Spark unionAll multiple dataframes

Question

For a set of dataframes

val df1 = sc.parallelize(1 to 4).map(i => (i,i*10)).toDF("id","x")
val df2 = sc.parallelize(1 to 4).map(i => (i,i*100)).toDF("id","y")
val df3 = sc.parallelize(1 to 4).map(i => (i,i*1000)).toDF("id","z")

to union all of them I do

df1.unionAll(df2).unionAll(df3)

Is there a more elegant and scalable way of doing this for any number of dataframes, for example from

Seq(df1, df2, df3) 

Answer 1

For pyspark you can do the following:

from functools import reduce
from pyspark.sql import DataFrame

dfs = [df1,df2,df3]
df = reduce(DataFrame.unionAll, dfs)

It's also worth nothing that the order of the columns in the dataframes should be the same for this to work. This can silently give unexpected results if you don't have the correct column orders!!

If you are using pyspark 2.3 or greater, you can use unionByName so you don't have to reorder the columns.

Answer 2

The simplest solution is to reduce with union (unionAll in Spark < 2.0):

val dfs = Seq(df1, df2, df3)
dfs.reduce(_ union _)

This is relatively concise and shouldn't move data from off-heap storage but extends lineage with each union requires non-linear time to perform plan analysis. what can be a problem if you try to merge large number of DataFrames.

You can also convert to RDDs and use SparkContext.union:

dfs match {
  case h :: Nil => Some(h)
  case h :: _   => Some(h.sqlContext.createDataFrame(
                     h.sqlContext.sparkContext.union(dfs.map(_.rdd)),
                     h.schema
                   ))
  case Nil  => None
}

It keeps lineage short analysis cost low but otherwise it is less efficient than merging DataFrames directly.