It is from stack overflow
StructType
I assume you start with some kind of flat schema like this:
root
|-- lat: double (nullable = false)
|-- long: double (nullable = false)
|-- key: string (nullable = false)
First lets create example data:
import org.apache.spark.sql.Row
import org.apache.spark.sql.functions.{col, udf}
import org.apache.spark.sql.types._
val rdd = sc.parallelize(
Row(52.23, 21.01, "Warsaw") :: Row(42.30, 9.15, "Corte") :: Nil)
val schema = StructType(
StructField("lat", DoubleType, false) ::
StructField("long", DoubleType, false) ::
StructField("key", StringType, false) ::Nil)
val df = sqlContext.createDataFrame(rdd, schema)
An easy way is to use an udf and case class:
case class Location(lat: Double, long: Double)
val makeLocation = udf((lat: Double, long: Double) => Location(lat, long))
val dfRes = df.
withColumn("location", makeLocation(col("lat"), col("long"))).
drop("lat").
drop("long")
dfRes.printSchema
and we get
root
|-- key: string (nullable = false)
|-- location: struct (nullable = true)
| |-- lat: double (nullable = false)
| |-- long: double (nullable = false)
A hard way is to transform your data and apply schema afterwards:
val rddRes = df.
map{case Row(lat, long, key) => Row(key, Row(lat, long))}
val schemaRes = StructType(
StructField("key", StringType, false) ::
StructField("location", StructType(
StructField("lat", DoubleType, false) ::
StructField("long", DoubleType, false) :: Nil
), true) :: Nil
)
sqlContext.createDataFrame(rddRes, schemaRes).show
and we get an expected output
+------+-------------+
| key| location|
+------+-------------+
|Warsaw|[52.23,21.01]|
| Corte| [42.3,9.15]|
+------+-------------+