I have a table in the snowflake with a time range from for example 2019.01 to 2020.01. An ID can appear multiple times (match with) on any of the dates.
For example:
my_table: two columns dddate and id
dddate
id
2019-02-03
607
2019-01-07
356
2019-08-06
491
2019-01-01
607
2019-12-17
529
2019-04-15
356
......
Is there a way I can find the total number of IDs that appeared at least one time in the current month that also appeared at least one time in the previous three months, and group by month to show each month's number count starting from 2019-04 (The first month that has previous three months data available in the table) until 2020-01.
I am thinking of some code like this:
WITH PREV_THREE AS (
SELECT
DATE_TRUNC('MONTH', dddate) AS MONTH,
ID AS CURR_ID
FROM my_table mt
INNER JOIN
(
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -1, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
UNION ALL
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -2, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
UNION ALL
(
SELECT
MONTH(DATEADD(DATE_TRUNC('MONTH', dddate), -3, GETDATE())) AS PREV_MONTH,
ID AS PREV_3_MON_ID
FROM my_table
)
) AS PREV_3_MON
ON mt.CURR_ID = PREV_3_MON.PREV_3_MON_ID
)
SELECT MONTH, COUNT(DISTINCT ID) AS COUNTER
FROM PREV_THREE
GROUP BY 1
ORDER BY 1
However, it somehow returns an error and doesn't seem working. Could anyone please help me with this? Thank you in advance!
You can use lag():
select distinct id
from (select t.*,
lag(dddate) over (partition by id order by dddate) as prev_dddate
from my_table t
) t
where dddate >= date_trunc('MONTH', current_date) and
prev_dddate < date_trunc('MONTH', current_date) and
prev_dddate >= date_trunc('MONTH', current_date) - interval '3 month';
You can do this for multiple months as:
select date_trunc('MONTH', dddate), count(distinct id)
from (select t.*,
lag(dddate) over (partition by id order by dddate) as prev_dddate
from my_table t
) t
where prev_dddate < date_trunc('MONTH', date_trunc('MONTH', dddate)) and
prev_dddate >= date_trunc('MONTH', date_trunc('MONTH', dddate)) - interval '3 month'
group by date_trunc('MONTH', dddate);
Even if an id appears multiple times in one month, one of those will be first and the lag() will identify the most recent previous month.
Related
I have this table for example:
Date
amount
2021-02-16T21:06:38
10
2021-02-16T21:07:01
5
2021-02-17T01:10:12
-1
2021-02-19T12:00:00
3
2021-02-24T12:00:00
20
2021-02-25T12:00:00
-1
I want the total amount of all previous weeks, per week. So the result in this case would be:
Date
amount
2021-02-15
0
2021-02-22
17
2021-03-01
36
Note: The dates are now the start of each week (Monday).
Any help would this would be greatly appreciated.
Try This:
select week_date, sum(amount) over (order by week_date )
from (
SELECT date(date_) + cast(abs(extract(dow FROM date_) -7 ) + 1 as int) "week_date",
sum(amount) "amount"
from example group by 1) t
DEMO
Above Query will cover only the week in which transaction records are there. If you want to cover all missing week then try below query:
with cte as (
SELECT date(date_) + cast(abs(extract(dow FROM date_) -7 ) + 1 as int) "week_date",
sum(amount) "amount"
from example group by 1
)
select
t1."Date",coalesce(sum(cte.amount) over (order by t1."Date"),0)
from cte right join
(select generate_series(min(week_date)- interval '1 week', max(week_date),interval '1 week') "Date" from cte) t1 on cte.week_date=t1."Date"
DEMO
Use generate_series() to generate the dates you want. Then use left join to bring in the data and aggregate with a cumulative sum:
select gs.week,
coalesce(sum(e.amount), 0) as week_amount,
sum(coalesce(sum(e.amount), 0)) over (order by gs.week) as running_amount
from generate_series('2021-02-15'::date, '2021-03-01'::date, interval '1 week') gs(week) left join
example e
on e.date < gs.week and
e.date >= gs.week - interval '1 week'
group by gs.week
order by gs.week;
Here is a db<>fiddle.
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;
The best way to explain what I need is showing, so, here it is:
Currently I have this query
select
date_
,count(*) as count_
from table
group by date_
which returns me the following database
Now I need to get a new column, that shows me the count off all the previous 7 days, considering the row date_.
So, if the row is from day 29/06, I have to count all ocurrencies of that day ( my query is already doing it) and get all ocurrencies from day 22/06 to 29/06
The result should be something like this:
If you have values for all dates, without gaps, then you can use window functions with a rows frame:
select
date,
count(*) cnt
sum(count(*)) over(order by date rows between 7 preceding and current row) cnt_d7
from mytable
group by date
order by date
you can try something like this:
select
date_,
count(*) as count_,
(select count(*)
from table as b
where b.date_ <= a.date_ and b.date_ > a.date - interval '7 days'
) as count7days_
from table as a
group by date_
If you have gaps, you can do a more complicated solution where you add and subtract the values:
with t as (
select date_, count(*) as count_
from table
group by date_
union all
select date_ + interval '8 day', -count(*) as count_
from table
group by date_
)
select date_,
sum(sum(count_)) over (order by date_ rows between unbounded preceding and current row) - sum(count_)
from t;
The - sum(count_) is because you do not seem to want the current day in the cumulated amount.
You can also use the nasty self-join approach . . . which should be okay for 7 days:
with t as (
select date_, count(*) as count_
from table
group by date_
)
select t.date_, t.count_, sum(tprev.count_)
from t left join
t tprev
on tprev.date_ >= t.date_ - interval '7 day' and
tprev.date_ < t.date_
group by t.date_, t.count_;
The performance will get worse and worse as "7" gets bigger.
Try with subquery for the new column:
select
table.date_ as groupdate,
count(table.date_) as date_count,
(select count(table.date_)
from table
where table.date_ <= groupdate and table.date_ >= groupdate - interval '7 day'
) as total7
from table
group by groupdate
order by groupdate
I am trying to count unique users on a monthly basis that were not present in the previous month. So if a user has a record for January and then another one for February, then I would only count January for that user.
user_id time
a1 1/2/17
a1 2/10/17
a2 2/18/17
a4 2/5/17
a5 3/25/17
My results should look like this
Month User Count
January 1
February 2
March 1
I'm not really familiar with BigQuery, but here's how I would solve the problem using TSQL. I imagine that you'd be able to use similar logic in BigQuery.
1). Order the data by user_id first, and then time. In TSQL, you can accomplish this with the following and store it in a common table expression, which you will query in the step after this.
;WITH cte AS
(
select ROW_NUMBER() OVER (PARTITION BY [user_id] ORDER BY [time]) AS rn,*
from dbo.employees
)
2). Next query for only the rows with rn = 1 (the first occurrence for a particular user) and group by the month.
select DATENAME(month, [time]) AS [Month], count(*) AS user_count
from cte
where rn = 1
group by DATENAME(month, [time])
This is assuming that 2017 is the only year you're dealing with. If you're dealing with more than one year, you probably want step #2 to look something like this:
select year([time]) as [year], DATENAME(month, [time]) AS [month],
count(*) AS user_count
from cte
where rn = 1
group by year([time]), DATENAME(month, [time])
First aggregate by the user id and the month. Then use lag() to see if the user was present in the previous month:
with du as (
select date_trunc(time, month) as yyyymm, user_id
from t
group by date_trunc(time, month)
)
select yyyymm, count(*)
from (select du.*,
lag(yyyymm) over (partition by user_id order by yyyymm) as prev_yyyymm
from du
) du
where prev_yyyymm is not null or
prev_yyyymm < date_add(yyyymm, interval 1 month)
group by yyyymm;
Note: This uses the date functions, but similar functions exist for timestamp.
The way I understood question is - to exclude user to be counted in given month only if same user presented in previous month. But if same user present in few months before given, but not in previous - user should be counted.
If this is correct - Try below for BigQuery Standard SQL
#standardSQL
SELECT Year, Month, COUNT(DISTINCT user_id) AS User_Count
FROM (
SELECT *,
DATE_DIFF(time, LAG(time) OVER(PARTITION BY user_id ORDER BY time), MONTH) AS flag
FROM (
SELECT
user_id,
DATE_TRUNC(PARSE_DATE('%x', time), MONTH) AS time,
EXTRACT(YEAR FROM PARSE_DATE('%x', time)) AS Year,
FORMAT_DATE('%B', PARSE_DATE('%x', time)) AS Month
FROM yourTable
GROUP BY 1, 2, 3, 4
)
)
WHERE IFNULL(flag, 0) <> 1
GROUP BY Year, Month, time
ORDER BY time
you can test / play with above using below example with dummy data from your question
#standardSQL
WITH yourTable AS (
SELECT 'a1' AS user_id, '1/2/17' AS time UNION ALL
SELECT 'a1', '2/10/17' UNION ALL
SELECT 'a2', '2/18/17' UNION ALL
SELECT 'a4', '2/5/17' UNION ALL
SELECT 'a5', '3/25/17'
)
SELECT Year, Month, COUNT(DISTINCT user_id) AS User_Count
FROM (
SELECT *,
DATE_DIFF(time, LAG(time) OVER(PARTITION BY user_id ORDER BY time), MONTH) AS flag
FROM (
SELECT
user_id,
DATE_TRUNC(PARSE_DATE('%x', time), MONTH) AS time,
EXTRACT(YEAR FROM PARSE_DATE('%x', time)) AS Year,
FORMAT_DATE('%B', PARSE_DATE('%x', time)) AS Month
FROM yourTable
GROUP BY 1, 2, 3, 4
)
)
WHERE IFNULL(flag, 0) <> 1
GROUP BY Year, Month, time
ORDER BY time
The output is
Year Month User_Count
2017 January 1
2017 February 2
2017 March 1
Try this query:
SELECT
t1.d,
count(DISTINCT t1.user_id)
FROM
(
SELECT
EXTRACT(MONTH FROM time) AS d,
--EXTRACT(MONTH FROM time)-1 AS d2,
user_id
FROM nbitra.tmp
) t1
LEFT JOIN
(
SELECT
EXTRACT(MONTH FROM time) AS d,
user_id
FROM nbitra.tmp
) t2
ON t1.d = t2.d+1
WHERE
(
t1.user_id <> t2.user_id --User is in previous month
OR t2.user_id IS NULL --To handle january, since there is no previous month to compare to
)
GROUP BY t1.d;
I have a table, lets say "Records" with structure:
id date
-- ----
1 2012-08-30
2 2012-08-29
3 2012-07-25
I need to write an SQL query in PostgreSQL to get record_id for MIN date in each month.
month record_id
----- ---------
8 2
7 3
as we see 2012-08-29 < 2012-08-30 and it is 8 month, so we should show record_id = 2
I tried something like this,
SELECT
EXTRACT(MONTH FROM date) as month,
record_id,
MIN(date)
FROM Records
GROUP BY 1,2
but it shows 3 records.
Can anybody help?
SELECT DISTINCT ON (EXTRACT(MONTH FROM date))
id,
date
FROM Records1
ORDER BY EXTRACT(MONTH FROM date),date
SQLFiddle http://sqlfiddle.com/#!12/76ca2/3
UPD: This query:
1) Orders the records by month and date
2) For every month picks the first record (the first record has MIN(date) because of ordering)
Details here http://www.postgresql.org/docs/current/static/sql-select.html#SQL-DISTINCT
This will return multiples if you have duplicate minimum dates:
Select
minbymonth.Month,
r.record_id
From (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
) minbymonth
Inner Join
records r
On minbymonth.date = r.date
Order By
1;
Or if you have CTEs
With MinByMonth As (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
)
Select
m.Month,
r.record_id
From
MinByMonth m
Inner Join
Records r
On m.date = r.date
Order By
1;
http://sqlfiddle.com/#!1/2a054/3
select extract(month from date)
, record_id
, date
from
(
select
record_id
, date
, rank() over (partition by extract(month from date) order by date asc) r
from records
) x
where r=1
order by date
SQL Fiddle
select distinct on (date_trunc('month', date))
date_trunc('month', date) as month,
id,
date
from records
order by 1, 3 desc
I think you need use sub-query, something like this:
SELECT
EXTRACT(MONTH FROM r.date) as month,
r.record_id
FROM Records as r
INNER JOIN (
SELECT
EXTRACT(MONTH FROM date) as month,
MIN(date) as mindate
FROM Records
GROUP BY EXTRACT(MONTH FROM date)
) as sub on EXTRACT(MONTH FROM r.date) = sub.month and r.date = sub.mindate