Why doesn't dataset's foreach method require an encoder, but map does?











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I have two datasets: Dataset[User] and Dataset[Book] where both User and Book are case classes. I join them like this:



val joinDS = ds1.join(ds2, "userid")



If I try to map over each element in joinDS, the compiler complains that an encoder is missing:




not enough arguments for method map: (implicit evidence$46: org.apache.spark.sql.Encoder[Unit])org.apache.spark.sql.Dataset[Unit].
Unspecified value parameter evidence$46.
Unable to find encoder for type stored in a Dataset.



But the same error does not occur if I use foreach instead of map. Why doesn't foreach require an encoder as well? I have imported all implicits from the spark session already, so why does map require an encoder at all, when the dataset is a result of joining two datasets containing case classes)? Also, what type of dataset do I get from that join? Is it a Dataset[Row], or something else?










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  • Pretty sure you can't encode Unit.
    – erip
    Nov 8 at 18:15















up vote
1
down vote

favorite












I have two datasets: Dataset[User] and Dataset[Book] where both User and Book are case classes. I join them like this:



val joinDS = ds1.join(ds2, "userid")



If I try to map over each element in joinDS, the compiler complains that an encoder is missing:




not enough arguments for method map: (implicit evidence$46: org.apache.spark.sql.Encoder[Unit])org.apache.spark.sql.Dataset[Unit].
Unspecified value parameter evidence$46.
Unable to find encoder for type stored in a Dataset.



But the same error does not occur if I use foreach instead of map. Why doesn't foreach require an encoder as well? I have imported all implicits from the spark session already, so why does map require an encoder at all, when the dataset is a result of joining two datasets containing case classes)? Also, what type of dataset do I get from that join? Is it a Dataset[Row], or something else?










share|improve this question






















  • Pretty sure you can't encode Unit.
    – erip
    Nov 8 at 18:15













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have two datasets: Dataset[User] and Dataset[Book] where both User and Book are case classes. I join them like this:



val joinDS = ds1.join(ds2, "userid")



If I try to map over each element in joinDS, the compiler complains that an encoder is missing:




not enough arguments for method map: (implicit evidence$46: org.apache.spark.sql.Encoder[Unit])org.apache.spark.sql.Dataset[Unit].
Unspecified value parameter evidence$46.
Unable to find encoder for type stored in a Dataset.



But the same error does not occur if I use foreach instead of map. Why doesn't foreach require an encoder as well? I have imported all implicits from the spark session already, so why does map require an encoder at all, when the dataset is a result of joining two datasets containing case classes)? Also, what type of dataset do I get from that join? Is it a Dataset[Row], or something else?










share|improve this question













I have two datasets: Dataset[User] and Dataset[Book] where both User and Book are case classes. I join them like this:



val joinDS = ds1.join(ds2, "userid")



If I try to map over each element in joinDS, the compiler complains that an encoder is missing:




not enough arguments for method map: (implicit evidence$46: org.apache.spark.sql.Encoder[Unit])org.apache.spark.sql.Dataset[Unit].
Unspecified value parameter evidence$46.
Unable to find encoder for type stored in a Dataset.



But the same error does not occur if I use foreach instead of map. Why doesn't foreach require an encoder as well? I have imported all implicits from the spark session already, so why does map require an encoder at all, when the dataset is a result of joining two datasets containing case classes)? Also, what type of dataset do I get from that join? Is it a Dataset[Row], or something else?







scala apache-spark






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asked Nov 8 at 17:51









vaer-k

3,49342031




3,49342031












  • Pretty sure you can't encode Unit.
    – erip
    Nov 8 at 18:15


















  • Pretty sure you can't encode Unit.
    – erip
    Nov 8 at 18:15
















Pretty sure you can't encode Unit.
– erip
Nov 8 at 18:15




Pretty sure you can't encode Unit.
– erip
Nov 8 at 18:15












1 Answer
1






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4
down vote



accepted










TL;DR Encoder is required to transform the outcome to the internal Spark SQL format and there is no need for that in case of foreach (or any other sink).



Just take a look at the signatures. map is



def map[U](func: (T) ⇒ U)(implicit arg0: Encoder[U]): Dataset[U] 


so in plain words it transforms records from T to U and then uses the Encoder of U to transform the result to internal representation.



foreach from the other hand, is



def foreach(f: (T) ⇒ Unit): Unit 


In other words it doesn't expect any result. Since there is no result to be stored, Encoder is just obsolete.






share|improve this answer























  • I see. I thought it needed to encode the input
    – vaer-k
    Nov 8 at 18:20










  • @vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
    – Alexey Romanov
    Nov 9 at 8:21













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










TL;DR Encoder is required to transform the outcome to the internal Spark SQL format and there is no need for that in case of foreach (or any other sink).



Just take a look at the signatures. map is



def map[U](func: (T) ⇒ U)(implicit arg0: Encoder[U]): Dataset[U] 


so in plain words it transforms records from T to U and then uses the Encoder of U to transform the result to internal representation.



foreach from the other hand, is



def foreach(f: (T) ⇒ Unit): Unit 


In other words it doesn't expect any result. Since there is no result to be stored, Encoder is just obsolete.






share|improve this answer























  • I see. I thought it needed to encode the input
    – vaer-k
    Nov 8 at 18:20










  • @vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
    – Alexey Romanov
    Nov 9 at 8:21

















up vote
4
down vote



accepted










TL;DR Encoder is required to transform the outcome to the internal Spark SQL format and there is no need for that in case of foreach (or any other sink).



Just take a look at the signatures. map is



def map[U](func: (T) ⇒ U)(implicit arg0: Encoder[U]): Dataset[U] 


so in plain words it transforms records from T to U and then uses the Encoder of U to transform the result to internal representation.



foreach from the other hand, is



def foreach(f: (T) ⇒ Unit): Unit 


In other words it doesn't expect any result. Since there is no result to be stored, Encoder is just obsolete.






share|improve this answer























  • I see. I thought it needed to encode the input
    – vaer-k
    Nov 8 at 18:20










  • @vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
    – Alexey Romanov
    Nov 9 at 8:21















up vote
4
down vote



accepted







up vote
4
down vote



accepted






TL;DR Encoder is required to transform the outcome to the internal Spark SQL format and there is no need for that in case of foreach (or any other sink).



Just take a look at the signatures. map is



def map[U](func: (T) ⇒ U)(implicit arg0: Encoder[U]): Dataset[U] 


so in plain words it transforms records from T to U and then uses the Encoder of U to transform the result to internal representation.



foreach from the other hand, is



def foreach(f: (T) ⇒ Unit): Unit 


In other words it doesn't expect any result. Since there is no result to be stored, Encoder is just obsolete.






share|improve this answer














TL;DR Encoder is required to transform the outcome to the internal Spark SQL format and there is no need for that in case of foreach (or any other sink).



Just take a look at the signatures. map is



def map[U](func: (T) ⇒ U)(implicit arg0: Encoder[U]): Dataset[U] 


so in plain words it transforms records from T to U and then uses the Encoder of U to transform the result to internal representation.



foreach from the other hand, is



def foreach(f: (T) ⇒ Unit): Unit 


In other words it doesn't expect any result. Since there is no result to be stored, Encoder is just obsolete.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 8 at 18:21

























answered Nov 8 at 18:19









user10465355

66729




66729












  • I see. I thought it needed to encode the input
    – vaer-k
    Nov 8 at 18:20










  • @vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
    – Alexey Romanov
    Nov 9 at 8:21




















  • I see. I thought it needed to encode the input
    – vaer-k
    Nov 8 at 18:20










  • @vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
    – Alexey Romanov
    Nov 9 at 8:21


















I see. I thought it needed to encode the input
– vaer-k
Nov 8 at 18:20




I see. I thought it needed to encode the input
– vaer-k
Nov 8 at 18:20












@vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
– Alexey Romanov
Nov 9 at 8:21






@vaer-k Then it would need Encoder[T], not Encoder[U] (or both).
– Alexey Romanov
Nov 9 at 8:21




















 

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