Dear Stack Overflow community,
I am looking for the patient id where the two consecutive dates after the very first one are less than 7 days.
So differences between 2nd and 1st date <= 7 days
and differences between 3rd and 2nd date <= 7 days
Example:
ID Date
1 9/8/2014
1 9/9/2014
1 9/10/2014
2 5/31/2014
2 7/20/2014
2 9/8/2014
For patient 1, the two dates following it are less than 7 days apart.
For patient 2 however, the following date are more than 7 days apart (50 days).
I am trying to write an SQL query that just output the patient id "1".
Thanks for your help :)
You want to use lead(), but this is complicated because you want this only for the first three rows. I think I would go for:
select t.*
from (select t.*,
lead(date, 1) over (partition by id order by date) as next_date,
lead(date, 2) over (partition by id order by date) as next_date_2,
row_number() over (partition by id order by date) as seqnum
from t
) t
where seqnum = 1 and
next_date <= date + interval '7' day and
next_date2 <= next_date + interval '7' day;
You can try using window function lag()
select * from
(
select id,date,lag(date) over(order by date) as prevdate
from tablename
)A where datediff(day,date,prevdate)<=7
Related
I need to get a series of data that will give me consecutive dates that are at least 14 days apart.
For example:
userid
date
1
1/1/2022
1
1/5/2022
1
1/31/2022
1
2/22/2022
Expected Output:
userid
date
1
1/1/2022
1
1/31/2022
1
2/22/2022
I am stuck at how do i remove 1/5/2022 from the data? I not even sure which function i can try to use in postgresql.
TIA
You can use the LAG window functions to check for consecutive dates. After that, you can check whether your date is bigger than your previous date + 14 days.
WITH cte AS (
SELECT *, LAG(date_) OVER(PARTITION BY userid ORDER BY date_) AS prevdate
FROM tab
)
SELECT userid, date_
FROM cte
WHERE date_ > prevdate + INTERVAL '14 day' OR prevdate IS NULL
Check the demo here.
I am looking at Sales Rates by month, and was able to query the 1st table. I am quite new to PostgreSQL and am trying to figure out how I can query the second (I had to do the 2nd one in Excel for now)
I have the current Sales Rate and I would like to compare it to the Sales Rate 1 and 2 months ago, as an averaged rate.
I am not asking for an answer how exactly to solve it because this is not the point of getting better, but just for hints for functions to use that are specific to PostgreSQL. What I am trying to calculate is the 2 month average in the 2nd table based on the lagged values of the 2nd table. Thanks!
Here is the query for the 1st table:
with t1 as
(select date,
count(sales)::numeric/count(poss_sales) as SR_1M_before
from data
where date between '2019-07-01' and '2019-11-30'
group by 1),
t2 as
(select date,
count(sales)::numeric/count(poss_sales) as SR_2M_before
from data
where date between '2019-07-01' and '2019-10-31'
group by 1)
select t0.date,
count(t0.sales)::numeric/count(t0.poss_sales) as Sales_Rate
t1.SR_1M_before,
t2.SR_2M_before
from data as t0
left join t1 on t0.date=t1.date
left join t2 on t0.date=t1.date
where date between '2019-07-01' and '2019-12-31'
group by 1,3,4
order by 1;
As commented by a_horse_with_no_name, you can use window functions to take the average of the two previous monthes with a range clause:
select
date,
count(sales)::numeric/count(poss_sales) as Sales_Rate,
avg(count(sales)::numeric/count(poss_sales)) over(
order by date
rows between '2 month' preceding and '1 month' preceding
) Sales_Rate,
count(sales)::numeric/count(poss_sales) as Sales_Rate
- avg(count(sales)::numeric/count(poss_sales)) over(
order by date
rows between '2 month' preceding and '1 month' preceding
) PercentDeviation
from data
where date between '2019-07-01' and '2019-12-31'
group by date
order by date;
Your data is a bit confusing -- it would be less confusing if you had decimal places (that is, 58% being the average of 57% and 58% is not obvious).
Because you want to have NULL values on the first two rows, I'm going to calculate the values using sum() and count():
with q as (
<whatever generates the data you have shown>
)
select q.*,
(sum(sales_rate) over (order by date
rows between 2 preceding and 1 preceding
) /
nullif(count(*) over (order by date
rows between 2 preceding and 1 preceding
)
) as two_month_average
from q;
You could also express this using case and avg():
select q.*,
(case when row_number() over (order by date) > 2)
then avg(sales_rate) over (order by date
rows between 2 preceding and 1 preceding
)
end) as two_month_average
from q;
Let's say I have a dataset with two columns: ID and timestamp. My goal is to count return IDs that have at least n timestamps in any 30 day window.
Here is an example:
ID Timestamp
1 '2019-01-01'
2 '2019-02-01'
3 '2019-03-01'
1 '2019-01-02'
1 '2019-01-04'
1 '2019-01-17'
So, let's say I want to return a list of IDs that have 3 timestamps in any 30 day window.
Given above, my resultset would just be ID = 1. I'm thinking some kind of windowing function would accomplish this, but I'm not positive.
Any chance you could help me write a query that accomplishes this?
A relatively simple way to do this involves lag()/lead():
select t.*
from (select t.*,
lead(timestamp, 2) over (partition by id order by timestamp) as timestamp_2
from t
) t
where datediff(day, timestamp, timestamp_2) <= 30;
The lag() looks at the third timestamp in a series. The where checks if this is within 30 days of the original one. The result is rows where this occurs.
If you just want the ids, then:
select distinct id
from (select t.*,
lead(timestamp, 2) over (partition by id order by timestamp) as timestamp_2
from t
) t
where datediff(day, timestamp, timestamp_2) <= 30;
I'm pretty new to SQL and have this problem:
I have a filled table with a date column and other not interesting columns.
date | name | name2
2015-03-20 | peter | pan
2015-03-20 | john | wick
2015-03-18 | harry | potter
What im doing right now is counting everything for a date
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
what i want to do now is counting the resulting lines and only returning them if there are less then 10 resulting lines.
What i tried so far is surrounding the whole query with a temp table and the counting everything which gives me the number of resulting lines (yeah)
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select count(*)
from temp_count
What is still missing the check if the number is smaller then 10.
I was searching in this Forum and came across some "having" structs to use, but that forced me to use a "group by", which i can't.
I was thinking about something like this :
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
having count(*) < 10
maybe im too tired to think of an easy solution, but i can't solve this so far
Edit: A picture for clarification since my english is horrible
http://imgur.com/1O6zwoh
I want to see the 2 columned results ONLY IF there are less then 10 rows overall
I think you just need to move your having clause to the inner query so that it is paired with the GROUP BY:
with temp_count (date, counter) as
(
select date, count(*)
from testtable
where date >= current date - 10 days
group by date
having count(*) < 10
)
select *
from temp_count
If what you want is to know whether the total # of records (after grouping), are returned, then you could do this:
with temp_count (date, counter) as
(
select date, counter=count(*)
from testtable
where date >= current date - 10 days
group by date
)
select date, counter
from (
select date, counter, rseq=row_number() over (order by date)
from temp_count
) x
group by date, counter
having max(rseq) >= 10
This will return 0 rows if there are less than 10 total, and will deliver ALL the results if there are 10 or more (you can just get the first 10 rows if needed with this also).
In your temp_count table, you can filter results with the WHERE clause:
with temp_count (date, counter) as
(
select date, count(distinct date)
from testtable
where date >= current date - 10 days
group by date
)
select *
from temp_count
where counter < 10
Something like:
with t(dt, rn, cnt) as (
select dt, row_number() over (order by dt) as rn
, count(1) as cnt
from testtable
where dt >= current date - 10 days
group by dt
)
select dt, cnt
from t where 10 >= (select max(rn) from t);
will do what you want (I think)
I am sorry if the question is silly because i'm new to SQL Server. I want to select top 5 records for each day of specified month.
e.g.
top 5 records for day 1 in month september
top 5 records for day 2 in month september
top 5 records for day 3 in month september
.
.
top 5 records for day 31 in month september
and show these all records as a one result.
Let's say you're checking speeding records for the month June 2012, and you wanted the top 5 speeds (by speed desc).
SELECT *
FROM (
SELECT *, RowNum = Row_number() over (partition by Cast(EventTime as Date)
order by Speed desc)
FROM Events
WHERE EventTime >= '20120601'
AND EventTime < '20120701'
) X
WHERE RowNum <= 5
Try this one,
WITH TopFiveRecords
AS
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY dayColumn ORDER BY colName DESC) RN
FROM tableName
)
SELECT *
FROM TopFiveRecords
WHERE RN <= 5
-- AND date condition here ....
dayColumn the column that contains the date of the month
colName the column to be sorted