was a good reason, but still it
breaks my code and I don't have a workaround :)
Here are the details / steps to reproduce
**Worked in 1.2.1** (without any deprecation warnings)
val sqlContext = new HiveContext(sc)
import sqlContext._
val jsonRDD = sqlContext.jsonFile(jsonFilePath)
jsonRDD.registerTempTable("jsonTable")
val jsonResult = sql(s"select * from jsonTable")
val foo = jsonResult.zipWithUniqueId().map {
case (Row(...), uniqueId) => // do something useful
...
}
foo.registerTempTable("...")
**Stopped working in 1.3.0** (simply does not compile, and all I did was change
to 1.3)
jsonResult.zipWithUniqueId() //since RDDApi doesn't implement that method
**Not working workaround:**
although this might give me an RDD[Row]:
jsonResult.map(identity).zipWithUniqueId()
now this won't work as `RDD[Row]` does not have a `registerTempTable` method of
course
foo.registerTempTable("...")
(see related SO question:
http://stackoverflow.com/questions/29243186/is-this-a-regression-bug-in-spark-1-3)
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