Spark foldbykey doubt

Question

val a= spark.sparkContext.parallelize(Array(("a",1),("a",2),("b",2)))
val b =a.foldByKey(1)(_+_)

scala> b.collect
res2: Array[(String, Int)] = Array((b,3), (a,5))

Can someone tell me why a value is 5 not 4?

Answer 1

Please have a look below for your reference.

(a,1) (a,2) => foldByKey(1)(_+_) => (a,1+1)+(a,2+1) => 2+3 = 5

(b,2) => foldByKey(1)(_+_) => (b,2+1) = 3

According to that logic, the value is 5. 

Comments

val a= spark.sparkContext.parallelize(Array(("a",1),("a",2),("b",2),("a",2)))

a: org.apache.spark.rdd.RDD[(String, Int)] = ParallelCollectionRDD[0] at parallelize at <console>:23

scala> val b =a.foldByKey(1)(_+_).collect

b: Array[(String, Int)] = Array((b,3), (a,7))

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scala> val a= spark.sparkContext.parallelize(Array(("a",1),("a",2),("b",2),("b",3),("a",5)))

a: org.apache.spark.rdd.RDD[(String, Int)] = ParallelCollectionRDD[9] at parallelize at <console>:23

scala> val b =a.foldByKey(1)(_+_)

b: org.apache.spark.rdd.RDD[(String, Int)] = ShuffledRDD[10] at foldByKey at <console>:25

scala> b.collect

res6: Array[(String, Int)] = Array((b,6), (a,10))

Q> Can anyone clarify me how the result is  Array((b,6), (a,10)) instead of  Array((b,7), (a,11))?

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Hey, @Sitaram,

According to calculations, the result will be Array((b,6),(a,10), if you follow the above answer.