If you have table like this:
Name
Data type
UserID
INT
StartDate
DATETIME
EndDate
DATETIME
With data like this:
UserID
StartDate
EndDate
21
2021-01-02 00:00:00
2021-01-02 23:59:59
21
2021-01-03 00:00:00
2021-01-04 15:42:00
24
2021-01-02 00:00:00
2021-01-06 23:59:59
And you want to calculate number of users that is represented on each day in a week with a result like this:
Year
Week
NumberOfTimes
2021
1
8
2021
2
10
2021
3
4
Basically I want to to a Select like this:
SELECT YEAR(dateColumn) AS yearname, WEEK(dateColumn)as week name, COUNT(somecolumen)
GROUP BY YEAR(dateColumn) WEEK(dateColumn)
The problem I have is the start and end date if the date goes over several days I want it to counted each day. Preferably I don't want the same user counted twice each day. There are millions of rows that are constantly being deleted and added so speed is key.
The database is MS-SQL 2019
I would suggest a recursive CTE:
with cte as (
select userid, startdate, enddate
from t
union all
select userid, startdate,
enddate
from cte
where startdate < enddate and
week(startdate) <> week(enddate)
)
select year(startdate), week(startdate), count(*)
from cte
group by year(startdate), week(startdate)
option (maxrecursion 0);
The CTE expands the data by adding 7 days to each row. This should be one day per week.
There is a little logic in the second part to handle the situation where the enddate ends in the same week as the last start date. The above solution assumes that the dates are all in the same year -- which seems quite reasonable given the sample data. There are other ways to prevent this problem.
You need to cross-join each row with the relevant dates.
Create a calendar table with columns of years and weeks, include a start and end date of the week. See here for an example of how to create one, and make sure you index those columns.
Then you can cross-join like this
SELECT
YEAR(dateColumn) AS yearname,
WEEK(dateColumn)as weekname,
COUNT(somecolumen)
FROM Table t
JOIN CalendarWeek c ON c.StartDate >= t.StartDate AND c.EndDate <= t.EndDate
GROUP BY YEAR(dateColumn), WEEK(dateColumn)
Related
For many years I've been collecting data and I'm interested in knowing the historic counts of IDs that appeared in the last 30 days. The source looks like this
id
dates
1
2002-01-01
2
2002-01-01
3
2002-01-01
...
...
3
2023-01-10
If I wanted to know the historic count of ids that appeared in the last 30 days I would do something like this
with total_counter as (
select id, count(id) counts
from source
group by id
),
unique_obs as (
select id
from source
where dates >= DATEADD(Day ,-30, current_date)
group by id
)
select count(distinct(id))
from unique_obs
left join total_counter
on total_counter.id = unique_obs.id;
The problem is that this results would return a single result for today's count as provided by current_date.
I would like to see a table with such counts as if for example I had ran this analysis yesterday, and the day before and so on. So the expected result would be something like
counts
date
1235
2023-01-10
1234
2023-01-09
1265
2023-01-08
...
...
7383
2022-12-11
so for example, let's say that if the current_date was 2023-01-10, my query would've returned 1235.
If you need a distinct count of Ids from the 30 days up to and including each date the below should work
WITH CTE_DATES
AS
(
--Create a list of anchor dates
SELECT DISTINCT
dates
FROM source
)
SELECT COUNT(DISTINCT s.id) AS "counts"
,D.dates AS "date"
FROM CTE_DATES D
LEFT JOIN source S ON S.dates BETWEEN DATEADD(DAY,-29,D.dates) AND D.dates --30 DAYS INCLUSIVE
GROUP BY D.dates
ORDER BY D.dates DESC
;
If the distinct count didnt matter you could likely simplify with a rolling sum, only hitting the source table once:
SELECT S.dates AS "date"
,COUNT(1) AS "count_daily"
,SUM("count_daily") OVER(ORDER BY S.dates DESC ROWS BETWEEN CURRENT ROW AND 29 FOLLOWING) AS "count_rolling" --assumes there is at least one row for every day.
FROM source S
GROUP BY S.dates
ORDER BY S.dates DESC;
;
This wont work though if you have gaps in your list of dates as it'll just include the latest 30 days available. In which case the first example without distinct in the count will do the trick.
SELECT count(*) AS Counts
dates AS Date
FROM source
WHERE dates >= DATEADD(DAY, -30, CURRENT_DATE)
GROUP BY dates
ORDER BY dates DESC
The question I have is very similar to the question here, but I am using Presto SQL (on aws athena) and couldn't find information on loops in presto.
To reiterate the issue, I want the query that:
Given table that contains: Day, Number of Items for this Day
I want: Day, Average Items for Last 7 Days before "Day"
So if I have a table that has data from Dec 25th to Jan 25th, my output table should have data from Jan 1st to Jan 25th. And for each day from Jan 1-25th, it will be the average number of items from last 7 days.
Is it possible to do this with presto?
maybe you can try this one
calendar Common Table Expression (CTE) is used to generate dates between two dates range.
with calendar as (
select date_generated
from (
values (sequence(date'2021-12-25', date'2022-01-25', interval '1' day))
) as t1(date_array)
cross join unnest(date_array) as t2(date_generated)),
temp CTE is basically used to make a date group which contains last 7 days for each date group.
temp as (select c1.date_generated as date_groups
, format_datetime(c2.date_generated, 'yyyy-MM-dd') as dates
from calendar c1, calendar c2
where c2.date_generated between c1.date_generated - interval '6' day and c1.date_generated
and c1.date_generated >= date'2021-12-25' + interval '6' day)
Output for this part:
date_groups
dates
2022-01-01
2021-12-26
2022-01-01
2021-12-27
2022-01-01
2021-12-28
2022-01-01
2021-12-29
2022-01-01
2021-12-30
2022-01-01
2021-12-31
2022-01-01
2022-01-01
last part is joining day column from your table with each date and then group it by the date group
select temp.date_groups as day
, avg(your_table.num_of_items) avg_last_7_days
from your_table
join temp on your_table.day = temp.dates
group by 1
You want a running average (AVG OVER)
select
day, amount,
avg(amount) over (order by day rows between 6 preceding and current row) as avg_amount
from mytable
order by day
offset 6;
I tried many different variations of getting the "running average" (which I now know is what I was looking for thanks to Thorsten's answer), but couldn't get the output I wanted exactly with my other columns (that weren't included in my original question) in the table, but this ended up working:
SELECT day, <other columns>, avg(amount) OVER (
PARTITION BY <other columns>
ORDER BY date(day) ASC
ROWS 6 PRECEDING) as avg_7_days_amount FROM table ORDER BY date(day) ASC
Using SQL Server, I need to fetch the date ranges for every week by week number for a whole year. The first day of week needs to be Monday, unless the week starts in another year, then I need it to start at the first day of the year.
For Example,
if I call the select with the year 2021 as the parameter, I would like the output to be something like that :
WeekNumbers: DateStart: DateEnd:
1 2021-01-01 2021-01-03
2 2021-01-04 2021-01-10
...
...
52 2021-12-20 2021-12-26
53 2021-12-27 2021-12-31
Please notice the first and the last week number of the year of the example.
Thanks
There are a lot of ways to solve this problem. A brute force approach generates all the dates using a some method -- this uses a recursive CTE -- and then assigns the weeks based on a window function counting the Mondays.
with cte as (
select datefromparts(2021, 1, 1) as dte
union all
select dateadd(day, 1, dte)
from cte
where dte < datefromparts(2021, 12, 31)
)
select weeknum, min(dte), max(dte)
from (select cte.*,
sum(case when datename(weekday, dte) = 'Monday' then 1 else 0 end) over (order by dte) as weeknum
from cte
) cte
group by weeknum
option (maxrecursion 0);
This is fine for a one-off query, but isn't the most efficient method to generate the weeks. It is, however, a fun illustration of recursive CTEs. That said, you should probably really have a calendar table of some sort that encapsulates this information as a table.
Suppose you have a table like:
id subscription_start subscription_end segment
1 2016-12-01 2017-02-01 87
2 2016-12-01 2017-01-24 87
...
And wish to generate a temporary table with months.
One way would be to encode the month date as:
with months as (
select
'2016-12-01' as 'first',
'2016-12-31' as 'last'
union
select
'2017-01-01' as 'first',
'2017-01-31' as 'last'
...
) select * from months;
So that I have an output table like:
first_day last_day
2017-01-01 2017-01-31
2017-02-01 2017-02-31
2017-03-01 2017-03-31
I would like to generate a temporary table with a custom interval (above), without manually encoding all the dates.
Say the interval is of 12 months, for each year, for as many years there are in the db.
I'd like to have general approach to compute the months table with the same output as above.
Or, one may adjust the range to a custom interval (months split an year in 12 parts, but one may want to split a time in a custom interval of days).
To start, I was thinking to use recursive query like:
with months(id, first_day, last_day, month) as (
select
id,
first_day,
last_day,
0
where
subscriptions.first_day = min(subscriptions.first_day)
union all
select
id,
first_day,
last_day,
months.month + 1
from
subscriptions
left join months on cast(
strftime('%m', datetime(subscriptions.subscription_start)) as int
) = months.month
where
months.month < 13
)
select
*
from
months
where
month = 1;
but it does not do what I'd expect: here I was attempting to select the first row from the table with the minimum date, and populate a table at interval of months, ranging from 1 to 12. For each month, I was comparing the string date field of my table (e.g. 2017-03-01 = 3 is march).
The query above does work and also seems a bit complicated, but for the sake of learning, which alternative would you propose to create a temporary table months without manually coding the intervals ?
Suppose I have the following table:
Id Visitors Date
------------------------------
1 100 '2017-01-01'
2 200 '2017-01-02'
3 150 '2017-01-03'
I want a query to provide the average of a range of records for the last 12 months.
For one record I know that it would be like :
select avg(Visitors), Date
from Visitors_table
where Date between '2018-01-01' and '2017-01-01'
However, I need to do that for a range of dates and multiple records.
I know that Union will solve it, but if the range is one year for example It is not optimized to use 365 union
Get the dates from 1 year ago to current date:
SELECT
Date,
AVG(Visitors) AS avgvisitors,
FROM Visitors_table
WHERE Date > dateadd(year, -1, getdate())
GROUP BY Date
ORDER BY Date;
Since you need to group by date.