Removing Unnecessary JSON fields using SPARK (SQL) - json

Question

I'm a new spark user currently playing around with Spark and some big data and I have a question related to Spark SQL or more formally the SchemaRDD. I'm reading a JSON file containing data about some weather forecasts and I'm not really interested in all of the fields that I have ... I only want 10 fields out of 50+ fields returned for each record. Is there a way (similar to filter) that I can use to specify the names of some fields that I want remove from spark.
Just a small descriptive example. Consider I have the Schema "Person" with 3 fields "Name", "Age", and "Gender" and I'm not interested in the "Age" field and wold like to remove it. Can I use spark some how to do that. ? Thanks

Answer 1

If you are using Spark 1.2, you can do the following (using Scala)...
If you already know what fields you want to use, you can construct the schema for these fields and apply this schema to the JSON dataset. Spark SQL will return a SchemaRDD. Then, you can register it and query it as a table. Here is a snippet...
// sc is an existing SparkContext.
val sqlContext = new org.apache.spark.sql.hive.HiveContext(sc)
// The schema is encoded in a string
val schemaString = "name gender"
// Import Spark SQL data types.
import org.apache.spark.sql._
// Generate the schema based on the string of schema
val schema =
StructType(
schemaString.split(" ").map(fieldName => StructField(fieldName, StringType, true)))
// Create the SchemaRDD for your JSON file "people" (every line of this file is a JSON object).
val peopleSchemaRDD = sqlContext.jsonFile("people.txt", schema)
// Check the schema of peopleSchemaRDD
peopleSchemaRDD.printSchema()
// Register peopleSchemaRDD as a table called "people"
peopleSchemaRDD.registerTempTable("people")
// Only values of name and gender fields will be in the results.
val results = sqlContext.sql("SELECT * FROM people")
When you look at the schema of peopleSchemaRDD (peopleSchemaRDD.printSchema()), you will only see name and gender field.
Or, if you want to explore the dataset and determine what fields you want after you see all fields, you can ask Spark SQL to infer the schema for you. Then, you can register the SchemaRDD as a table and use projection to remove unneeded fields. Here is a snippet...
// Spark SQL will infer the schema of the given JSON file.
val peopleSchemaRDD = sqlContext.jsonFile("people.txt")
// Check the schema of peopleSchemaRDD
peopleSchemaRDD.printSchema()
// Register peopleSchemaRDD as a table called "people"
peopleSchemaRDD.registerTempTable("people")
// Project name and gender field.
sqlContext.sql("SELECT name, gender FROM people")

Answer 2

You can specify what fields you would like to have in the schemaRDD. Below is an example. Create a case class, with only the fields that you need. Read the data into an rdd, then specify the only the fileds that you need(in the same order as you have specified the schema in the case class).
Sample Data: People.txt
foo,25,M
bar,24,F
Code:
case class Person(name: String, gender: String)
val people = sc.textFile("People.txt").map(_.split(",")).map(p => Person(p(0), p(2)))
people.registerTempTable("people")