Havnar Havnar - 1 year ago 102
Scala Question

Spark map dataframe using the dataframe's schema

I have a dataframe, created from a JSON object. I can query this dataframe and write it to parquet.

Since I infer the schema, I don't necessarily know what's in the dataframe.

Is there a way to the the column names out or map the dataframe using its own schema?

// The results of SQL queries are DataFrames and support all the normal RDD operations.
// The columns of a row in the result can be accessed by field index:
df.map(t => "Name: " + t(0)).collect().foreach(println)

// or by field name:
df.map(t => "Name: " + t.getAs[String]("name")).collect().foreach(println)

// row.getValuesMap[T] retrieves multiple columns at once into a Map[String, T]
df.map(_.getValuesMap[Any](List("name", "age"))).collect().foreach(println)
// Map("name" -> "Justin", "age" -> 19)

I would want to do something like

df.map (_.getValuesMap[Any](ListAll())).collect().foreach(println)
// Map ("name" -> "Justin", "age" -> 19, "color" -> "red")

without knowing the actual amount or names of the columns.

Answer Source

Well, you can but result is rather useless:

val df = Seq(("Justin", 19, "red")).toDF("name", "age", "color")

def getValues(row: Row, names: Seq[String]) = names.map(
  name => name -> row.getAs[Any](name)

val names = df.columns
df.rdd.map(getValues(_, names)).first

// scala.collection.immutable.Map[String,Any] = 
//   Map(name -> Justin, age -> 19, color -> red)

To get something actually useful one would a proper mapping between SQL types and Scala types. It is not hard in simple cases but it is hard in general. For example there is built-in type which can be used to represent an arbitrary struct. This can be done using a little bit of meta-programming but arguably it is not worth all the fuss.

Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download