Use rlike with regex column in spark 1.5.1 - scala

I want to filter dataframe based on applying regex values in one of the columns to another column.
Example:
Id Column1 RegexColumm
1 Abc A.*
2 Def B.*
3 Ghi G.*
The result of filtering dataframe using RegexColumm should give rows with id 1 and 3.
Is there a way to do this in spark 1.5.1? Don't want to use UDF as this might cause scalability issues, looking for spark native api.

You can convert df -> rdd then by traversing through row we can match the regex and filter out only the matching data without using any UDF.
Example:
import org.apache.spark.sql.types._
import org.apache.spark.sql._
import org.apache.spark.sql.functions._
df.show()
//+---+-------+--------+
//| id|column1|regexCol|
//+---+-------+--------+
//| 1| Abc| A.*|
//| 2| Def| B.*|
//| 3| Ghi| G.*|
//+---+-------+--------+
//creating new schema to add new boolean field
val sch = StructType(df.schema.fields ++ Array(StructField("bool_col", BooleanType, false)))
//convert df to rdd and match the regex using .map
val rdd = df.rdd.map(row => {
val regex = row.getAs[String]("regexCol")
val bool = row.getAs[String]("column1").matches(regex)
val bool_col = s"$bool".toBoolean
val newRow = Row.fromSeq(row.toSeq ++ Array(bool_col))
newRow
})
//convert rdd to dataframe filter out true values for bool_col
val final_df = sqlContext.createDataFrame(rdd, sch).where(col("bool_col")).drop("bool_col")
final_df.show(10)
//+---+-------+--------+
//| id|column1|regexCol|
//+---+-------+--------+
//| 1| Abc| A.*|
//| 3| Ghi| G.*|
//+---+-------+--------+
UPDATE:
Instead of .map we can use .mapPartition (map vs mapPartiiton):
val rdd = df.rdd.mapPartitions(
partitions => {
partitions.map(row => {
val regex = row.getAs[String]("regexCol")
val bool = row.getAs[String]("column1").matches(regex)
val bool_col = s"$bool".toBoolean
val newRow = Row.fromSeq(row.toSeq ++ Array(bool_col))
newRow
})
})

scala> val df = Seq((1,"Abc","A.*"),(2,"Def","B.*"),(3,"Ghi","G.*")).toDF("id","Column1","RegexColumm")
df: org.apache.spark.sql.DataFrame = [id: int, Column1: string ... 1 more field]
scala> val requiredDF = df.filter(x=> x.getAs[String]("Column1").matches(x.getAs[String]("RegexColumm")))
requiredDF: org.apache.spark.sql.Dataset[org.apache.spark.sql.Row] = [id: int, Column1: string ... 1 more field]
scala> requiredDF.show
+---+-------+-----------+
| id|Column1|RegexColumm|
+---+-------+-----------+
| 1| Abc| A.*|
| 3| Ghi| G.*|
+---+-------+-----------+
You can use like above, I think this is what you are lioking for. Please do let me know if it helps you.

Related

How to add a new column to my DataFrame such that values of new column are populated by some other function in scala?

myFunc(Row): String = {
//process row
//returns string
}
appendNewCol(inputDF : DataFrame) : DataFrame ={
inputDF.withColumn("newcol",myFunc(Row))
inputDF
}
But no new column got created in my case. My myFunc passes this row to a knowledgebasesession object and that returns a string after firing rules. Can I do it this way? If not, what is the right way? Thanks in advance.
I saw many StackOverflow solutions using expr() sqlfunc(col(udf(x)) and other techniques but here my newcol is not derived directly from existing column.
Dataframe:
import org.apache.spark.sql.Row
import org.apache.spark.sql.types.{StringType, StructField, StructType}
val myFunc = (r: Row) => {r.getAs[String]("col1") + "xyz"} // example transformation
val testDf = spark.sparkContext.parallelize(Seq(
(1, "abc"), (2, "def"), (3, "ghi"))).toDF("id", "col1")
testDf.show
val rddRes = testDf
.rdd
.map{x =>
val y = myFunc (x)
Row.fromSeq (x.toSeq ++ Seq(y) )
}
val newSchema = StructType(testDf.schema.fields ++ Array(StructField("col2", dataType =StringType, nullable =false)))
spark.sqlContext.createDataFrame(rddRes, newSchema).show
Results:
+---+----+
| id|col1|
+---+----+
| 1| abc|
| 2| def|
| 3| ghi|
+---+----+
+---+----+------+
| id|col1| col2|
+---+----+------+
| 1| abc|abcxyz|
| 2| def|defxyz|
| 3| ghi|ghixyz|
+---+----+------+
With Dataset:
case class testData(id: Int, col1: String)
case class transformedData(id: Int, col1: String, col2: String)
val test: Dataset[testData] = List(testData(1, "abc"), testData(2, "def"), testData(3, "ghi")).toDS
val transformedData: Dataset[transformedData] = test
.map { x: testData =>
val newCol = x.col1 + "xyz"
transformedData(x.id, x.col1, newCol)
}
transformedData.show
As you can see datasets is more readable, plus provides strong type casting.
Since I'm unaware of your spark version, providing both solutions here. However if you're using spark v>=1.6, you should look into Datasets. Playing with rdd is fun, but can quickly devolve into longer job runs and a host of other issues that you wont foresee

Add new column containing an Array of column names sorted by the row-wise values

Given a dataFrame with a few columns, I'm trying to create a new column containing an array of these columns' names sorted by decreasing order, based on the row-wise values of these columns.
| a | b | c | newcol|
|---|---|---|-------|
| 1 | 4 | 3 |[b,c,a]|
| 4 | 1 | 3 |[a,c,b]|
---------------------
The names of the columns are stored in a var names:Array[String]
What approach should I go for?
Using UDF is most simple way to achieve custom tasks here.
val df = spark.createDataFrame(Seq((1,4,3), (4,1,3))).toDF("a", "b", "c")
val names=df.schema.fieldNames
val sortNames = udf((v: Seq[Int]) => {v.zip(names).sortBy(_._1).map(_._2)})
df.withColumn("newcol", sortNames(array(names.map(col): _*))).show
Something like this can be an approach using Dataset:
case class Element(name: String, value: Int)
case class Columns(a: Int, b: Int, c: Int, elements: Array[String])
def function1()(implicit spark: SparkSession) = {
import spark.implicits._
val df0: DataFrame =
spark.createDataFrame(spark.sparkContext
.parallelize(Seq(Row(1, 2, 3), Row(4, 1, 3))),
StructType(Seq(StructField("a", IntegerType, false),
StructField("b", IntegerType, false),
StructField("c", IntegerType, false))))
val df1 = df0
.flatMap(row => Seq(Columns(row.getAs[Int]("a"),
row.getAs[Int]("b"),
row.getAs[Int]("c"),
Array(Element("a", row.getAs[Int]("a")),
Element("b", row.getAs[Int]("b")),
Element("c", row.getAs[Int]("c"))).sortBy(-_.value).map(_.name))))
df1
}
def main(args: Array[String]) : Unit = {
implicit val spark = SparkSession.builder().master("local[1]").getOrCreate()
function1().show()
}
gives:
+---+---+---+---------+
| a| b| c| elements|
+---+---+---+---------+
| 1| 2| 3|[a, b, c]|
| 4| 1| 3|[b, c, a]|
+---+---+---+---------+
Try something like this:
val sorted_column_names = udf((column_map: Map[String, Int]) =>
column_map.toSeq.sortBy(- _._2).map(_._1)
)
df.withColumn("column_map", map(lit("a"), $"a", lit("b"), $"b", lit("c"), $"c")
.withColumn("newcol", sorted_column_names($"column_map"))

How to append collection as new column to DataFrame with many columns?

I'd like to append (add) a new column to an existing dataframe with multiple columns.
val a = Seq(
("10", "MILLER", "1300", "2017-11-03"),
("30", "Martin", "1250", "2017-11-21")).toDF("dept_no","emp_name","sal","date")
scala> a.show
+-------+--------+----+----------+
|dept_no|emp_name| sal| date|
+-------+--------+----+----------+
| 10| MILLER|1300|2017-11-03|
| 30| Martin|1250|2017-11-21|
+-------+--------+----+----------+
With the above dataframe I'd like to add every element of a collection (be it a regular Scala collection or another single-column DataFrame), e.g.
val lst = List("10", "Susan")
How to add the elements of lst above to the rows of a dataframe (one element per row)?
Let's convert lst to a DataFrame:
val lst = List("10", "Susan").toDF
You can use zip method of RDD:
import org.apache.spark.sql.Row
val data = a.rdd.zip(lst.rdd).map { case (l, r) => Row.fromSeq(l.toSeq ++ r.toSeq) }
import org.apache.spark.sql.types.StructType
val schema = StructType(a.schema.fields ++ lst.schema.fields)
val solution = spark.createDataFrame(data, schema)
scala> solution.show
+-------+--------+----+----------+-----+
|dept_no|emp_name| sal| date|value|
+-------+--------+----+----------+-----+
| 10| MILLER|1300|2017-11-03| 10|
| 30| Martin|1250|2017-11-21|Susan|
+-------+--------+----+----------+-----+

Spark: reduce/aggregate by key

I am new to Spark and Scala, so I have no idea how this kind of problem is called (which makes searching for it pretty hard).
I have data of the following structure:
[(date1, (name1, 1)), (date1, (name1, 1)), (date1, (name2, 1)), (date2, (name3, 1))]
In some way, this has to be reduced/aggregated to:
[(date1, [(name1, 2), (name2, 1)]), (date2, [(name3, 1)])]
I know how to do reduceByKey on a list of key-value pairs, but this particular problem is a mystery to me.
Thanks in advance!
My data, but here goes, step-wise:
val rdd1 = sc.makeRDD(Array( ("d1",("A",1)), ("d1",("A",1)), ("d1",("B",1)), ("d2",("E",1)) ),2)
val rdd2 = rdd1.map(x => ((x._1, x._2._1), x._2._2))
val rdd3 = rdd2.groupByKey
val rdd4 = rdd3.map{
case (str, nums) => (str, nums.sum)
}
val rdd5 = rdd4.map(x => (x._1._1, (x._1._2, x._2))).groupByKey
rdd5.collect
returns:
res28: Array[(String, Iterable[(String, Int)])] = Array((d2,CompactBuffer((E,1))), (d1,CompactBuffer((A,2), (B,1))))
Better approach avoiding groupByKey is as follows:
val rdd1 = sc.makeRDD(Array( ("d1",("A",1)), ("d1",("A",1)), ("d1",("B",1)), ("d2",("E",1)) ),2)
val rdd2 = rdd1.map(x => ((x._1, x._2._1), (x._2._2))) // Need to add quotes around V part for reduceByKey
val rdd3 = rdd2.reduceByKey(_+_)
val rdd4 = rdd3.map(x => (x._1._1, (x._1._2, x._2))).groupByKey // Necessary Shuffle
rdd4.collect
As I stated in the columns it can be done with DataFrames for structured data, so run this below:
// This above should be enough.
import org.apache.spark.sql.expressions._
import org.apache.spark.sql.functions._
val rddA = sc.makeRDD(Array( ("d1","A",1), ("d1","A",1), ("d1","B",1), ("d2","E",1) ),2)
val dfA = rddA.toDF("c1", "c2", "c3")
val dfB = dfA
.groupBy("c1", "c2")
.agg(sum("c3").alias("sum"))
dfB.show
returns:
+---+---+---+
| c1| c2|sum|
+---+---+---+
| d1| A| 2|
| d2| E| 1|
| d1| B| 1|
+---+---+---+
But you can do this to approximate the above of the CompactBuffer above.
import org.apache.spark.sql.functions.{col, udf}
case class XY(x: String, y: Long)
val xyTuple = udf((x: String, y: Long) => XY(x, y))
val dfC = dfB
.withColumn("xy", xyTuple(col("c2"), col("sum")))
.drop("c2")
.drop("sum")
dfC.printSchema
dfC.show
// Then ... this gives you the CompactBuffer answer but from a DF-perspective
val dfD = dfC.groupBy(col("c1")).agg(collect_list(col("xy")))
dfD.show
returns - some renaming req'd and possible sorting:
---+----------------+
| c1|collect_list(xy)|
+---+----------------+
| d2| [[E, 1]]|
| d1|[[A, 2], [B, 1]]|
+---+----------------+

Spark - How to convert map function output (Row,Row) tuple to one Dataframe

I need to write one scenario in Spark using Scala API.
I am passing a user defined function to a Dataframe which processes each row of data frame one by one and returns tuple(Row, Row). How can i change RDD ( Row, Row) to Dataframe (Row)? See below code sample -
**Calling map function-**
val df_temp = df_outPut.map { x => AddUDF.add(x,date1,date2)}
**UDF definition.**
def add(x: Row,dates: String*): (Row,Row) = {
......................
........................
var result1,result2:Row = Row()
..........
return (result1,result2)
Now df_temp is a RDD(Row1, Row2). my requirement is to make it one RDD or Dataframe by breaking tuple elements to 1 record of RDD or Dataframe
RDD(Row). Appreciate your help.
You can use flatMap to flatten your Row tuples, say if we start from this example rdd:
rddExample.collect()
// res37: Array[(org.apache.spark.sql.Row, org.apache.spark.sql.Row)] = Array(([1,2],[3,4]), ([2,1],[4,2]))
val flatRdd = rddExample.flatMap{ case (x, y) => List(x, y) }
// flatRdd: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[45] at flatMap at <console>:35
To convert it to data frame.
import org.apache.spark.sql.types.{StructType, StructField, IntegerType}
val schema = StructType(StructField("x", IntegerType, true)::
StructField("y", IntegerType, true)::Nil)
val df = sqlContext.createDataFrame(flatRdd, schema)
df.show
+---+---+
| x| y|
+---+---+
| 1| 2|
| 3| 4|
| 2| 1|
| 4| 2|
+---+---+