Convert HiveQL into Spark Scala

Question

I want to convert HiveQL query with window function into Scala Spark query... but I'm constantly receiving the same exception.

Problem context: mytable consists of category and product fields. I want to get list with top N frequent product for each category. DF below is a HiveContext object

Original query (works correctly):

SELECT category, product, freq FROM (
    SELECT category, product, COUNT(*) AS freq, 
    ROW_NUMBER() OVER (PARTITION BY category ORDER BY COUNT(*) DESC) as seqnum
    FROM mytable GROUP BY category, product) ci 
WHERE seqnum <= 10;

What I have now (partially converted, doesn't work):

val w = row_number().over(Window.partitionBy("category").orderBy(count("*").desc))
val result = df.select("category", "product").groupBy("category", "product").agg(count("*").as("freq"))
val new_res = result.withColumn("seqNum", w).where(col("seqNum") <= 10).drop("seqNum")

Constantly receiving the following exception:

Exception in thread "main" org.apache.spark.sql.AnalysisException: expression 'category' is neither present in the group by, nor is it an aggregate function. Add to group by or wrap in first() (or first_value) if you don't care which value you get.;

What can be wrong here?

10465355 10465355 · Accepted Answer · 2019-01-17T18:05:57

Your mistake is to use aggregate in the orderBy clause:

.orderBy(count("*").desc)

If written like that, expression introduces new aggregate expression. Instead you should reference existing aggregate by name:

.orderBy("freq")

So your code should look like:

val w = row_number().over(
  Window.partitionBy("category").orderBy("freq"))
val result = df.select("category", "product")
  .groupBy("category", "product")
  .agg(count("*").as("freq"))
val new_res = result
  .withColumn("seqNum", w).where(col("seqNum") <= 10)
  .drop("seqNum")

Convert HiveQL into Spark Scala

1 Answers