SQL query challenge - find top frequent items in columns and summarize result to a pivot table -


i looking query following transformation.

basically want find top 3 frequent sell_country , top 3 frequent category, on per website, per day bases. (for example, website 1, date 6-5-2017, there 2*us, 1*jp , 1*uk sell_country, therefore top1_sell_country us, , jp , uk going top2_sell_country , top3_sell_country. same idea category column)

my current solution involves many subqueries, works, feel complicated. interested in how sql master in elegant way.

currently know how uses

from enter image description here

to enter image description here

i in 3 steps:

  1. group country , rank count
  2. group category , rank count
  3. blend results using conditional aggregate (which place values in necessary cells because result of case value , many null values, min() outputs value)

like this:

with countries (     select *, row_number() on (partition website,date order count desc)     (         select          website         ,date::date         ,sell_country         ,count(1)         your_table         group 1,2,3     ) ) ,categories (     select *, row_number() on (partition website,date order count desc)     (         select          website         ,date::date         ,category         ,count(1)         your_table         group 1,2,3     ) ) select  website ,date ,coalesce(min(case when t1.row_number=1 t1.sell_country end),'na') top1_sell_country ,coalesce(min(case when t1.row_number=2 t1.sell_country end),'na') top2_sell_country ,coalesce(min(case when t1.row_number=3 t1.sell_country end),'na') top3_sell_country ,coalesce(min(case when t2.row_number=1 t2.category end),'na') top1_sell_category ,coalesce(min(case when t2.row_number=2 t2.category end),'na') top2_sell_category ,coalesce(min(case when t2.row_number=3 t2.category end),'na') top3_sell_category countries t1  full join categories t2 using (website,date) group 1,2 order 1,2

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