Not serializable result: org.apache.hadoop.io.IntWritable when reading Sequence File with Spark / Scala









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0
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Reading a sequence file with Int and String logically,



then if I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x.toString(), y.toString().split("/")(0), y.toString().split("/")(1))
.collect


this is ok as the IntWritable is converted to String.



If I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x, y.toString().split("/")(0), y.toString().split("/")(1))
.collect


then I get this error immediately:



org.apache.spark.SparkException: Job aborted due to stage failure: Task 5.0 in stage 42.0 (TID 692) had a not serializable result: org.apache.hadoop.io.IntWritable


Underlying reason is not really clear - serialization, but why so difficult? This is another type of serialization aspect I note. Also it is only noted at run-time.










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  • 1




    Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
    – cricket_007
    Nov 10 at 14:10










  • So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
    – thebluephantom
    Nov 10 at 14:11






  • 1




    No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
    – cricket_007
    Nov 10 at 14:13










  • I took the get() out in this example. Ok, gotcha!
    – thebluephantom
    Nov 10 at 14:14










  • You may as well make an answer
    – thebluephantom
    Nov 10 at 14:14














up vote
0
down vote

favorite












Reading a sequence file with Int and String logically,



then if I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x.toString(), y.toString().split("/")(0), y.toString().split("/")(1))
.collect


this is ok as the IntWritable is converted to String.



If I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x, y.toString().split("/")(0), y.toString().split("/")(1))
.collect


then I get this error immediately:



org.apache.spark.SparkException: Job aborted due to stage failure: Task 5.0 in stage 42.0 (TID 692) had a not serializable result: org.apache.hadoop.io.IntWritable


Underlying reason is not really clear - serialization, but why so difficult? This is another type of serialization aspect I note. Also it is only noted at run-time.










share|improve this question



















  • 1




    Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
    – cricket_007
    Nov 10 at 14:10










  • So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
    – thebluephantom
    Nov 10 at 14:11






  • 1




    No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
    – cricket_007
    Nov 10 at 14:13










  • I took the get() out in this example. Ok, gotcha!
    – thebluephantom
    Nov 10 at 14:14










  • You may as well make an answer
    – thebluephantom
    Nov 10 at 14:14












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Reading a sequence file with Int and String logically,



then if I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x.toString(), y.toString().split("/")(0), y.toString().split("/")(1))
.collect


this is ok as the IntWritable is converted to String.



If I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x, y.toString().split("/")(0), y.toString().split("/")(1))
.collect


then I get this error immediately:



org.apache.spark.SparkException: Job aborted due to stage failure: Task 5.0 in stage 42.0 (TID 692) had a not serializable result: org.apache.hadoop.io.IntWritable


Underlying reason is not really clear - serialization, but why so difficult? This is another type of serialization aspect I note. Also it is only noted at run-time.










share|improve this question















Reading a sequence file with Int and String logically,



then if I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x.toString(), y.toString().split("/")(0), y.toString().split("/")(1))
.collect


this is ok as the IntWritable is converted to String.



If I do this:



val sequence_data = sc.sequenceFile("/seq_01/seq-directory/*", classOf[IntWritable], classOf[Text])
.mapcase (x, y) => (x, y.toString().split("/")(0), y.toString().split("/")(1))
.collect


then I get this error immediately:



org.apache.spark.SparkException: Job aborted due to stage failure: Task 5.0 in stage 42.0 (TID 692) had a not serializable result: org.apache.hadoop.io.IntWritable


Underlying reason is not really clear - serialization, but why so difficult? This is another type of serialization aspect I note. Also it is only noted at run-time.







apache-spark hadoop serialization sequencefile






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edited Nov 10 at 14:11









cricket_007

75.9k1042106




75.9k1042106










asked Nov 10 at 14:06









thebluephantom

2,0632823




2,0632823







  • 1




    Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
    – cricket_007
    Nov 10 at 14:10










  • So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
    – thebluephantom
    Nov 10 at 14:11






  • 1




    No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
    – cricket_007
    Nov 10 at 14:13










  • I took the get() out in this example. Ok, gotcha!
    – thebluephantom
    Nov 10 at 14:14










  • You may as well make an answer
    – thebluephantom
    Nov 10 at 14:14












  • 1




    Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
    – cricket_007
    Nov 10 at 14:10










  • So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
    – thebluephantom
    Nov 10 at 14:11






  • 1




    No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
    – cricket_007
    Nov 10 at 14:13










  • I took the get() out in this example. Ok, gotcha!
    – thebluephantom
    Nov 10 at 14:14










  • You may as well make an answer
    – thebluephantom
    Nov 10 at 14:14







1




1




Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
– cricket_007
Nov 10 at 14:10




Well, Strings are subclasses from Serializable, IntWritable is not. hadoop.apache.org/docs/r2.7.4/api/org/apache/hadoop/io/…
– cricket_007
Nov 10 at 14:10












So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
– thebluephantom
Nov 10 at 14:11




So reading a sequence file means all aspects should be set to String and then processed and converted from there onwards?
– thebluephantom
Nov 10 at 14:11




1




1




No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
– cricket_007
Nov 10 at 14:13




No, you can x.get(), then Integers are serializable as well (and less overhead of serializing strings)
– cricket_007
Nov 10 at 14:13












I took the get() out in this example. Ok, gotcha!
– thebluephantom
Nov 10 at 14:14




I took the get() out in this example. Ok, gotcha!
– thebluephantom
Nov 10 at 14:14












You may as well make an answer
– thebluephantom
Nov 10 at 14:14




You may as well make an answer
– thebluephantom
Nov 10 at 14:14












1 Answer
1






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1
down vote



accepted










If the goal is to just get an Integer value, you would need to call a get on the writable



.map{case (x, y) => (x.get()


And then the JVM handles serialization of the Integer object rather than not knowing how to process a IntWritable because it doesn't implement the Serializable interface



String does implement Serializable






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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    If the goal is to just get an Integer value, you would need to call a get on the writable



    .map{case (x, y) => (x.get()


    And then the JVM handles serialization of the Integer object rather than not knowing how to process a IntWritable because it doesn't implement the Serializable interface



    String does implement Serializable






    share|improve this answer
























      up vote
      1
      down vote



      accepted










      If the goal is to just get an Integer value, you would need to call a get on the writable



      .map{case (x, y) => (x.get()


      And then the JVM handles serialization of the Integer object rather than not knowing how to process a IntWritable because it doesn't implement the Serializable interface



      String does implement Serializable






      share|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        If the goal is to just get an Integer value, you would need to call a get on the writable



        .map{case (x, y) => (x.get()


        And then the JVM handles serialization of the Integer object rather than not knowing how to process a IntWritable because it doesn't implement the Serializable interface



        String does implement Serializable






        share|improve this answer












        If the goal is to just get an Integer value, you would need to call a get on the writable



        .map{case (x, y) => (x.get()


        And then the JVM handles serialization of the Integer object rather than not knowing how to process a IntWritable because it doesn't implement the Serializable interface



        String does implement Serializable







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 10 at 14:16









        cricket_007

        75.9k1042106




        75.9k1042106



























             

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