Count the number of transactions per month for an individual group by date Hive - sql

Question

I have a table of customer transactions where each item purchased by a customer is stored as one row. So, for a single transaction there can be multiple rows in the table. I have another col called visit_date.
There is a category column called cal_month_nbr which ranges from 1 to 12 based on which month transaction occurred.
The data looks like below
Id visit_date Cal_month_nbr
---- ------ ------
1 01/01/2020 1
1 01/02/2020 1
1 01/01/2020 1
2 02/01/2020 2
1 02/01/2020 2
1 03/01/2020 3
3 03/01/2020 3
first
I want to know how many times customer visits per month using their visit_date
i.e i want below output
id cal_month_nbr visit_per_month
--- --------- ----
1 1 2
1 2 1
1 3 1
2 2 1
3 3 1
and what is the avg frequency of visit per ids
ie.
id Avg_freq_per_month
---- -------------
1 1.33
2 1
3 1
I tried with below query but it counts each item as one transaction
select avg(count_e) as num_visits_per_month,individual_id
from
(
select r.individual_id, cal_month_nbr, count(*) as count_e
from
ww_customer_dl_secure.cust_scan
GROUP by
r.individual_id, cal_month_nbr
order by count_e desc
) as t
group by individual_id
I would appreciate any help, guidance or suggestions

Answer 1

You can divide the total visits by the number of months:
select individual_id,
count(*) / count(distinct cal_month_nbr)
from ww_customer_dl_secure.cust_scan c
group by individual_id;
If you want the average number of days per month, then:
select individual_id,
count(distinct visit_date) / count(distinct cal_month_nbr)
from ww_customer_dl_secure.cust_scan c
group by individual_id;
Actually, Hive may not be efficient at calculating count(distinct), so multiple levels of aggregation might be faster:
select individual_id, avg(num_visit_days)
from (select individual_id, cal_month_nbr, count(*) as num_visit_days
from (select distinct individual_id, visit_date, cal_month_nbr
from ww_customer_dl_secure.cust_scan c
) iv
group by individual_id, cal_month_nbr
) ic
group by individual_id;