I have a dataset like this called data_per_day
instructional_day
points
2023-01-24
2
2023-01-23
2
2023-01-20
1
2023-01-19
0
and so on. the table shows weekdays (days minus holidays and weekends) and the number of points someone has earned. 1 is the start of a streak and 0 is the end of a streak. 2 is max points after a streak has started.
I need to find how long is the latest streak. so in this case the result should be 3
I created a recursive cte but the query returns 2 as the streak count because i'm using lag mechanism with days. instead I need to adjust so that the instructional days are used rather than all dates.
RECURSIVE cte AS (
SELECT
student_unique_id,
instructional_day,
points,
1 AS cnt
FROM
`data_per_day`
WHERE
instructional_day = DATE_ADD(CURRENT_DATE('America/Chicago'), INTERVAL -1 DAY)
UNION ALL
SELECT
a.student_unique_id,
a.instructional_day,
a.points,
c.cnt+1
FROM (
SELECT
*
FROM
`data_per_day`
WHERE
points > 0 ) a
INNER JOIN
cte c
ON
a.student_unique_id = c.student_unique_id
AND a.instructional_day = c.instructional_day - INTERVAL '1' day )
SELECT
student_unique_id,
MAX(cnt) AS streak
FROM
cte --
WHERE
student_unique_id = "419"
GROUP BY
student_unique_id
How do I adjust the query?
This is not a trivial coding exercise, so I won't actually write the code and provide it.
What you have here is a gaps and islands question. You want to identify the largest "island" of days with points within a date range. Depending upon what dates are contained in your data, you may need to generate a list of sequential dates that meet your criteria.
One problem I see is that you are trying to combine the steps to generate the date range (the recursive CTE) with the points. You'll need to separate those steps.
Define the date range.
Generate the dates within the range.
Filter the dates with isweekday = 'no' and isholiday = 'no'. You will probably want to add a row number during this step.
[left] join the dates to your data, including coalesce(points, 0)
Filter the data to points > 0.
Identify the islands.
Identify the largest island per student.
Related
I'm trying to get the day difference between 2 dates in Impala but I need to exclude weekends.
I know it should be something like this but I'm not sure how the weekend piece would go...
DATEDIFF(resolution_date,created_date)
Thanks!
One approach at such task is to enumerate each and every day in the range, and then filter out the week ends before counting.
Some databases have specific features to generate date series, while in others offer recursive common-table-expression. Impala does not support recursive queries, so we need to look at alternative solutions.
If you have a table wit at least as many rows as the maximum number of days in a range, you can use row_number() to offset the starting date, and then conditional aggregation to count working days.
Assuming that your table is called mytable, with column id as primary key, and that the big table is called bigtable, you would do:
select
t.id,
sum(
case when dayofweek(dateadd(t.created_date, n.rn)) between 2 and 6
then 1 else 0 end
) no_days
from mytable t
inner join (select row_number() over(order by 1) - 1 rn from bigtable) n
on t.resolution_date > dateadd(t.created_date, n.rn)
group by id
I have a table of records ordered by date. There is a maximum of 1 record per day, but some days there is no record (weekends and bank holidays).
When I query a record by date, if no record exists for that day I am interested in the previous record by date. Eg:
SELECT * FROM rates WHERE date <= $mydate ORDER BY date DESC LIMIT 1;
Given a list of dates, how would I construct a query to return multiple records matching the exact or closest previous record for each date? Is this possible to achieve in a single query?
The array of dates may be spread over a large time frame but I wouldn't necessarily want every record in the entire time span (eg query 20 dates spread over a year long time span).
You can construct the dates as a derived table and then use SQL logic. A lateral join is convenient:
select v.dte, r.*
from (values ($date1), ($date2), ($date3)
) v(dte) left join lateral
(select r.*
from rates r
where r.date <= v.dte
order by r.date desc
limit 1
) r
on 1=1;
You might find it useful to use an array to pass in the dates using an array and using unnest() on that array.
Consider this list of dates as timestamptz:
I grouped the dates by hand using colors: every group is separated from the next by a gap of at least 2 minutes.
I'm trying to measure how much a given user studied, by looking at when they performed an action (the data is when they finished studying a sentence.) e.g.: on the yellow block, I'd consider the user studied in one sitting, from 14:24 till 14:27, or roughly 3 minutes in a row.
I see how I could group these dates with a programming language by going through all of the dates and looking for the gap between two rows.
My question is: how would go about grouping dates in this way with Postgres?
(Looking for 'gaps' on Google or SO brings too many irrelevant results; I think I'm missing the vocabulary for what I'm trying to do here.)
SELECT done, count(*) FILTER (WHERE step) OVER (ORDER BY done) AS grp
FROM (
SELECT done
, lag(done) OVER (ORDER BY done) <= done - interval '2 min' AS step
FROM tbl
) sub
ORDER BY done;
The subquery sub returns step = true if the previous row is at least 2 min away - sorted by the timestamp column done itself in this case.
The outer query adds a rolling count of steps, effectively the group number (grp) - combining the aggregate FILTER clause with another window function.
fiddle
Related:
Query to find all timestamps more than a certain interval apart
How to label groups in postgresql when group belonging depends on the preceding line?
Select longest continuous sequence
Grouping or Window
About the aggregate FILTER clause:
Aggregate columns with additional (distinct) filters
Conditional lead/lag function PostgreSQL?
Building up on Erwin's answer, here is the full query for tallying up the amount of time people spent on those sessions/islands:
My data only shows when people finished reviewing something, not when they started, which means we don't know when a session truly started; and some islands only have one timestamp in them (leading to a 0-duration estimate.) I'm accounting for both by calculating the average review time and adding it to the total duration of islands.
This is likely very idiosyncratic to my use case, but I learned a thing or two in the process, so maybe this will help someone down the line.
-- Returns estimated total study time and average time per review, both in seconds
SELECT (EXTRACT( EPOCH FROM logged) + countofislands * avgreviewtime) as totalstudytime, avgreviewtime -- add total logged time to estimate for first-review-in-island and 1-review islands
FROM
(
SELECT -- get the three key values that will let us calculate total time spent
sum(duration) as logged
, count(island) as countofislands
, EXTRACT( EPOCH FROM sum(duration) FILTER (WHERE duration != '00:00:00'::interval) )/( sum(reviews) FILTER (WHERE duration != '00:00:00'::interval) - count(reviews) FILTER (WHERE duration != '00:00:00'::interval)) as avgreviewtime
FROM
(
SELECT island, age( max(done), min(done) ) as duration, count(island) as reviews -- calculate the duration of islands
FROM
(
SELECT done, count(*) FILTER (WHERE step) OVER (ORDER BY done) AS island -- give a unique number to each island
FROM (
SELECT -- detect the beginning of islands
done,
(
lag(done) OVER (ORDER BY done) <= done - interval '2 min'
) AS step
FROM review
WHERE clicker_id = 71 AND "done" > '2015-05-13' AND "done" < '2015-05-13 15:00:00' -- keep the queries small and fast for now
) sub
ORDER BY done
) grouped
GROUP BY island
) sessions
) summary
I work for a sports film analysis company. We have teams with unique team IDs and I would like to find the number of consecutive weeks they have uploaded film to our site moving backwards from today. Each upload also has its own row in a separate table that I can join on teamid and has a unique date of when it was uploaded. So far I put together a simple query that pulls each unique DATEDIFF(week) value and groups on teamid.
Select teamid, MAX(weekdiff)
(Select teamid, DATEDIFF(week, dateuploaded, GETDATE()) as weekdiff
from leroy_events
group by teamid, weekdiff)
What I am given is a list of teamIDs and unique weekly date differences. I would like to then find the max for each teamID without breaking an increment of 1. For example, if my data set is:
Team datediff
11453 0
11453 1
11453 2
11453 5
11453 7
11453 13
I would like the max value for team: 11453 to be 2.
Any ideas would be awesome.
I have simplified your example assuming that I already have a table with weekdiff column. That would be what you're doing with DATEDIFF to calculate it.
First, I'm using LAG() window function to assign previous value (in ordered set) of a weekdiff to the current row.
Then, using a WHERE condition I'm retrieving max(weekdiff) value that has a previous value which is current_value - 1 for consecutive weekdiffs.
Data:
create table leroy_events ( teamid int, weekdiff int);
insert into leroy_events values (11453,0),(11453,1),(11453,2),(11453,5),(11453,7),(11453,13);
Code:
WITH initial_data AS (
Select
teamid,
weekdiff,
lag(weekdiff,1) over (partition by teamid order by weekdiff) as lag_weekdiff
from
leroy_events
)
SELECT
teamid,
max(weekdiff) AS max_weekdiff_consecutive
FROM
initial_data
WHERE weekdiff = lag_weekdiff + 1 -- this insures retrieving max() without breaking your consecutive increment
GROUP BY 1
SQLFiddle with your sample data to see how this code works.
Result:
teamid max_weekdiff_consecutive
11453 2
You can use SQL window functions to probe relationships between rows of the table. In this case the lag() function can be used to look at the previous row relative to a given order and grouping. That way you can determine whether a given row is part of a group of consecutive rows.
You still need overall to aggregate or filter to reduce the number of rows for each group of interest (i.e. each team) to 1. It's convenient in this case to aggregate. Overall, it might look like this:
select
team,
case min(datediff)
when 0 then max(datediff)
else -1
end as max_weeks
from (
select
team,
datediff,
case
when (lag(datediff) over (partition by team order by datediff) != datediff - 1)
then 0
else 1
end as is_consec
from diffs
) cd
where is_consec = 1
group by team
The inline view just adds an is_consec column to the data, marking whether each row is part of a group of consecutive rows. The outer query filters on that column (you cannot filter directly on a window function), and chooses the maximum datediff from the remaining rows for each team.
There are a few subtleties there:
The case expression in the inline view is written as it is to exploit the fact that the lag() computed for the first row of each partition will be NULL, which does not evaluate unequal (nor equal) to any value. Thus the first row in each partition is always marked consecutive.
The case testing min(datediff) in the outer select clause picks up teams that have no record with datediff = 0, and assigns -1 to column max_weeks for them.
It would also have been possible to mark rows non-consecutive if the first in their group did not have datediff = 0, but then you would lose such teams from the results altogether.
I have basic knowledge of SQL and have a question:
I am trying to select data from a time series (date and windspeed). I want to select the original wind speed value if it lies between hours 7 and 21. If the hour is outside this range I would like to assign the wind speed to the previous wind speed at hour 21. There is also a concern that there is the occasional point where hour 21 does not exist and would like to assign the windspeed as hour 20... 19 etc until it finds the next available hour.
SELECT
date,
CASE WHEN DATEPART(HH,date) < 7 OR DATEPART(HH,date) > 21
THEN '<WIND SPEED AT HOUR 21> ELSE <WIND SPEED> END AS ModifiedWindspeed
,WindSpeed, winddirection
from TerrainCorrectedHourlyWind w
This might make things clearer. If the hour is in the specified range, select windspeed. If not then select the wind speed from the prior day at 21 hours.
Though you've tagged the question mysql, I'm guessing this is actually SQL Server because of the DATEPART() function used. Try the following, which uses an OUTER APPLY to get your alternate value:
SELECT Date
, CASE
WHEN DATEPART(HOUR, Date)BETWEEN 7 AND 21 THEN w.WindSpeed
ELSE m.WindSpeed
END AS ModifiedWindSpeed
, w.WindSpeed
, w.WindDirection
FROM TerrainCorrectedHourlyWind AS w
OUTER APPLY(SELECT TOP 1 WindSpeed
FROM TerrainCorrectedHourlyWind
WHERE DATEPART(HOUR, Date)BETWEEN 7 AND 21
AND Date < w.Date
ORDER BY Date DESC)AS m;
Just to explain what this is doing--the OUTER APPLY will get the single most recent record (TOP 1 and ORDER BY Date DESC) for dates prior to the record in question (Date < w.Date) as well as within the hours specified. The CASE near the top chooses whether to use the current value or this alternate one based on the hour.