How do I get a SQL row_number equivalent for a Spark RDD?

The row_number() over (partition by ... order by ...) functionality was added to Spark 1.4. This answer uses PySpark/DataFrames.

Create a test DataFrame:

from pyspark.sql import Row, functions as F

testDF = sc.parallelize(
    (Row(k="key1", v=(1,2,3)),
     Row(k="key1", v=(1,4,7)),
     Row(k="key1", v=(2,2,3)),
     Row(k="key2", v=(5,5,5)),
     Row(k="key2", v=(5,5,9)),
     Row(k="key2", v=(7,5,5))
    )
).toDF()

Add the partitioned row number:

from pyspark.sql.window import Window

(testDF
 .select("k", "v",
         F.rowNumber()
         .over(Window
               .partitionBy("k")
               .orderBy("k")
              )
         .alias("rowNum")
        )
 .show()
)

+----+-------+------+
|   k|      v|rowNum|
+----+-------+------+
|key1|[1,2,3]|     1|
|key1|[1,4,7]|     2|
|key1|[2,2,3]|     3|
|key2|[5,5,5]|     1|
|key2|[5,5,9]|     2|
|key2|[7,5,5]|     3|
+----+-------+------+

This is an interesting problem you're bringing up. I will answer it in Python but I'm sure you will be able to translate seamlessly to Scala.

Here is how I would tackle it:

1- Simplify your data:

temp2 = temp1.map(lambda x: (x[0],(x[1],x[2],x[3])))

temp2 is now a "real" key-value pair. It looks like that:

[
((3, 4), (5, 5, 5)),  
((3, 4), (5, 5, 9)),   
((3, 4), (7, 5, 5)),   
((1, 2), (1, 2, 3)),  
((1, 2), (1, 4, 7)),   
((1, 2), (2, 2, 3))

]

2- Then, use the group-by function to reproduce the effect of the PARTITION BY:

temp3 = temp2.groupByKey()

temp3 is now a RDD with 2 rows:

[((1, 2), <pyspark.resultiterable.ResultIterable object at 0x15e08d0>),  
 ((3, 4), <pyspark.resultiterable.ResultIterable object at 0x15e0290>)]

3- Now, you need to apply a rank function for each value of the RDD. In python, I would use the simple sorted function (the enumerate will create your row_number column):

 temp4 = temp3.flatMap(lambda x: tuple([(x[0],(i[1],i[0])) for i in enumerate(sorted(x[1]))])).take(10)

Note that to implement your particular order, you would need to feed the right "key" argument (in python, I would just create a lambda function like those:

lambda tuple : (tuple[0],-tuple[1],tuple[2])

At the end (without the key argument function, it looks like that):

[
((1, 2), ((1, 2, 3), 0)), 
((1, 2), ((1, 4, 7), 1)), 
((1, 2), ((2, 2, 3), 2)), 
((3, 4), ((5, 5, 5), 0)), 
((3, 4), ((5, 5, 9), 1)), 
((3, 4), ((7, 5, 5), 2))

]

Hope that helps!

Good luck.