Spark: convert a CSV to RDD[Row] - scala

Question

I have a .csv file, which contains 258 columns in following structure.
["label", "index_1", "index_2", ... , "index_257"]
Now I wanna transform this .csv file to a RDD[Row]:
val data_csv = sc.textFile("~/test.csv")
val rowRDD = data_csv.map(_.split(",")).map(p => Row( p(0), p(1).trim, p(2).trim))
If I do the transform in this way, I have to write down 258 columns specifically. So I tried:
val rowRDD = data_csv.map(_.split(",")).map(p => Row( _ => p(_).trim))
and
val rowRDD = data_csv.map(_.split(",")).map(p => Row( x => p(x).trim))
But these two also not working and report error:
error: missing parameter type for expanded function ((x$2) => p(x$2).trim)
Can anyone tell me how to do this transform? Thanks a lot.

Answer 1

you should use sqlContext instead of sparkContext as
val df = sqlContext.read
.format("com.databricks.spark.csv")
.option("header", true)
.load(("~/test.csv")
this will create dataframe. calling .rdd on df should give you RDD[Row]
val rdd = df.rdd

Answer 2

Rather reading as a textFile read CSV files with the spark-csv
In your case
val df = sqlContext.read
.format("com.databricks.spark.csv")
.option("header", "true") // Use first line of all files as header
.option("inferSchema", "true") // Automatically infer data types
.option("quote", "\"") //escape the quotes
.option("ignoreLeadingWhiteSpace", true) // escape space before your data
.load("cars.csv")
This loads data as a dataframe, now you can easily change it to RDD.
Hope this helps!

Answer 3

Apart from the other answers that are correct, the correct way to do what you're trying to do is to use Row.fromSeq inside the map function.
val rdd = sc.parallelize(Array((1 to 258).toArray, (1 to 258).toArray) )
.map(Row.fromSeq(_))
This will turn your rdd to type Row:
Array[org.apache.spark.sql.Row] = Array([1,2,3,4,5,6,7,8,9,10...