I have a List
data. Something like:
[[dev, engg, 10000], [kar
You can create DataFrame from List
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 data = new ArrayList();
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(col("salary")).cast("integer").as("salary"));