How to create a Row from a List or Array in Spark using Scala

Something like the following should work:

import org.apache.spark.sql._

def f(n: List[Int], s: String) : Row =
  Row.fromSeq(s.split(",").zipWithIndex.collect{case (a,b) if n.contains(b) => a}.toSeq)

You are missing creation of the StructField and StructType. Refer to the official guide http://spark.apache.org/docs/latest/sql-programming-guide.html, part Programmatically Specifying the Schema

I'm not a Scala specialist, but in Python it would look like this:

from pyspark.sql import *
sqlContext = SQLContext(sc)

input = [1,2]

def parse(line):
    global input
    l = line.split(',')
    res = [l[0]]
    for ind in input:
        res.append(l[ind])
    return res

csv  = sc.textFile("file:///tmp/inputfile.csv")
rows = csv.map(lambda x: parse(x))

fieldnum = len(input) + 1
fields = [StructField("col"+str(i), StringType(), True) for i in range(fieldnum)]
schema = StructType(fields)

csvWithSchema = sqlContext.applySchema(rows, schema)
csvWithSchema.registerTempTable("test")
sqlContext.sql("SELECT * FROM test").collect()

In short, you should not directly convert them to Row objects, just leave as RDD and apply schema to it with applySchema

You can also try:

    Row.fromSeq(line(0).toString ++ line(1).toDouble ++ line(2).toDouble ++ line.slice(2, line.size).map(value => value.toString))