I have a postgres table "Generation" with half-hourly timestamps spanning 2009 - present with energy data:
I need to aggregate (average) the data across different intervals from specific timepoints, for example data from 2021-01-07T00:00:00.000Z for one year at 7 day intervals, or 3 months at 1 day interval or 7 days at 1h interval etc. date_trunc() partly solves this, but rounds the weeks to the nearest monday e.g.
SELECT date_trunc('week', "DATETIME") AS week,
count(*),
AVG("GAS") AS gas,
AVG("COAL") AS coal
FROM "Generation"
WHERE "DATETIME" >= '2021-01-07T00:00:00.000Z' AND "DATETIME" <= '2022-01-06T23:59:59.999Z'
GROUP BY week
ORDER BY week ASC
;
returns the first time series interval as 2021-01-04 with an incorrect count:
week count gas coal
"2021-01-04 00:00:00" 192 18291.34375 2321.4427083333335
"2021-01-11 00:00:00" 336 14477.407738095239 2027.547619047619
"2021-01-18 00:00:00" 336 13947.044642857143 1152.047619047619
****EDIT: the following will return the correct weekly intervals by checking the start date relative to the nearest monday / start of week, and adjusts the results accordingly:
WITH vars1 AS (
SELECT '2021-01-07T00:00:00.000Z'::timestamp as start_time,
'2021-01-28T00:00:00.000Z'::timestamp as end_time
),
vars2 AS (
SELECT
((select start_time from vars1)::date - (date_trunc('week', (select start_time from vars1)::timestamp))::date) as diff
)
SELECT date_trunc('week', "DATETIME" - ((select diff from vars2) || ' day')::interval)::date + ((select diff from vars2) || ' day')::interval AS week,
count(*),
AVG("GAS") AS gas,
AVG("COAL") AS coal
FROM "Generation"
WHERE "DATETIME" >= (select start_time from vars1) AND "DATETIME" < (select end_time from vars1)
GROUP BY week
ORDER BY week ASC
returns..
week count gas coal
"2021-01-07 00:00:00" 336 17242.752976190477 2293.8541666666665
"2021-01-14 00:00:00" 336 13481.497023809523 1483.0565476190477
"2021-01-21 00:00:00" 336 15278.854166666666 1592.7916666666667
And then for any daily or hourly (swap out day with hour) intervals you can use the following:
SELECT date_trunc('day', "DATETIME") AS day,
count(*),
AVG("GAS") AS gas,
AVG("COAL") AS coal
FROM "Generation"
WHERE "DATETIME" >= '2022-01-07T00:00:00.000Z' AND "DATETIME" < '2022-01-10T23:59:59.999Z'
GROUP BY day
ORDER BY day ASC
;
In order to select the complete week, you should change the WHERe-clause to something like:
WHERE "DATETIME" >= date_trunc('week','2021-01-07T00:00:00.000Z'::timestamp)
AND "DATETIME" < (date_trunc('week','2022-01-06T23:59:59.999Z'::timestamp) + interval '7' day)::date
This will effectively get the records from January 4,2021 until (and including ) January 9,2022
Note: I changed <= to < to stop the end-date being included!
EDIT:
when you want your weeks to start on January 7, you can always group by:
(date_part('day',(d-'2021-01-07'))::int-(date_part('day',(d-'2021-01-07'))::int % 7))/7
(where d is the column containing the datetime-value.)
see: dbfiddle
EDIT:
This will get the list from a given date, and a specified interval.
see DBFIFFLE
WITH vars AS (
SELECT
'2021-01-07T00:00:00.000Z'::timestamp AS qstart,
'2022-01-06T23:59:59.999Z'::timestamp AS qend,
7 as qint,
INTERVAL '1 DAY' as qinterval
)
SELECT
(select date(qstart) FROM vars) + (SELECT qinterval from vars) * ((date_part('day',("DATETIME"-(select date(qstart) FROM vars)))::int-(date_part('day',("DATETIME"-(select date(qstart) FROM vars)))::int % (SELECT qint FROM vars)))::int) AS week,
count(*),
AVG("GAS") AS gas,
AVG("COAL") AS coal
FROM "Generation"
WHERE "DATETIME" >= (SELECT qstart FROM vars) AND "DATETIME" <= (SELECT qend FROM vars)
GROUP BY week
ORDER BY week
;
I added the WITH vars to do the variable stuff on top and no need to mess with the rest of the query. (Idea borrowed here)
I only tested with qint=7,qinterval='1 DAY' and qint=14,qinterval='1 DAY' (but others values should work too...)
Using the function EXTRACT you may calculate the difference in days, weeks and hours between your timestamp ts and the start_date as follows
Difference in Days
extract (day from ts - start_date)
Difference in Weeks
Is the difference in day divided by 7 and truncated
trunc(extract (day from ts - start_date)/7)
Difference in Hours
Is the difference in day times 24 + the difference in hours of the day
extract (day from ts - start_date)*24 + extract (hour from ts - start_date)
The difference can be used in GROUP BY directly. E.g. for week grouping the first group is difference 0, i.e. same week, the next group with difference 1, the next week, etc.
Sample Example
I'm using a CTE for the start date to avoid multpile copies of the paramater
with start_time as
(select DATE'2021-01-07' as start_ts),
prep as (
select
ts,
extract (day from ts - (select start_ts from start_time)) day_diff,
trunc(extract (day from ts - (select start_ts from start_time))/7) week_diff,
extract (day from ts - (select start_ts from start_time)) *24 + extract (hour from ts - (select start_ts from start_time)) hour_diff,
value
from test_table
where ts >= (select start_ts from start_time)
)
select week_diff, avg(value)
from prep
group by week_diff order by 1
Related
For a given date I want to add business days to it. For example, if today is 10-17-2022 and I have a field that is 8 business days. How can I add 8 business days to 10-17-2022 which would be 10-27-2022.
Current Data:
BUSINESS_DAYS
Date
8
10-11-2022
10
10-13-2022
9
10-12-2022
Desired Output Data
BUSINESS_DAYS
Date
FINAL_DATE
8
10-11-2022
10-21-2022
10
10-13-2022
10-27-2022
9
10-12-2022
10-25-2022
As you can see we are skipping all weekends. We can ignore holidays for now.
Update:
Using
The suggest logic I got the following answer. I changed the names up.
I used:
DATE_ADD(A.PO_SENT_DATE , INTERVAL
(CAST(PREDICTED_LEAD_TIME AS INT64)
+ (date_diff(A.PO_SENT_DATE , DATE_ADD(A.PO_SENT_DATE , INTERVAL CAST(PREDICTED_LEAD_TIME AS INT64) DAY), week)* 2))
DAY) as FINAL_DATE
Update2: Using the following:
DATE_ADD(`Date`, INTERVAL
(BUSINESS_DAYS
+ (date_diff( DATE_ADD(`Date`, INTERVAL BUSINESS_DAYS DAY),`Date`, week) * 2))
DAY) as FINAL_DATE
There are instances where the result falls on the weekend. See screenshot below. 10-22-2022 falls on a Saturday.
Consider below simple solution
select *,
( select day
from unnest(generate_date_array(date, date + (div(business_days, 5) + 1) * 7)) day
where not extract(dayofweek from day) in (1, 7)
qualify row_number() over(order by day) = business_days + 1
) final_date
from your_table
if applied to sample data in your question
with your_table as (
select 8 business_days, date '2022-10-11' date union all
select 10, '2022-10-13' union all
select 9, '2022-10-12'
)
output is
The solution from #mikhailberlyant is really really cool, and very innovative. However if you have a lot of rows in your table and value of "business_days" column varies a lot, query will be less efficient especially for larger "business_days" values as implementation needs to generate entire range of array for each row, unnest it, and then do manipulation in that array.
This might help you do calculation without any array business:
select day, add_days as add_business_days,
DATE_ADD(day, INTERVAL cast(add_days +2*ceil((add_days -(5-(
(case when EXTRACT(DAYOFWEEK FROM day) = 7 then 1 else EXTRACT(DAYOFWEEK FROM day) end)
-1)))/5)+(case when EXTRACT(DAYOFWEEK FROM day) = 7 then 1 else 0 end) as int64) DAY) as final_day
from
(select parse_date('%Y-%m-%d', "2022-10-11") as day, 8 as add_days)
I want to get data from last month day by day, I can get the last 30 days but I just want the month as it may be less or more than 30 days,
this is the query for getting the last 30 days
SELECT Trunc(timestamp),
Count(*)
FROM table1
WHERE Trunc(timestamp) > Trunc(sysdate - 30)
GROUP BY Trunc(timestamp)
ORDER BY 1;
Also, I am using it in a shell script if I can make a variable in the script and put it the query
To get data from the start of the current month until today:
SELECT TRUNC(timestamp) AS day,
COUNT(*)
FROM table1
WHERE timestamp >= TRUNC(SYSDATE, 'MM')
AND timestamp < TRUNC(SYSDATE) + INTERVAL '1' DAY
GROUP BY TRUNC(timestamp)
ORDER BY day
To get data from the same day last month until today:
SELECT TRUNC(timestamp) AS day,
COUNT(*)
FROM table1
WHERE timestamp >= ADD_MONTHS(TRUNC(SYSDATE), -1)
AND timestamp < TRUNC(SYSDATE) + INTERVAL '1' DAY
GROUP BY TRUNC(timestamp)
ORDER BY day
db<>fiddle here
I have a input table with three columns :
id => string
start_date => timestamptz
end_date => timestamptz
I want to get the average duration in seconds (end_date - start_date) per week-day number over records.
My problem is : If I have a record where interval between start_date and end_date is 4 days, I want to get the result per day, not only at the start_date or end_date, and if I have no records between 3 weeks for example, take no value for a weekday as 'zero' value in the average.
Example :
id
start_date
end_date
1 (Friday to Sunday)
2021-03-12T01:00:00.000Z
2021-03-14T01:00:00.000Z
2 (Friday)
2021-03-12T01:00:00.000Z
2021-03-12T05:00:00.000Z
3 (Wed.)
2021-03-03T16:00:00.000Z
2021-03-03T17:00:00.000Z
Expected result (european weekday here for example, sunday is 7) :
weekday
avg_duration_seconds
1
0
2
0
3
1800
4
0
5
48600
6
86400
7
3600
Thank's for your help !
Note: the following works on Postgres as you tagged that as well. I have no idea if this works on CockroachDB as well.
You can "expand" the start/end timestamps to days by using generate_series(). To calculate the effective duration on each day, the full days need to be treated differently than the partial days at the start and end. Once those timestamps are calculated it's easy to get the duration per day. The do a left join on all weekdays and group by them:
select x.weekday,
avg(extract(epoch from real_end - real_start)) as duration
from generate_series(1,7) as x(weekday)
left join (
select t.id,
extract(isodow from g.dt) as weekday,
case
when start_date < g.dt then date_trunc('day', g.dt)
else start_date
end as real_start,
case
when end_date::date > g.dt then date_trunc('day', g.dt::date + 1)
else end_date
end as real_end
from the_table t
cross join generate_series(start_date, end_date, interval '1 day') as g(dt)
) t on x.weekday = t.weekday
group by x.weekday
order by x.weekday;
I am not 100% my expressions for "real_start" and "real_end" cover all corner cases, but it should be enough to get you started.
This gives a slightly different result than your expected one, because you have the weekdays wrong for 2021-03-02 and 2021-03-11.
Online example
I need to know how many entries appear in my DB for the past 7 days with a timestamp between 23:00 & 01:00...
The Issue I have is the timestamp goes across 2 days and unsure if this is even possible in the one query.
So far I have come up with the below:
select trunc(timestamp) as DTE, extract(hour from timestamp) as HR, count(COLUMN) as Total
from TABLE
where trunc(timestamp) >= '12-NOV-19' and
extract(hour from timestamp) in ('23','00','01')
group by trunc(timestamp), extract(hour from timestamp)
order by 1,2 desc;
The result I am hoping for is something like this:
DTE | Total
20-NOV-19 5
19-NOV-19 4
18-NOV-19 4
17-NOV-19 6
Many thanks
Filter on the day first comparing it to TRUNC( SYSDATE ) - INTERVAL '7' DAY and then consider the hours by comparing the timestamp to itself truncated back to midnight with an offset of a number of hours.
select trunc(timestamp) as DTE,
extract(hour from timestamp) as HR,
count(COLUMN) as Total
from TABLE
WHERE timestamp >= TRUNC( SYSDATE ) - INTERVAL '7' DAY
AND ( timestamp <= TRUNC( timestamp ) + INTERVAL '01:00' HOUR TO MINUTE
OR timestamp >= TRUNC( timestamp ) + INTERVAL '23:00' HOUR TO MINUTE
)
group by trunc(timestamp), extract(hour from timestamp)
order by DTE, HR desc;
Subtract or add an hour to derive the date. I'm not sure what date you want to assign to each period, but the idea is:
select trunc(timestamp - interval '1' hour) as DTE,
count(*) as Total
from t
where trunc(timestamp - interval '1' hour) >= DATE '2019-11-12' and
extract(hour from timestamp) in (23, 0)
group by trunc(timestamp - interval '1' hour)
order by 1 desc;
Note: If you want times between 11:00 p.m. and 1:00 a.m., then you want the hour to be 23 or 0.
I've been trying for hours now to write a date_trunc statement to be used in a group by where my week starts on a Friday and ends the following Thursday.
So something like
SELECT
DATE_TRUNC(...) sales_week,
SUM(sales) sales
FROM table
GROUP BY 1
ORDER BY 1 DESC
Which would return the results for the last complete week (by those standards) as 09-13-2019.
You can subtract 4 days and then add 4 days:
SELECT DATE_TRUNC(<whatever> - INTERVAL '4 DAY') + INTERVAL '4 DAY' as sales_week,
SUM(sales) as sales
FROM table
GROUP BY 1
ORDER BY 1 DESC
The expression
select current_date - cast(cast(7 - (5 - extract(dow from current_date)) as text) || ' days' as interval);
should always give you the previous Friday's date.
if by any chance you might have gaps in data (maybe more granular breakdowns vs just per week), you can generate a set of custom weeks and left join to that:
drop table if exists sales_weeks;
create table sales_weeks as
with
dates as (
select generate_series('2019-01-01'::date,current_date,interval '1 day')::date as date
)
,week_ids as (
select
date
,sum(case when extract('dow' from date)=5 then 1 else 0 end) over (order by date) as week_id
from dates
)
select
week_id
,min(date) as week_start_date
,max(date) as week_end_date
from week_ids
group by 1
order by 1
;