Dataframe from List in Java

Question

• Spark Version : 1.6.2
• Java Version: 7

I have a List data. Something like:

[[dev, engg, 10000], [kar         


        

Answer 1

Task can be completed without JSON, on Scala:

val data = List("dev, engg, 10000", "karthik, engg, 20000")
val intialRdd = sparkContext.parallelize(data)
val splittedRDD = intialRdd.map(current => {
  val array = current.split(",")
  (array(0), array(1), array(2))
})
import sqlContext.implicits._
val dataframe = splittedRDD.toDF("name", "degree", "salary")
dataframe.show()

Output is:

+-------+------+------+
|   name|degree|salary|
+-------+------+------+
|    dev|  engg| 10000|
|karthik|  engg| 20000|
+-------+------+------+

Note: (array(0), array(1), array(2)) is a Scala Tuple

Answer 2

DataFrame createNGramDataFrame(JavaRDD<String> lines) {
 JavaRDD<Row> rows = lines.map(new Function<String, Row>(){
    private static final long serialVersionUID = -4332903997027358601L;

    @Override
    public Row call(String line) throws Exception {
        return RowFactory.create(line.split("\\s+"));
    }
 });
 StructType schema = new StructType(new StructField[] {
        new StructField("words",
                DataTypes.createArrayType(DataTypes.StringType), false,
                Metadata.empty()) });
 DataFrame wordDF = new SQLContext(jsc).createDataFrame(rows, schema);
 // build a bigram language model
 NGram transformer = new NGram().setInputCol("words")
        .setOutputCol("ngrams").setN(2);
 DataFrame ngramDF = transformer.transform(wordDF);
 ngramDF.show(10, false);
 return ngramDF;
}

Answer 3

You can create DataFrame from List<String> and then use selectExpr and split to get desired DataFrame.

public class SparkSample{
public static void main(String[] args) {
    SparkConf conf = new SparkConf().setAppName("SparkSample").setMaster("local[*]");
    JavaSparkContext jsc = new JavaSparkContext(conf);
    SQLContext sqc = new SQLContext(jsc);
    // sample data
    List<String> data = new ArrayList<String>();
    data.add("dev, engg, 10000");
    data.add("karthik, engg, 20000");
    // DataFrame
    DataFrame df = sqc.createDataset(data, Encoders.STRING()).toDF();
    df.printSchema();
    df.show();
    // Convert
    DataFrame df1 = df.selectExpr("split(value, ',')[0] as name", "split(value, ',')[1] as degree","split(value, ',')[2] as salary");
    df1.printSchema();
    df1.show(); 
   }
}

You will get below output.

root
 |-- value: string (nullable = true)

+--------------------+
|               value|
+--------------------+
|    dev, engg, 10000|
|karthik, engg, 20000|
+--------------------+

root
 |-- name: string (nullable = true)
 |-- degree: string (nullable = true)
 |-- salary: string (nullable = true)

+-------+------+------+
|   name|degree|salary|
+-------+------+------+
|    dev|  engg| 10000|
|karthik|  engg| 20000|
+-------+------+------+

The sample data you have provided has empty spaces. If you want to remove space and have the salary type as "integer" then you can use trim and cast function like below.

df1 = df1.select(trim(col("name")).as("name"),trim(col("degree")).‌​as("degree"),trim(co‌​l("salary")).cast("i‌​nteger").as("salary"‌​));