Hi I'm trying to create a usage profile, for each hour of the week from three months of data in Postgres.
The raw data is 90 days of sensor_id, timestamp, value and the table should have these columns:
sensor_id, day_of_week, hour_of_day, avg_1hour, max_1hour, min_1hour, p95_1hour, max_2hour, avg_2hour min_2hour, p95_2hour, avg_3hour, max_3hour, min_3hour, p95_3hour
The *_1hour columns are the result of aggregate functions for each hour_of_day, day_of_week pair over data within that hour. This is not so bad, and I believe this query generates the desired result.
Select
sensor_id,
extract(dow from ts) as day_of_week,
extract(hour from ts) as hour_of_day,
avg(val) as avg_1hour,
PERCENTILE_CONT(.95) within group (order by val) as p95_1hour,
max(val) as max_1hour,
min(val) as min_1hour
from timeseries_data
group by hour_of_day, day_of_week
where ts between current_date - interval '1 day' and current_date - interval '91 day'
order by day_of_week, hour_of_day asc
For example avg_1hour should have a row
where day_of_week is 1 (Monday) and hour_of_day would be 6 (5am), and then avg_1hour would be the average of every reading at Monday, 5am for the last 90 days.
The *_2hour and *_3hour columns are harder for me.
For the same day, hour pair on the *_2hour columns, for example there would be a row where day_of_week is 1 (Monday), hour_of_day would be 6 (5am), but include the prior hour, so avg_2hour would be the average of vals from all rows where day_of_week is 1 (Monday), and hour_of_day would be 6 (5am) or 5 (4am).
avg_3hour would be the average of vals from all rows where day_of_week is 1 (Monday), and hour_of_day would be 6 (5am), 5 (4am) or 4 (3am).
This is running on a TimescaleDB server with Postgres 13.3.
Thanks in advance.
Welcome, Gregory! Have you checked the window function?
You can create window in different timeframes to project the groups.
Timescale also offers counter-aggregations that can be used with continuous aggregates to save aggregated data that can be reused in bigger timeframes without needing to recalculate everything.
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 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
I work with a hotel client where they have a BigQuery database which has hotel booking data. I've shared the relevant columns in the image below which list the names of each hotel, the arrival date of the guest, the departure date, and the revenue generated from the each booking:
My problem statement is that I have to showcase how many rooms have been booked, and how much revenue has been made for each hotel every month where my final grid would look similar to this:
The important points to remember are:
the depart_dt - arrival_dt are the number of nights that the guest is staying
the Rez_rate_total / (depart_dt - arrival_dt) is the revenue made per night
My problem here is trying to figure out how to change the start date and end date columns into groups of months. The challenge comes when a guest arrives in one month and leaves in the next month. For example, Row 5 in the original data has the guest coming in on 18th July and leaving on 1st Aug - so 13 days of his stay and 13 days of revenue has to be included in July and 1 day has to be included in August.
I haven't used SQL in a while so this is as far as I got:
WITH
temp_table AS (
SELECT
hotel_long_nm,
arrival_dt,
depart_dt,
DATE_DIFF(depart_dt, arrival_dt, day) AS room_nights,
rez_rate_total
FROM
`DATABASE.analytics.bookings` )
SELECT
*
FROM
temp_table
Any help would be greatly appreciated!
Consider the following approach:
with bookings as (
select hotel_long_nm, date(arrival_dt) as arrival_dt, date(depart_dt) as depart_dt, rez_rate_total from project.dataset.bookings
),
tmp as (
-- expose the dates in the reservation (excluding last day of reservation)
select *, generate_date_array(arrival_dt,date_sub(depart_dt, interval 1 day)) as stay_dates from bookings
),
calc as (
-- unnest and calculate the daily rate
select
hotel_long_nm,
stay_dt,
1 as stay_nights,
rez_rate_total/array_length(stay_dates) as rez_rate_daily
from tmp
left join unnest(stay_dates) as stay_dt
),
agg as (
-- aggregate to the year-month level
select
date_trunc(stay_dt, month) as year_month,
hotel_long_nm,
sum(stay_nights) as room_nights,
round(sum(rez_rate_daily),2) as rez_rate_total
from calc
group by 1,2
)
select * from agg
order by hotel_long_nm, year_month
You can consider this approach, following this logic.
Validate if both dates are in the same month
If are not in the same month, i get the final date of the month of
arrival date and subtract both dates
I get the first date of the month of the depart date and subtract
and subtract both dates
In this code you can see an example:
SELECT
/*arrival date*/
CURRENT_DATE() AS the_arival,
/*depart_dt*/
DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) AS the_depart,
/*total of night between arrival date and depart date*/
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) , CURRENT_DATE(), DAY) AS total_room_nights,
/* validate if the dates are in the same month or different month if equal 0 same month if >0 another month */
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY) , CURRENT_DATE(), MONTH) AS Same_Month,/*1 no and 0 yes/
/*in this case are in different month*/
/*I get the final date of the arrival month and subtract with the arrival date*/
DATE_DIFF(DATE_SUB(DATE_TRUNC(DATE_ADD(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), INTERVAL 1 MONTH), MONTH), INTERVAL 1 DAY),DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), DAY) as total_room_nights_first_mont,
/*I get the initial date of the depart month and subtract with the depart date i add +1 because is the night between last day of the mont and first day of the next month*/
DATE_DIFF(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY),DATE_TRUNC(DATE_ADD(CURRENT_DATE(), INTERVAL 30 DAY), MONTH), DAY)+1 as total_room_nights_second_month
You can see more information about the date function.Click 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
column created_at type timestamp without timezone.
I need to get delta in minutes between current date and column created_at
Query:
select id, created_at,
extract(minutes from (CURRENT_TIMESTAMP) - created_at) as delta
from shop_order order by created_at
And here result:
Why in record with id = 20 the delta is 19 ?
It's difference is 3 DAYS. Why show only 19 minutes?
An interval (which is the result of subtracting two timestamp) consists of several "parts" (similar to a date) and extract only extracts the named part, not the representation of that interval for that unit. If the result of the subtraction is e.g. 3 days 19 minutes extract will return 19 minutes - similar to the way extract(year ...) or extract(month ...) work.
You can extract the number of seconds and then divide that by 60 to get the total duration in minutes:
select id,
created_at,
extract(epoch from CURRENT_TIMESTAMP - created_at) / 60 as delta
from shop_order order
by created_at