I have a table with 17,000 records that is ordered by time spaced in 15 minute intervals. The time values loop back onto themselves every 24 hours, so for example, I could have 100 records that are all at 1 AM, just on different days. I want to create a 'average day' by taking those 100 records at 1 am and finding the average of them for the averaged 1 am.
I don't know how to format the table to make it show up nicely here.
I'm assuming you want to calculate the average value per time interval regardless of the day in a query. You could use this SQL to group your table by Time interval only (assuming that it's separate from the date field), and average whichever fields you want to average. Do not select or group by the date field, just select and group by the time field.
SELECT TimeField
, AVG([Field1ToAverage])
, AVG([Field2ToAverage])
FROM MyTable
GROUP BY TimeField;
If the date and time fields are stored together in the same column, you will have to extract the time only:
SELECT TimeValue([DateTimeField])
, AVG([Field1ToAverage])
, AVG([Field2ToAverage])
FROM MyTable
GROUP BY TimeValue([DateTimeField]);
Related
I need to find the result of a calculation that is nothing more than the average time in days from creation to completion of a task.
In this case, using a Redshift database (looker).
I have two dates (2022/10/01 to 2022/10/21) and I need to find the average day of execution of the creation of an object from start to finish.
Previously, I was able to calculate the totals of objects created per day, but I can't bring up the average:
SELECT created::date, count(n1pk_package_id)
FROM dbt_dw.base_package
WHERE fk_company_id = 245821 and created >= '2022-10-01' and created < '2022-10-22'
GROUP BY created::date
ORDER BY created DESC
I'm not able to do the opposite way of the count to bring the average of the range of days.
Assumption:
There is a created column in your table
You want to know the 'average' of the created column
You could extract the number of days that each date is different from a base date, and then use that to determine the 'average date'. It would be something like this:
select
date '2022-10-01' + interval '1 day' * int(avg(created - date '2022-10-01'))
from table
It subtracts a date (any date will do) from created, finds the average of that value against all desired rows, converts it to days and adds it back to that same date.
Need help to get the data of particular format
We have a table which have a data which of production now we need to select the data of each day with particular time period which is differentiate between three shift A,B,C.
In our table we have a datetime column which capture's each seconds data now that data we need in shiftwise like 6am to 2pm is of A shift production count and 2pm to 10pm of shift B and 10pm to 6 am of shift C.
here i am getting the data for single day where i have written the below query which is working good.
select distinct(count(PRD_SERIAL_NUMBER)),(select convert(date,getdate())) as date,'B' as shift_name
from table_name
where status=02
and LAST_UPDATED_DATE
between (SELECT FORMAT(GETDATE(),'yyyy-MM-dd 14:01:00.000')) and
(SELECT FORMAT(GETDATE()-26,'yyyy-MM-dd 22:01:00.000'))
refer below output image 1
Here i am getting the count for single day and for upcoming days i have solution but now the question arise is i have a past 4 Month data which i need to get in datewise and shiftwise count and for the column prd_serial_number have duplicate entries so it should be in distinct.
please refer below image 2 for required output format
I am looking for a query which gives me the daily playing time. The start (first_date) and end date(last_update) are given as shown in the Table. The following query gives me the sum of playing time on given date. How can I extend it to get a table from first day to last day and plot the query data in it and show 0 on dates when no game is played.
SELECT startTime, SUM(duration) as sum
FROM myTable
WHERE startTime = endTime
GROUP BY startTime
To show date when no one play you will need create a table days with a date field day so you could do a left join. (100 years is only 36500 rows).
Using select Generate days from date range
This use store procedure in MSQL
I will assume if a play pass the midnight a new record begin. So I could simplify my code and remove the time from datetime field
SELECT d.day, SUM(duration) as sum
FROM
days d
left join myTable m
on CONVERT(date, m.starttime) = d.day
GROUP BY d.day
If I understand correctly, you could try:
SELECT SUM(duration) AS duration, date
FROM myTable
WHERE date <= 20140430
AND date => 20140401
GROUP BY date
This would get the total time played for each date between april 1 and april 30
As far as showing 0 for dates not in the table, I don't know.
Also, the table you posted doesn't show a duration column, but the query you posted does, so I went ahead and used it.
I have a table that lists activity for people and start / end timed for activity.
How do I get total amount of records for each person?
SELECT NAME,
--sum(startDT- endDT) AS minutes -- stuck here
FROM TABLE1
GROUP BY NAME
You're subtracting end time from start time, which will produce a negative value - try flipping those around (subtract start time from end time). The following will give you the number of records and the total elapsed time for each NAME:
SELECT NAME,
COUNT(*) AS "Records for NAME",
TO_CHAR(NUMTODSINTERVAL(SUM(END_DATE_TIME - START_DATE_TIME), 'DAY')) AS MINUTES
FROM TABLE1
GROUP BY NAME
SQLFiddle here
Share and enjoy.
Assuming that startDT and endDT are both of type date, you were really close. Subtracting two dates gives a difference in days. Multiply by 24 to get a difference in hours and again by 60 to get minutes
SELECT NAME,
sum(endDT - startDT)*24*60 AS minutes -- stuck here
FROM TABLE1
GROUP BY NAME
Assuming that your differences aren't always an exactly even number of minutes, you'll either get a non-integer result (e.g. 12.5 for 12 minutes 30 seconds) here or you'll want to either round or trunc the sum to get an integer number of minutes.
I need to show distinct users per week. I have a date-visit column, and a user id, it is a big table with 1 billion rows.
I can change the date column from the CSVs to year,month, day columns. but how do I deduce the week from that in the query.
I can calculate the week from the CSV, but this is a big process step.
I also need to show how many distinct users visit day after day, looking for workaround as there is no date type.
any ideas?
To get the week of year number:
SELECT STRFTIME_UTC_USEC(TIMESTAMP('2015-5-19'), '%W')
20
If you have your date as a timestamp (i.e microseconds since the epoch) you can use the UTC_USEC_TO_DAY/UTC_USEC_TO_WEEK functions. Alternately, if you have an iso-formatted date string (e.g. "2012/03/13 19:00:06 -0700") you can call PARSE_UTC_USEC to turn the string into a timestamp and then use that to get the week or day.
To see an example, try:
SELECT LEFT((format_utc_usec(day)),10) as day, cnt
FROM (
SELECT day, count(*) as cnt
FROM (
SELECT UTC_USEC_TO_DAY(PARSE_UTC_USEC(created_at)) as day
FROM [publicdata:samples.github_timeline])
GROUP BY day
ORDER BY cnt DESC)
To show week, just change UTC_USEC_TO_DAY(...) to UTC_USEC_TO_WEEK(..., 0) (the 0 at the end is to indicate the week starts on Sunday). See the documentation for the above functions at https://developers.google.com/bigquery/docs/query-reference for more information.