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Count Distinct with PostgreSQL

Tags:

sql

postgresql

I am trying to change the count DISTINCT portion of the query below, because the current query returns WINDOW definition is not supported. What is most seamless way to do this calc using PostgreSQL? Thank you.

SELECT transaction_date, brand, Description, Amount, newval
into temp.table
FROM (SELECT transaction_date, brand, Description, Amount,
         (Amount / count(DISTINCT unique_mem_id) over (partition by to_char(transaction_date, 'YYYY-MM'),brand
         )) as newval
  FROM source.table
 )
WHERE DESCRIPTION iLIKE '%Criteria%';
like image 593
ZJAY Avatar asked Oct 29 '22 17:10

ZJAY


1 Answers

Due to the use-case it seems better to split the code.

Create a table month_brand based on the following query:

select      to_char(transaction_date, 'YYYY-MM')    as yyyymm
           ,brand
           ,count (distinct unique_mem_id)          as count_distinct_unique_mem_id

from        source.table

group by    yyyymm
           ,brand
;

Join month_brand with your source table:

select      t.transaction_date, t.brand, t.Description, t.Amount, t.Amount / m.count_distinct_unique_mem_id as newval

from                    source.table    as t

            join        month_brand     as m

            on          m.yyyymm = to_char(t.transaction_date, 'YYYY-MM')    

where       t.description ilike '%Criteria%'
;

Instead of count(distinct ...), 2 phase solution:

  1. give row numbers to the duplicated unique_mem_id
  2. count only the unique_mem_id with row_number = 1

select  *

into     temp.table

from   (SELECT  transaction_date, brand, Description, Amount, 
                (Amount / count(case rn when 1 then unique_mem_id end) over (partition by to_char(transaction_date, 'YYYY-MM'),brand)) as newval

        FROM    (SELECT     transaction_date, brand, Description, Amount,unique_mem_id
                            row_numner () over (partition by to_char(transaction_date, 'YYYY-MM'),brand,unique_mem_id) as rn

                 FROM       source.table
                 )
        )

WHERE   DESCRIPTION iLIKE '%Criteria%'
;
like image 125
David דודו Markovitz Avatar answered Nov 15 '22 06:11

David דודו Markovitz