I have a data set with user ids that have made purchases over time. I would like to show a YTD distinct count of users that have made a purchase, partitioned by State and Country. The output would have 4 columns: Country, State, Year, Month, YTD Count of Distinct Users with purchase activity.
Is there a way to do this? The following code works when I exclude the month from the view and do a distinct count:
Select Year, Country, State,
COUNT(DISTINCT (CASE WHEN ActiveUserFlag > 0 THEN MBR_ID END)) AS YTD_Active_Member_Count
From MemberActivity
Where Month <= 5
Group By 1,2,3;
The issue occurs when the user has purchases across multiple months, because I can’t aggregate at a monthly level then sum, because it duplicates user counts.
I need to see the YTD count for each month of the year, for trending purposes.
Return each member only once for the first month they make a purchase, count by month and then apply a Cumulative Sum:
select Year, Country, State, month,
sum(cnt)
over (partition by Year, Country, State
order by month
rows unbounded preceding) AS YTD_Active_Member_Count
from
(
Select Year, Country, State, month,
COUNT(*) as cnt -- 1st purchses per month
From
( -- this assumes there's at least one new active member per year/month/country
-- otherwise there would be mising rows
Select *
from MemberActivity
where ActiveUserFlag > 0 -- only active members
and Month <= 5
-- and year = 2019 -- seems to be for this year only
qualify row_number() -- only first purchase per member/year
over (partition by MBR_ID, year
order by month --? probably there's a purchase_date) = 1
) as dt
group by 1,2,3,4
) as dt
;
Count users in the first month they appear:
select Country, State, year, month,
sum(case when ActiveUserFlag > 0 and seqnum = 1 then 1 else 0 end) as YTD_Active_Member_Count
from (select ma.*,
row_number() over (partition by year order by month) as seqnum
from MemberActivity ma
) ma
where Month <= 5
group by Country, State, year, month;
Related
I need to write SQL query to pull the single, highest-earning day for a certain brand of each quarter of 2018. I have the following but it does not pull a singular day - it pulls the highest earnings for each day.
select distinct quarter, order_event_date, max(gc) as highest_day_gc
from (
select sum(commission) as cm, order_date,
extract(quarter from order__date) as quarter
from order_table
where advertiser_id ='123'
and event_year='2018'
group by 3,2
)
group by 1,2
order by 2 DESC
You can use window functions to find the highest earning day per quarter by using rank().
select rank() over (partition by quarter order by gc desc) as rank, quarter, order_event_date, gc
from (select sum(gross_commission) gc,
order_event_date,
extract(quarter from order_event_date) quarter
from order_aggregation
where advertiser_id = '123'
and event_year = '2018'
group by order_event_date, quarter) a
You could create the query above as view and filter it by using where rank = 1.
You could add the LIMIT clause at the end of the sentence. Also, change the las ORDER BY clause to ORDER BY highest_day_gc. Something like:
SELECT DISTINCT quarter
,order_event_date
,max(gc) as highest_day_gc
FROM (SELECT sum(gross_commission) as gc
,order_event_date
,extract(quarter from order_event_date) as quarter
FROM order_aggregation
WHERE advertiser_id ='123'
AND event_year='2018'
GROUP BY 3,2) as subquery
GROUP BY 1,2
ORDER BY 3 DESC
LIMIT 1
I'm working in SQL Workbench.
I'd like to track every time a unique customer clicks the new feature in trailing 30 days, displayed week over week. An example of the data output would be as follows:
Week 51: Reflects usage through the end of week 51 (Dec 20th) - 30 days. aka Nov 20-Dec 20th
Week 52: Reflects usage through the end of week 52 (Dec 31st) - 30 days. aka Dec 1 - Dec 31st.
Say there are 22MM unique customer clicks that occurred from Nov 20-Dec 20th. Week 51 data = 22MM.
Say there are 25MM unique customer clicks that occurred from Dec 1-Dec 31st. Week 52 data = 25MM. The customer uniqueness is only relevant to that particular week. Aka, if a customer clicks twice in Week 51 they're only counted once. If they click once in Week 51 and once in Week 52, they are counted once in each week.
Here is what I have so far:
select
min_e_date
,sum(count(*)) over (order by min_e_date rows between unbounded preceding and current row) as running_distinct_customers
from (select customer_id, min(DATE_TRUNC('week', event_date)) as min_e_date
from final
group by 1
) c
group by
min_e_date
I don't think a rolling count is the right way to go. As I add in additional parameters (country, subscription), the rolling count doesn't distinguish between them - the figures just get added to the prior row.
Any suggestions are appreciated!
edit Additional data below. Data collection begins on 11/23. No data precedes that date.
You can get the count of distinct customers per week like so:
select date_trunc('week', event_date) as week_start,
count(distinct customer_id) cnt
from final
group by 1
Now if you want a rolling sum of that count(say, the current week and the three preceding weeks), you can use window functions:
select date_trunc('week', event_date) as week_start,
count(distinct customer_id) cnt,
sum(count(distinct customer_id)) over(
order by date_trunc('week', event_date)
range between 3 week preceding and current row
) as rolling_cnt
from final
group by 1
Rolling distinct counts are quite difficult in RedShift. One method is a self-join and aggregation:
select t.date,
count(distinct case when tprev.date >= t.date - interval '6 day' then customer_id end) as trailing_7,
count(distinct customer_id) as trailing_30
from t join
t tprev
on tprev.date >= t.date - interval '29 day' and
tprev.date <= t.date
group by t.date;
If you can get this to work, you can just select every 7th row to get the weekly values.
EDIT:
An entirely different approach is to use aggregation and keep track of when customers enter and end time periods of being counted. This is a pain with two different time frames. Here is what it looks like for one.
The idea is to
Create an enter/exit record for each record being counted. The "exit" is n days after the enter.
Summarize these into periods of activity for each customer. So, there is one record with an enter and exit date. This is a type of gaps-and-islands problem.
Unpivot this result to count +1 for a customer being counted and -1 for a customer not being counted.
Do a cumulative sum of this count.
The code looks something like this:
with cd as (
select customer_id, date,
lead(date) over (partition by customer_id order by date) as next_date,
sum(sum(inc)) over (partition by customer_id order by date) as cnt
from ((select t.customer_id, t.date, 1 as inc
from t
) union all
(select t.customer_id, t.date + interval '7 day', -1
from t
)
) tt
),
cd2 as (
select customer_id, min(date) as enter_date, max(date) as exit_date
from (select cd.*,
sum(case when cnt = 0 then 1 else 0 end) over (partition by customer_id order by date) as grp
from (select cd.*,
lag(cnt) over (partition by customer_id order by date) as prev_cnt
from cd
) cd
) cd
group by customer_id, grp
having max(cnt) > 0
)
select dte, sum(sum(inc)) over (order by dte)
from ((select customer_id, enter_date as dte, 1 as inc
from cd2
) union all
(select customer_id, exit_date as dte, -1 as inc
from cd2
)
) cd2
group by dte;
I have daily city level data with some counts. I have to aggregate this data at monthly level(1st day of each month) and then create lag variables based on last 1 week from 1st day of month.
I have used following code to create lag variables for last 1 month using (after aggregating data at monthly level ( with 1st date of month)
sum(count) over (partition by City order by month_date rows between 1 preceding and 1 preceding) as last_1_month_count
Is there a way to aggregate data at monthly level and create lag variables based on last 7,14,21,28 days using window function?
you can use this L
select
CITY
, month(Date)
, year(date)
, sum(count)
from table1
where date < Datediff(days , 7 , getdate())
group by
City
, month(Date)
, year(date)
I think you're looking for something like this. The first cte summarizes city counts to the day, week, month, year. The second summarizes the counts to the week, month, year. To group sales by weeks starting from the 1st day it uses the DAY function along with YEAR and MONTH. Since DAY returns and integer, groups of distinct weeks can be created by dividing by 7, i.e. DAY(day_dt)/7.
One way to get the prior week sales would be to join the week sales summary cte to itself where the week is offset by -1. Since the prior week might possible have 0 sales it seems safer to LEFT JOIN than to use LAG imo
with
day_sales_cte(city, day_dt, yr, mo, wk, sum_count) as (
select city, day_dt, year(day_dt), month(day_cte), day(day_dt)/7, sum([count]) sum_counts
from city_level_data
group by city, day_dt, year(day_dt), month(day_cte), day(day_dt)/7)
wk_sales_cte(city, yr, mo, wk, sum_count) as (
select city, yr, mo, wk, sum(sum_counts) sum_counts
from sales_cte
group by city, yr, mo, wk)
select ws.*, ws2.sum_sales prior_wk_sales
from wk_sales_cte ws
left join wk_sales_cte ws2 on ws.city=ws2.city
and ws.yr=ws2.yr
and ws.mo=ws2.mo
and ws.wk=ws.wk-1;
I have a table with columns month, name and transaction_id. I would like to count the number of transactions per month and name. However, for each month I want to have the top N names with the highest transaction counts.
The following query groups by month and name. However the LIMIT is applied to the complete result and not per month:
SELECT
month,
name,
COUNT(*) AS transaction_count
FROM my_table
GROUP BY month, name
ORDER BY month, transaction_count DESC
LIMIT N
Does anyone have an idea how I can get the top N results per month?
Use row_number():
SELECT month, name, transaction_count
FROM (SELECT month, name, COUNT(*) AS transaction_count,
ROW_NUMBER() OVER (PARTITION BY month ORDER BY COUNT(*) DESC) as seqnum
FROM my_table
GROUP BY month, name
) mn
WHERE seqnum <= N
ORDER BY month, transaction_count DESC
Here's what I'm trying to achieve. Basically I have a relation in which I have a count for each ID and month. I'd like to sum the counts for each id by month (middle table in the picture) and then from that find the maximum value from all those sums by month, and show the month, id, and the maximum value in that order. Here's what I've got so far:
SELECT month, MAX(summed_counts) AS maximum_result
FROM
(SELECT month, id, SUM(counts) AS summed_counts
FROM info WHERE year=2017 GROUP BY month, id)
AS final_result GROUP BY month ORDER BY month ASC;
However as soon as I add id it no longer works:
SELECT month, id, MAX(summed_counts) AS maximum_result
FROM
(SELECT month, id, SUM(counts) AS summed_counts
FROM info WHERE year=2017 GROUP BY month, id)
AS final_result GROUP BY month, id ORDER BY month ASC;
Any suggestions?
Try this (MS SQL):
select distinct month,
(select top 1 SUM(counts)
FROM info info_detail
WHERE year=2017 and info_detail.month=info.month
GROUP BY id
order by SUM(counts) desc
) as max_value,
(select top 1 id
FROM info info_detail
WHERE year=2017 and info_detail.month=info.month
GROUP BY id
order by SUM(counts) desc
) as max_value_id
from info
where year=2017
ORDER BY month