There are two rdds.
val pairRDD1 = sc.parallelize(List( ("cat",2), ("girl", 5), ("book", 4),("Tom", 12)))
val pairRDD2 = sc.parallelize(List( ("cat",2), ("cup", 5), ("mouse", 4),("girl", 12)))
And then I will do this join operation.
val kk = pairRDD1.fullOuterJoin(pairRDD2).collect
it shows like that:
kk: Array[(String, (Option[Int], Option[Int]))] = Array((book,(Some(4),None)), (Tom,(Some(12),None)), (girl,(Some(5),Some(12))), (mouse,(None,Some(4))), (cup,(None,Some(5))), (cat,(Some(2),Some(2))))
if i would like to fill the NONE by 0 and transform Option[int] to Int.what should I code?Thanks!
You can use mapValues on kk as follows (note this is before the collect):
pairRDD1.fullOuterJoin(pairRDD2).mapValues(pair => (pair._1.getOrElse(0), pair._2.getOrElse(0)))
You might have to do this before collect in an RDD, otherwise you could do:
kk.map { case (k, pair) => (k, (pair._1.getOrElse(0), pair._2.getOrElse(0))) }
Based on commnets in first answer, if you are fine using DataFrames, you can do with dataframes with any number of columns.
val ss = SparkSession.builder().master("local[*]").getOrCreate()
val sc = ss.sparkContext
import ss.implicits._
val pairRDD1 = sc.parallelize(List(("cat", 2,9999), ("girl", 5,8888), ("book", 4,9999), ("Tom", 12,6666)))
val pairRDD2 = sc.parallelize(List(("cat", 2,9999), ("cup", 5,7777), ("mouse", 4,3333), ("girl", 12,1111)))
val df1 = pairRDD1.toDF
val df2 = pairRDD2.toDF
val joined = df1.join(df2, df1.col("_1") === df2.col("_1"),"fullouter")
joined.show()
Here _1,_2 e.t.c are default column names provided by Spark. But, if you wish to have proper names you can change it as you wish.
Result:
+----+----+----+-----+----+----+
| _1| _2| _3| _1| _2| _3|
+----+----+----+-----+----+----+
|girl| 5|8888| girl| 12|1111|
| Tom| 12|6666| null|null|null|
| cat| 2|9999| cat| 2|9999|
|null|null|null| cup| 5|7777|
|null|null|null|mouse| 4|3333|
|book| 4|9999| null|null|null|
+----+----+----+-----+----+----+
Related
So, I have 2 lists in Spark(scala). They both contain the same number of values. The first list a contains all strings and the second list b contains all Long's.
a: List[String] = List("a", "b", "c", "d")
b: List[Long] = List(17625182, 17625182, 1059731078, 100)
I also have a schema defined as follows:
val schema2=StructType(
Array(
StructField("check_name", StringType, true),
StructField("metric", DecimalType(38,0), true)
)
)
What is the best way to convert my lists to a single dataframe, that has schema schema2 and the columns are made from a and b respectively?
You can create an RDD[Row] and convert to Spark dataframe with the given schema:
val df = spark.createDataFrame(
sc.parallelize(a.zip(b).map(x => Row(x._1, BigDecimal(x._2)))),
schema2
)
df.show
+----------+----------+
|check_name| metric|
+----------+----------+
| a| 17625182|
| b| 17625182|
| c|1059731078|
| d| 100|
+----------+----------+
Using Dataset:
import spark.implicits._
case class Schema2(a: String, b: Long)
val el = (a zip b) map { case (a, b) => Schema2(a, b)}
val df = spark.createDataset(el).toDF()
I am new to Scala and Spark .
There are 2 RDDs like
RDD_A= (keyA,5),(KeyB,10)
RDD_B= (keyA,3),(KeyB,7)
how do I calculate : RDD_A-RDD_B so that I get (keyA,2),(KeyB,3)
I tried subtract and subtractByKey but I am unable to get similar output like above
Let's assume that each RDD has only one value with specified key:
val df =
Seq(
("A", 5),
("B", 10)
).toDF("key", "value")
val df2 =
Seq(
("A", 3),
("B", 7)
).toDF("key", "value")
You can merge these RDDs using union and perform the computation via groupBy as follows:
import org.apache.spark.sql.functions._
df.union(df2)
.groupBy("key")
.agg(first("value").minus(last("value")).as("value"))
.show()
will print:
+---+-----+
|key|value|
+---+-----+
| B| 3|
| A| 2|
+---+-----+
RDD solution for the question
Please find inline code comments for the explanation
object SubtractRDD {
def main(args: Array[String]): Unit = {
val spark = SparkSession.builder().master("local[*]").getOrCreate(); // Create Spark Session
val list1 = List(("keyA",5),("keyB",10))
val list2 = List(("keyA",3),("keyB",7))
val rdd1= spark.sparkContext.parallelize(list1) // convert list to RDD
val rdd2= spark.sparkContext.parallelize(list2)
val result = rdd1.join(rdd2) // Inner join RDDs
.map(x => (x._1, x._2._1 - x._2._2 )) // Combiner function for RDDs
.collectAsMap() // Collect result as Map
println(result)
}
}
I have a spark Dataframe like Below.I'm trying to split the column into 2 more columns:
date time content
28may 11am [ssid][customerid,shopid]
val personDF2 = personDF.withColumn("temp",split(col("content"),"\\[")).select(
col("*") +: (0 until 3).map(i => col("temp").getItem(i).as(s/col$i)): _*)
date time content col1 col2 col3
28may 11 [ssid][customerid,shopid] ssid customerid shopid
Assuming a String to represent an Array of Words. Got your request. You can optimize the number of dataframes as well to reduce load on system. If there are more than 9 cols etc. you may need to use c00, c01, etc. for c10 etc. Or just use integer as name for columns. leave that up to you.
import org.apache.spark.sql.functions._
import scala.collection.mutable.WrappedArray
// Set up data
val df = spark.sparkContext.parallelize(Seq(
("A", "[foo][customerid,shopid][Donald,Trump,Esq][single]"),
("B", "[foo]")
)).toDF("k", "v")
val df2 = df.withColumn("words_temp", regexp_replace($"v", lit("]"), lit("" )))
val df3 = df2.withColumn("words_temp2", regexp_replace($"words_temp", lit(","), lit("[" ))).drop("words_temp")
val df4 = df3.withColumn("words_temp3", expr("substring(words_temp2, 2, length(words_temp2))")).withColumn("cnt", expr("length(words_temp2)")).drop("words_temp2")
val df5 = df4.withColumn("words",split(col("words_temp3"),"\\[")).drop("words_temp3")
val df6 = df5.withColumn("num_words", size($"words"))
val df7 = df6.withColumn("v2", explode($"words"))
// Convert to Array of sorts via group by
val df8 = df7.groupBy("k")
.agg(collect_list("v2"))
// Convert to rdd Tuple and then find position so as to gen col names! That is the clue so as to be able to use pivot
val rdd = df8.rdd
val rdd2 = rdd.map(row => (row.getAs[String](0), row.getAs[WrappedArray[String]](1).toArray))
val rdd3 = rdd2.map { case (k, list) => (k, list.zipWithIndex) }
val df9 = rdd3.toDF("k", "v")
val df10 = df9.withColumn("vn", explode($"v"))
val df11 = df10.select($"k", $"vn".getField("_1"), concat(lit("c"),$"vn".getField("_2"))).toDF("k", "v", "c")
// Final manipulation
val result = df11.groupBy("k")
.pivot("c")
.agg(expr("coalesce(first(v),null)")) // May never occur in your case, just done for completeness and variable length cols.
result.show(100,false)
returns in this case:
+---+---+----------+------+------+-----+----+------+
|k |c0 |c1 |c2 |c3 |c4 |c5 |c6 |
+---+---+----------+------+------+-----+----+------+
|B |foo|null |null |null |null |null|null |
|A |foo|customerid|shopid|Donald|Trump|Esq |single|
+---+---+----------+------+------+-----+----+------+
Update:
Based on original title stating array of words. See other answer.
If new, then a few things here. Can also be done with dataset and map I assume. Here is a solution using DFs and rdd's. I might well investigate a complete DS in future, but this works for sure and at scale.
// Can amalgamate more steps
import org.apache.spark.sql.functions._
import scala.collection.mutable.WrappedArray
// Set up data
val df = spark.sparkContext.parallelize(Seq(
("A", Array(Array("foo", "bar"), Array("Donald", "Trump","Esq"), Array("single"))),
("B", Array(Array("foo2", "bar2"), Array("single2"))),
("C", Array(Array("foo3", "bar3", "x", "y", "z")))
)).toDF("k", "v")
// flatten via 2x explode, can be done more elegeantly with def or UDF, but keeping it simple here
val df2 = df.withColumn("v2", explode($"v"))
val df3 = df2.withColumn("v3", explode($"v2"))
// Convert to Array of sorts via group by
val df4 = df3.groupBy("k")
.agg(collect_list("v3"))
// Convert to rdd Tuple and then find position so as to gen col names! That is the clue so as to be able to use pivot
val rdd = df4.rdd
val rdd2 = rdd.map(row => (row.getAs[String](0), row.getAs[WrappedArray[String]](1).toArray))
val rdd3 = rdd2.map { case (k, list) => (k, list.zipWithIndex) }
val df5 = rdd3.toDF("k", "v")
val df6 = df5.withColumn("vn", explode($"v"))
val df7 = df6.select($"k", $"vn".getField("_1"), concat(lit("c"),$"vn".getField("_2"))).toDF("k", "v", "c")
// Final manipulation
val result = df7.groupBy("k")
.pivot("c")
.agg(expr("coalesce(first(v),null)")) // May never occur in your case, just done for completeness and variable length cols.
result.show(100,false)
returns in correct col order:
+---+----+----+-------+-----+----+------+
|k |c0 |c1 |c2 |c3 |c4 |c5 |
+---+----+----+-------+-----+----+------+
|B |foo2|bar2|single2|null |null|null |
|C |foo3|bar3|x |y |z |null |
|A |foo |bar |Donald |Trump|Esq |single|
+---+----+----+-------+-----+----+------+
I am using Spark 1.60 and Scala 2.10.5
I have a dataframe like this,
+------------------+
|id | needed |
+------------------+
|1 | 2 |
|1 | 0 |
|1 | 3 |
|2 | 0 |
|2 | 0 |
|3 | 1 |
|3 | 2 |
+------------------+
From this df I created an rdd like this,
val dfRDD = df.rdd
from my rdd, I want to group by id and count of needed is > 0.
((1, 2), (2,0), (3,2))
So, I tried like this,
val groupedDF = dfRDD.map(x =>(x(0), x(1) > 0)).count.redueByKey(_+_)
In this case, I am getting an error:
error: value > is not a member of any
I need that in rdd level. Any help to get my desired output would be great.
The problem is that in your map you're calling the apply method of Row, and as you can see in its scaladoc, that method returns Any - and as you can see for the error and from the scaladoc there is not such method < in Any
You can fix it using the getAs[T] method.
import org.apache.spark.rdd.RDD
import org.apache.spark.sql.SparkSession
val spark =
SparkSession
.builder
.master("local[*]")
.getOrCreate()
import spark.implicits._
val df =
List(
(1, 2),
(1, 0),
(1, 3),
(2, 0),
(2, 0),
(3, 1),
(3, 2)
).toDF("id", "needed")
val rdd: RDD[(Int, Int)] = df.rdd.map(row => (row.getAs[Int](fieldName = "id"), row.getAs[Int](fieldName = "needed")))
From there you can continue with the aggregation, you have a few mistakes in your logic.
First, you don't need the count call.
And second, if you need to count the amount of times "needed" was greater than one you can't do _ + _, because that is the sum of needed values.
val grouped: RDD[(Int, Int)] = rdd.reduceByKey { (acc, v) => if (v > 0) acc + 1 else acc }
val result: Array[(Int, Int)] = grouped.collect()
// Array((1,3), (2,0), (3,2))
PS: You should tell your professor to upgrade to Spark 2 and Scala 2.11 ;)
Edit
Using case classes in the above example.
final case class Data(id: Int, needed: Int)
val rdd: RDD[Data] = df.as[Data].rdd
val grouped: RDD[(Int, Int)] = rdd.map(d => d.id -> d.needed).reduceByKey { (acc, v) => if (v > 0) acc + 1 else acc }
val result: Array[(Int, Int)] = grouped.collect()
// Array((1,3), (2,0), (3,2))
There's no need to do the calculation at the rdd level. Aggregation with the data frame should work:
df.groupBy("id").agg(sum(($"needed" > 0).cast("int")).as("positiveCount")).show
+---+-------------+
| id|positiveCount|
+---+-------------+
| 1| 2|
| 3| 2|
| 2| 0|
+---+-------------+
If you have to work with RDD, use row.getInt or as #Luis' answer row.getAs[Int] to get the value with explicit type, and then do the comparison and reduceByKey:
df.rdd.map(r => (r.getInt(0), if (r.getInt(1) > 0) 1 else 0)).reduceByKey(_ + _).collect
// res18: Array[(Int, Int)] = Array((1,2), (2,0), (3,2))
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]]|
+---+----------------+