Sequential Dynamic filters on the same Spark Dataframe Column in Scala Spark - scala

Question

I have a column named root and need to filter dataframe based on the different values of a root column.
Suppose I have a values in root are parent,child or sub-child and I want to apply these filters dynamically through a variable.
val x = ("parent,child,sub-child").split(",")
x.map(eachvalue <- {
var df1 = df.filter(col("root").contains(eachvalue))
}
But when I am doing it, it always overwriting the DF1 instead, I want to apply all the 3 filters and get the result.
May be in future I may extend the list to any number of filter values and the code should work.
Thanks,
Bab

Answer 1

You should apply the subsequent filters to the result of the previous filter, not on df:
val x = ("parent,child,sub-child").split(",")
var df1 = df
x.map(eachvalue <- {
df1 = df1.filter(col("root").contains(eachvalue))
}
df1 after the map operation will have all filters applied to it.

Answer 2

Let's see an example with spark shell. Hope it helps you.
scala> import spark.implicits._
import spark.implicits._
scala> val df0 =
spark.sparkContext.parallelize(List(1,2,1,3,3,2,1)).toDF("number")
df0: org.apache.spark.sql.DataFrame = [number: int]
scala> val list = List(1,2,3)
list: List[Int] = List(1, 2, 3)
scala> val dfFiltered = for (number <- list) yield { df0.filter($"number" === number)}
dfFiltered: List[org.apache.spark.sql.Dataset[org.apache.spark.sql.Row]] = List([number: int], [number: int], [number: int])
scala> dfFiltered(0).show
+------+
|number|
+------+
| 1|
| 1|
| 1|
+------+
scala> dfFiltered(1).show
+------+
|number|
+------+
| 2|
| 2|
+------+
scala> dfFiltered(2).show
+------+
|number|
+------+
| 3|
| 3|
+------+

Answer 3

AFAIK isin can be used in this case below is the example.
import spark.implicits._
val colorStringArr = "red,yellow,blue".split(",")
val colorDF =
List(
"red",
"yellow",
"purple"
).toDF("color")
// to derive a column using a list
colorDF.withColumn(
"is_primary_color",
col("color").isin(colorStringArr: _*)
).show()
println( "if you don't want derived column and directly want to filter using a list with isin then .. ")
colorDF.filter(col("color").isin(colorStringArr: _*)).show
Result :
+------+----------------+
| color|is_primary_color|
+------+----------------+
| red| true|
|yellow| true|
|purple| false|
+------+----------------+
if you don't want derived column and directly want to filter using a list with isin then ....
+------+
| color|
+------+
| red|
|yellow|
+------+

Answer 4

One more way using array_contains and swapping the arguments.
scala> val x = ("parent,child,sub-child").split(",")
x: Array[String] = Array(parent, child, sub-child)
scala> val df = Seq(("parent"),("grand-parent"),("child"),("sub-child"),("cousin")).toDF("root")
df: org.apache.spark.sql.DataFrame = [root: string]
scala> df.show
+------------+
| root|
+------------+
| parent|
|grand-parent|
| child|
| sub-child|
| cousin|
+------------+
scala> df.withColumn("check", array_contains(lit(x),'root)).show
+------------+-----+
| root|check|
+------------+-----+
| parent| true|
|grand-parent|false|
| child| true|
| sub-child| true|
| cousin|false|
+------------+-----+
scala>

Answer 5

Here are my two cents
val filters = List(1,2,3)
val data = List(5,1,2,1,3,3,2,1,4)
val colName = "number"
val df = spark.
sparkContext.
parallelize(data).
toDF(colName).
filter(
r => filters.contains(r.getAs[Int](colName))
)
df.show()
which results in
+------+
|number|
+------+
| 1|
| 2|
| 1|
| 3|
| 3|
| 2|
| 1|
+------+