scala - Spark : How to union all dataframe in loop

Question

Is there a way to get the dataframe that union dataframe in loop?

This is a sample code:

var fruits = List(
  "apple"
  ,"orange"
  ,"melon"
) 

for (x <- fruits){         
  var df = Seq(("aaa","bbb",x)).toDF("aCol","bCol","name")
}

I would want to obtain some like this:

aCol | bCol | fruitsName
aaa,bbb,apple
aaa,bbb,orange
aaa,bbb,melon

Thanks again

Answer 1

You could created a sequence of DataFrame s and then use reduce :

val results = fruits.
  map(fruit => Seq(("aaa", "bbb", fruit)).toDF("aCol","bCol","name")).
  reduce(_.union(_))

results.show()

Answer 2

Steffen Schmitz's answer is the most concise one I believe. Below is a more detailed answer if you are looking for more customization (of field types, etc):

import org.apache.spark.sql.types.{StructType, StructField, StringType}
import org.apache.spark.sql.Row

//initialize DF
val schema = StructType(
  StructField("aCol", StringType, true) ::
  StructField("bCol", StringType, true) ::
  StructField("name", StringType, true) :: Nil)
var initialDF = spark.createDataFrame(sc.emptyRDD[Row], schema)

//list to iterate through
var fruits = List(
    "apple"
    ,"orange"
    ,"melon"
)

for (x <- fruits) {
  //union returns a new dataset
  initialDF = initialDF.union(Seq(("aaa", "bbb", x)).toDF)
}

//initialDF.show()

references:

• How to create an empty DataFrame with a specified schema?
• https://spark.apache.org/docs/2.0.1/api/java/org/apache/spark/sql/Dataset.html
• https://docs.databricks.com/spark/latest/faq/append-a-row-to-rdd-or-dataframe.html

Answer 3

If you have different/multiple dataframes you can use below code, which is efficient.

val newDFs = Seq(DF1,DF2,DF3)
newDFs.reduce(_ union _)

Answer 4

In a for loop:

val fruits = List("apple", "orange", "melon")

( for(f <- fruits) yield ("aaa", "bbb", f) ).toDF("aCol", "bCol", "name")

Answer 5

Well... I think your question is a bit mis-guided.

As per my limited understanding of whatever you are trying to do, you should be doing following,

val fruits = List(
  "apple",
  "orange",
  "melon"
)

val df = fruits
  .map(x => ("aaa", "bbb", x))
  .toDF("aCol", "bCol", "name")

And this should be sufficient.

Answer 6

you can first create a sequence and then use toDF to create Dataframe .

scala> var dseq : Seq[(String,String,String)] = Seq[(String,String,String)]()
dseq: Seq[(String, String, String)] = List()

scala> for ( x <- fruits){
     |  dseq = dseq :+ ("aaa","bbb",x)
     | }

scala> dseq
res2: Seq[(String, String, String)] = List((aaa,bbb,apple), (aaa,bbb,orange), (aaa,bbb,melon))

scala> val df = dseq.toDF("aCol","bCol","name")
df: org.apache.spark.sql.DataFrame = [aCol: string, bCol: string, name: string]

scala> df.show
+----+----+------+
|aCol|bCol|  name|
+----+----+------+
| aaa| bbb| apple|
| aaa| bbb|orange|
| aaa| bbb| melon|
+----+----+------+