Finding id's available in previous weeks but not in current week - sql

How to find if an id which was present in previous weeks but not available in current week on a rolling basis. For e.g
Week1 has id 1,2,3,4,5
Week2 has id 3,4,5,7,8
Week3 has id 1,3,5,10,11
So I found out that id 1 and 2 are missing in week 2 and id 2,4,7,8 are missing in week 3 from previous 2 weeks But how to do this on a rolling window for a large amount of data distributed over a period of 20+ years
Please find the sample dataset and expected output. I am expecting the output to be partitioned based on the week_end Date
Dataset
ID|WEEK_START|WEEK_END|APPEARING_DATE
7152|2015-12-27|2016-01-02|2015-12-27
8350|2015-12-27|2016-01-02|2015-12-27
7152|2015-12-27|2016-01-02|2015-12-29
4697|2015-12-27|2016-01-02|2015-12-30
7187|2015-12-27|2016-01-02|2015-01-01
8005|2015-12-27|2016-01-02|2015-12-27
8005|2015-12-27|2016-01-02|2015-12-29
6254|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-04
3339|2016-01-03|2016-01-09|2016-01-06
7834|2016-01-03|2016-01-09|2016-01-03
7962|2016-01-03|2016-01-09|2016-01-05
7152|2016-01-03|2016-01-09|2016-01-07
8350|2016-01-03|2016-01-09|2016-01-09
2403|2016-01-10|2016-01-16|2016-01-10
0157|2016-01-10|2016-01-16|2016-01-11
2228|2016-01-10|2016-01-16|2016-01-14
4697|2016-01-10|2016-01-16|2016-01-14
Excepted Output
Partition1: WEEK_END=2016-01-02
ID|MAX(LAST_APPEARING_DATE)
7152|2015-12-29
8350|2015-12-27
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
Partition1: WEEK_END=2016-01-09
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2015-12-30
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
Partition3: WEEK_END=2016-01-10
ID|MAX(LAST_APPEARING_DATE)
7152|2016-01-07
8350|2016-01-09
4697|2016-01-14
7187|2015-01-01
8005|2015-12-29
6254|2016-01-03
7962|2016-01-05
3339|2016-01-06
7834|2016-01-03
2403|2016-01-10
0157|2016-01-11
2228|2016-01-14

Please use below query,
select ID, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;
Or, including WEEK)END,
select ID, WEEK_END, MAX(APPEARING_DATE) from table_name
group by ID, WEEK_END;

You can use aggregation:
select t.*, max(week_end)
from t
group by id
having max(week_end) < '2016-01-02';
Adjust the date in the having clause for the week end that you want.
Actually, your question is a bit unclear. I'm not sure if a later week end would keep the row or not. If you want "as of" data, then include a where clause:
select t.id, max(week_end)
from t
where week_end < '2016-01-02'
group by id
having max(week_end) < '2016-01-02';
If you want this for a range of dates, then you can use a derived table:
select we.the_week_end, t.id, max(week_end)
from (select '2016-01-02' as the_week_end union all
select '2016-01-09' as the_week_end
) we cross join
t
where t.week_end < we.the_week_end
group by id, we.the_week_end
having max(t.week_end) < we.the_week_end;

Related

SQL - Get historic count of rows collected within a certain period by date

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

How to conditional SQL select

My table consists of user_id, revenue, publish_month columns.
Right now I use group_by user_id and sum(revenue) to get revenue for all individual users.
Is there a single SQL query I can use to query for user revenue across a time period conditionally? If for a specific user, there is a row for this month, I want to query for this month, last month and the month before. If there is not yet a row for this month, I want to query for last month and the two months before.
Any advice with which approach to take would be helpful. If I should be using cases, if-elses with exists or if this is do-able with a single SQL query?
UPDATE---since I did a bad job of describing the question, I've come to include some example data and expected results
Where current month is not present for user 33
Where current month is present
Assuming publish_month is a DATE datatype, this should get the most recent three months of data per user...
SELECT
user_id, SUM(revenue) as s_revenue
FROM
(
SELECT
user_id, revenue, publish_month,
MAX(publish_month) OVER (PARTITION BY user_id) AS user_latest_publish_month
FROM
yourtableyoudidnotname
)
summarised
WHERE
publish_month >= DATEADD(month, -2, user_latest_publish_month)
GROUP BY
user_id
If you want to limit that to the most recent 3 months out of the last 4 calendar months, just add AND publish_month >= DATEADD(month, -3, DATE_TRUNC(month, GETDATE()))
The ambiguity here is why it is important to include a Minimal Reproducible Example
With input data and require results, we could test our code against your requirements
If you're using strings for the publish_month, you shouldn't be, and should fix that with utmost urgency.
You can use a windowing function to "number" the months. In this way the most recent one will have a value of 1, the prior 2, and the one before 3. Then you can only select the items with a number of 3 or less.
Here is how:
SELECT user_id, revienue, publish_month,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY publish_month DESC) as RN
FROM yourtableyoudidnotname
now you just select the items with RN less than 3 and do your sum
SELECT user_id, SUM(revenue) as s_revenue
FROM (
SELECT user_id, revenue, publish_month,
ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY publish_month DESC) as RN
FROM yourtableyoudidnotname
) X
WHERE RN <= 3
GROUP BY user_id
You could also do this without a sub query if you use the windowing function for SUM and a range, but I think this is easier to understand.
From the comment -- there could be an issue if you have months from more than one year. To solve this make the biggest number in the order by always the most recent. so instead of
ORDER BY publish_month DESC
you would have
ORDER BY (100*publish_year)+publish_month DESC
This means more recent years will always have a higher number so january of 2023 will be 202301 while december of 2022 will be 202212. Since january is a bigger number it will get a row number of 1 and december will get a row number of 2.

Need to count unique transactions by month but ignore records that occur 3 days after 1st entry for that ID

I have a table with just two columns: User_ID and fail_date. Each time somebody's card is rejected they are logged in the table, their card is automatically tried again 3 days later, and if they fail again, another entry is added to the table. I am trying to write a query that counts unique failures by month so I only want to count the first entry, not the 3 day retries, if they exist. My data set looks like this
user_id fail_date
222 01/01
222 01/04
555 02/15
777 03/31
777 04/02
222 10/11
so my desired output would be something like this:
month unique_fails
jan 1
feb 1
march 1
april 0
oct 1
I'll be running this in Vertica, but I'm not so much looking for perfect syntax in replies. Just help around how to approach this problem as I can't really think of a way to make it work. Thanks!
You could use lag() to get the previous timestamp per user. If the current and the previous timestamp are less than or exactly three days apart, it's a follow up. Mark the row as such. Then you can filter to exclude the follow ups.
It might look something like:
SELECT month,
count(*) unique_fails
FROM (SELECT month(fail_date) month,
CASE
WHEN datediff(day,
lag(fail_date) OVER (PARTITION BY user_id,
ORDER BY fail_date),
fail_date) <= 3 THEN
1
ELSE
0
END follow_up
FROM elbat) x
WHERE follow_up = 0
GROUP BY month;
I'm not so sure about the exact syntax in Vertica, so it might need some adaptions. I also don't know, if fail_date actually is some date/time type variant or just a string. If it's just a string the date/time specific functions may not work on it and have to be replaced or the string has to be converted prior passing it to the functions.
If the data spans several years you might also want to include the year additionally to the month to keep months from different years apart. In the inner SELECT add a column year(fail_date) year and add year to the list of columns and the GROUP BY of the outer SELECT.
You can add a flag about whether this is a "unique_fail" by doing:
select t.*,
(case when lag(fail_date) over (partition by user_id order by fail_date) > fail_date - 3
then 0 else 1
end) as first_failure_flag
from t;
Then, you want to count this flag by month:
select to_char(fail_date, 'Mon'), -- should aways include the year
sum(first_failure_flag)
from (select t.*,
(case when lag(fail_date) over (partition by user_id order by fail_date) > fail_date - 3
then 0 else 1
end) as first_failure_flag
from t
) t
group by to_char(fail_date, 'Mon')
order by min(fail_date)
In a Derived Table, determine the previous fail_date (prev_fail_date), for a specific user_id and fail_date, using a Correlated subquery.
Using the derived table dt, Count the failure, if the difference of number of days between current fail_date and prev_fail_date is greater than 3.
DateDiff() function alongside with If() function is used to determine the cases, which are not repeated tries.
To Group By this result on Month, you can use MONTH function.
But then, the data can be from multiple years, so you need to separate them out yearwise as well, so you can do a multi-level group by, using YEAR function as well.
Try the following (in MySQL) - you can get idea for other RDBMS as well:
SELECT YEAR(dt.fail_date) AS year_fail_date,
MONTH(dt.fail_date) AS month_fail_date,
COUNT( IF(DATEDIFF(dt.fail_date, dt.prev_fail_date) > 3, user_id, NULL) ) AS unique_fails
FROM (
SELECT
t1.user_id,
t1.fail_date,
(
SELECT t2.fail_date
FROM your_table AS t2
WHERE t2.user_id = t1.user_id
AND t2.fail_date < t1.fail_date
ORDER BY t2.fail_date DESC
LIMIT 1
) AS prev_fail_date
FROM your_table AS t1
) AS dt
GROUP BY
year_fail_date,
month_fail_date
ORDER BY
year_fail_date ASC,
month_fail_date ASC

Smoothing out a result set by date

Using SQL I need to return a smooth set of results (i.e. one per day) from a dataset that contains 0-N records per day.
The result per day should be the most recent previous value even if that is not from the same day. For example:
Starting data:
Date: Time: Value
19/3/2014 10:01 5
19/3/2014 11:08 3
19/3/2014 17:19 6
20/3/2014 09:11 4
22/3/2014 14:01 5
Required output:
Date: Value
19/3/2014 6
20/3/2014 4
21/3/2014 4
22/3/2014 5
First you need to complete the date range and fill in the missing dates (21/3/2014 in you example). This can be done by either joining a calendar table if you have one, or by using a recursive common table expression to generate the complete sequence on the fly.
When you have the complete sequence of dates finding the max value for the date, or from the latest previous non-null row becomes easy. In this query I use a correlated subquery to do it.
with cte as (
select min(date) date, max(date) max_date from your_table
union all
select dateadd(day, 1, date) date, max_date
from cte
where date < max_date
)
select
c.date,
(
select top 1 max(value) from your_table
where date <= c.date group by date order by date desc
) value
from cte c
order by c.date;
May be this works but try and let me know
select date, value from test where (time,date) in (select max(time),date from test group by date);

Last day of the month with a twist in SQLPLUS

I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)