Hi Nathan,
#1
Spark SQL & DSL can satisfy your requirement. You can refer the following
code snippet:
jdata.select(Star(Node), 'seven.getField("mod"), 'eleven.getField("mod"))
You need to import org.apache.spark.sql.catalyst.analysis.Star in advance.
#2
After you make the transform above, you do not need to make SchemaRDD
manually.
Because that jdata.select() return a SchemaRDD and you can operate on it
directly.
For example, the following code snippet will return a new SchemaRDD with
longer Row:
val t1 = jdata.select(Star(Node), 'seven.getField("mod") +
'eleven.getField("mod") as 'mod_sum)
You can use t1.printSchema() to print the schema of this SchemaRDD and
check whether it satisfy your requirements.
2014-12-13 0:00 GMT+08:00 Nathan Kronenfeld <nkronenf...@oculusinfo.com>:
>
> (1) I understand about immutability, that's why I said I wanted a new
> SchemaRDD.
> (2) I specfically asked for a non-SQL solution that takes a SchemaRDD, and
> results in a new SchemaRDD with one new function.
> (3) The DSL stuff is a big clue, but I can't find adequate documentation
> for it
>
> What I'm looking for is something like:
>
> import org.apache.spark.sql._
>
>
> val sqlc = new SQLContext(sc)
> import sqlc._
>
>
> val data = sc.parallelize(0 to 99).map(n =>
> ("{\"seven\": {\"mod\": %d, \"times\": %d}, "+
> "\"eleven\": {\"mod\": %d, \"times\": %d}}").format(n % 7, n * 7, n
> % 11, n * 11))
> val jdata = sqlc.jsonRDD(data)
> jdata.registerTempTable("jdata")
>
>
> val sqlVersion = sqlc.sql("SELECT *, (seven.mod + eleven.mod) AS modsum
> FROM jdata")
>
>
> This sqlVersion works fine, but if I try to do the same thing with a
> programatic function, I'm missing a bunch of pieces:
>
> - I assume I'd need to start with something like:
> jdata.select('*, 'seven.mod, 'eleven.mod)
> and then get and process the last two elements. The problems are:
> - I can't select '* - there seems no way to get the complete row
> - I can't select 'seven.mod or 'eleven.mod - the symbol evaluation
> seems only one deep.
> - Assuming I could do that, I don't see a way to make the result into
> a SchemaRDD. I assume I would have to do something like:
> 1. take my row and value, and create a new, slightly longer row
> 2. take my old schema, and create a new schema with one more field
> at the end, named and typed appropriately
> 3. combine the two into a SchemaRDD
> I think I see how to do 3, but 1 and 2 elude me.
>
> Is there more complete documentation somewhere for the DSL portion? Anyone
> have a clue about any of the above?
>
>
>
> On Fri, Dec 12, 2014 at 6:01 AM, Yanbo Liang <yanboha...@gmail.com> wrote:
>
>> RDD is immutable so you can not modify it.
>> If you want to modify some value or schema in RDD, using map to generate
>> a new RDD.
>> The following code for your reference:
>>
>> def add(a:Int,b:Int):Int = {
>> a + b
>> }
>>
>> val d1 = sc.parallelize(1 to 10).map { i => (i, i+1, i+2) }
>> val d2 = d1.map { i => (i._1, i._2, add(i._1, i._2))}
>> d2.foreach(println)
>>
>>
>> Otherwise, if your self-defining function is straightforward and you can
>> represent it by SQL, using Spark SQL or DSL is also a good choice.
>>
>> case class Person(id: Int, score: Int, value: Int)
>>
>> val sqlContext = new org.apache.spark.sql.SQLContext(sc)
>>
>> import sqlContext._
>>
>> val d1 = sc.parallelize(1 to 10).map { i => Person(i,i+1,i+2)}
>> val d2 = d1.select('id, 'score, 'id + 'score)
>> d2.foreach(println)
>>
>>
>> 2014-12-12 14:11 GMT+08:00 Nathan Kronenfeld <nkronenf...@oculusinfo.com>
>> :
>>
>>> Hi, there.
>>>
>>> I'm trying to understand how to augment data in a SchemaRDD.
>>>
>>> I can see how to do it if can express the added values in SQL - just run
>>> "SELECT *,valueCalculation AS newColumnName FROM table"
>>>
>>> I've been searching all over for how to do this if my added value is a
>>> scala function, with no luck.
>>>
>>> Let's say I have a SchemaRDD with columns A, B, and C, and I want to add
>>> a new column, D, calculated using Utility.process(b, c), and I want (of
>>> course) to pass in the value B and C from each row, ending up with a new
>>> SchemaRDD with columns A, B, C, and D.
>>>
>>> Is this possible? If so, how?
>>>
>>> Thanks,
>>> -Nathan
>>>
>>> --
>>> Nathan Kronenfeld
>>> Senior Visualization Developer
>>> Oculus Info Inc
>>> 2 Berkeley Street, Suite 600,
>>> Toronto, Ontario M5A 4J5
>>> Phone: +1-416-203-3003 x 238
>>> Email: nkronenf...@oculusinfo.com
>>>
>>
>>
>
>
> --
> Nathan Kronenfeld
> Senior Visualization Developer
> Oculus Info Inc
> 2 Berkeley Street, Suite 600,
> Toronto, Ontario M5A 4J5
> Phone: +1-416-203-3003 x 238
> Email: nkronenf...@oculusinfo.com
>
