SQL- Avg value by date and ID - sql

I'm very new to SQL and am having trouble figuring out how to find some average values. Basically I have table with 3 columns, Date, ID, and Value. I'm trying to find the average for each ID on a daily basis. The actual table has thousands of entries with a varrying number of values logged for each ID on each day. Sample table below-
SAMPLE TABLE
Date ID Value
1-1-14 12:45 1 5
1-1-14 13:45 1 6
1-1-14 08:45 2 8
1-1-14 10:45 2 8
1-2-14 01:45 1 2
1-2-14 04:45 1 4
1-2-14 01:45 2 20
1-2-14 04:45 2 24
SAMPLE RESULTS
Date ID AvgValue
1-1-14 1 5.5
1-1-14 2 8
1-2-14 1 3
1-2-14 2 22
I would greatly appreciate any help! Thanks!

The basic query is simple:
select date, id, avg(value)
from your_table
group by date, id, avg
However, as you just want the date part of the datetime column you can cast/convert that to a narrower date type, and also, if the value that you're averaging is an int you might have to cast it to a floating point type first. For SQL Server the query could look like this:
select
cast(date as date) as date,
id,
avg(cast(value as decimal(10,5))) as avg
from table1
group by cast(date as date), id
order by 1
Sample SQL Fiddle

Related

Handling duplicates when rolling totals using OVER Partition by

I'm trying to get the rolling amount column totals for each date, from the 1st day of the month to whatever the date column value is, shown in the input table.
Output Requirements
Partition by the 'team' column
Restart rolling totals on the 1st of each month
Question 1
Is my below query correct to get my desired output requirements shown in Output Table below? It seems to work but I must confirm.
SELECT
*,
SUM(amount) OVER (
PARTITION BY
team,
month_id
ORDER BY
date ASC
) rolling_amount_total
FROM input_table;
Question 2
How can I handle duplicate dates, shown in the first 2 rows of Input Table? Whenever there is a duplicate date the amount is a duplicate as well. I see a solution here: https://stackoverflow.com/a/60115061/6388651 but no luck getting it to remove the duplicates. My non-working code example is below.
SELECT
*,
SUM(amount) OVER (
PARTITION BY
team,
month_id
ORDER BY
date ASC
) rolling_amount_total
FROM (
SELECT DISTINCT
date,
amount,
team,
month_id
FROM input_table
) t
Input Table
date
amount
team
month_id
2022-04-01
1
A
2022-04
2022-04-01
1
A
2022-04
2022-04-02
2
A
2022-04
2022-05-01
4
B
2022-05
2022-05-02
4
B
2022-05
Desired Output Table
date
amount
team
month_id
Rolling_Amount_Total
2022-04-01
1
A
2022-04
1
2022-04-02
2
A
2022-04
3
2022-05-01
4
B
2022-05
4
2022-05-02
4
B
2022-05
8
Q1. Your sum() over () is correct
Q2. Replace from input_table, in your first query, with :
from (select date, sum(amount) as amount, team, month_id
from input_table
group by date, team, month_id
) as t

SQL: How to group rows with the condition that sum of fields is limited to a certain value?

This is my table:
id user_id date balance
1 1 2016-05-10 10
2 1 2016-05-10 30
3 2 2017-04-24 5
4 2 2017-04-27 10
5 3 2017-11-10 40
I want to group the rows by user_id and sum the balance, but so that the sum is equal or less than 30. Moreover, I need to display the minimum date in the group. It should look like this:
id balance date_start
1-1 10 2016-05-10
1-2 30 2016-05-10
2-1 15 2017-04-24
Excuse for my language. Thanks.
You should be able to do so by using group by & having, here is an example of what may solve your case :
SELECT id, user_id, SUM(balance) as balance, data_start
FROM your_table
GROUP BY user_id
HAVING SUM(balance) >= 30
AND MIN(date_start)
This is a good way to do it with one query, but it is a complex query and you should be careful if using it on a very large tables.

Count over relative dates in Amazon Redshift

I have 2 tables on Amazon Redshift and they look as follows:
Tablename: Groups
GroupID Created
1 2016-08-04
2 2017-05-24
3 2017-06-12
Tablename: GroupActivities
GroupID CreationTime ActivityType
1 2016-08-13 Assign
1 2016-09-13 Assign
2 2017-05-25 Create
2 2017-05-27 Assign
3 2017-06-24 Create
3 2017-06-28 Assign
I would like to count the number of activities within each 30 day period from group creation. For example, I would like the output to be the following:
GroupID Period ActivityCount
1 Period1 1
1 Period2 1
2 Period1 2
3 Period1 2
I could do this if the dates were not relative, but I'm not sure how to achieve this when the dates are relative. Any help would be much appreciated.
TIA.
join the tables by group id, use integer division of date difference to identify the period and aggregate:
select
group_id
,'Period'||((a.creationtime::date-g.created::date)/30+1)::varchar as period
,count(1) as activity_count
from groups g
join activities a
on g.groupid=a.groupid
group by 1,2

SQL - Get column value based on another column's average between select rows

I've got a table something like..
[DateValueField][Hour][Value]
2014-09-01 1 200
...
2014-09-01 24 400
2014-09-02 1 220
...
2014-09-02 24 200
...
I need the same value for each DateValueField based on the average Value for Hour between 6-12 for example but have that display for all hours, not just 6-12. For instance...
[DateValueField][Hour][Value]
2014-09-01 1 300
...
2014-09-01 24 300
2014-09-02 1 190
...
2014-09-02 24 190
...
Query I'm trying is...
select DateValueField, Hour,
(select avg(Value) as Value from MyTable where Hour
between 6 and 12) as Value from MyTable
where DateValueField between '2014' and '2015'
group by DateValueField, Hour
order by DateValueField, Hour
But it gives me the Value as an average of ALL Values but I need it averaged out for that particular day between the hours I specify.
I'd appreciate some help/advice. Thanks!
You can use a derived table to get the average value between hours 6 and 12 grouped by date and then join that to your original table
select t1.DateValueField, t1.Hour, t2.avg_value
from MyTable t1
join (
select DateValueField, avg(Value) avg_value
from MyTable
where hour between 6 and 12
group by DateValueField
) t2 on t2.DateValueField = t1.DateValueField
order by t1.DateValueField, t1.Hour
Note: You may want to use a left join if some of your dates don't have values between hours 6 and 12 but you still want to retrieve all rows from MyTable.

How to aggregate 7 days in SQL

I was trying to aggregate a 7 days data for FY13 (starts on 10/1/2012 and ends on 9/30/2013) in SQL Server but so far no luck yet. Could someone please take a look. Below is my example data.
DATE BREAD MILK
10/1/12 1 3
10/2/12 2 4
10/3/12 2 3
10/4/12 0 4
10/5/12 4 0
10/6/12 2 1
10/7/12 1 3
10/8/12 2 4
10/9/12 2 3
10/10/12 0 4
10/11/12 4 0
10/12/12 2 1
10/13/12 2 1
So, my desired output would be like:
DATE BREAD MILK
10/1/12 1 3
10/2/12 2 4
10/3/12 2 3
10/4/12 0 4
10/5/12 4 0
10/6/12 2 1
Total 11 15
10/7/12 1 3
10/8/12 2 4
10/9/12 2 3
10/10/12 0 4
10/11/12 4 0
10/12/12 2 1
10/13/12 2 1
Total 13 16
--------through 9/30/2013
Please note, since FY13 starts on 10/1/2012 and ends on 9/30/2012, the first week of FY13 is 6 days instead of 7 days.
I am using SQL server 2008.
You could add a new computed column for the date values to group them by week and sum the other columns, something like this:
SELECT DATEPART(ww, DATEADD(d,-2,[DATE])) AS WEEK_NO,
SUM(Bread) AS Bread_Total, SUM(Milk) as Milk_Total
FROM YOUR_TABLE
GROUP BY DATEPART(ww, DATEADD(d,-2,[DATE]))
Note: I used DATEADD and subtracted 2 days to set the first day of the week to Monday based on your dates. You can modify this if required.
Use option with GROUP BY ROLLUP operator
SELECT CASE WHEN DATE IS NULL THEN 'Total' ELSE CONVERT(nvarchar(10), DATE, 101) END AS DATE,
SUM(BREAD) AS BREAD, SUM(MILK) AS MILK
FROM dbo.test54
GROUP BY ROLLUP(DATE),(DATENAME(week, DATE))
Demo on SQLFiddle
Result:
DATE BREAD MILK
10/01/2012 1 3
10/02/2012 2 4
10/03/2012 2 3
10/04/2012 0 4
10/05/2012 4 0
10/06/2012 2 1
Total 11 15
10/07/2012 1 3
10/08/2012 4 7
10/10/2012 0 4
10/11/2012 4 0
10/12/2012 2 1
10/13/2012 2 1
Total 13 16
You are looking for a rollup. In this case, you will need at least one more column to group by to do your rollup on, the easiest way to do that is to add a computed column that groups them into weeks by date.
Take a lookg at: Summarizing Data Using ROLLUP
Here is the general idea of how it could be done:
You need a derived column for each row to determine which fiscal week that record belongs to. In general you could subtract that record's date from 10/1, get the number of days that have elapsed, divide by 7, and floor the result.
Then you can GROUP BY that derived column and use the SUM aggregate function.
The biggest wrinkle is that 6 day week you start with. You may have to add some logic to make sure that the weeks start on Sunday or whatever day you use but this should get you started.
The WITH ROLLUP suggestions above can help; you'll need to save the data and transform it as you need.
The biggest thing you'll need to be able to do is identify your weeks properly. If you don't have those loaded into tables already so you can identify them, you can build them on the fly. Here's one way to do that:
CREATE TABLE #fy (fyear int, fstart datetime, fend datetime);
CREATE TABLE #fylist(fyyear int, fydate DATETIME, fyweek int);
INSERT INTO #fy
SELECT 2012, '2011-10-01', '2012-09-30'
UNION ALL
SELECT 2013, '2012-10-01', '2013-09-30';
INSERT INTO #fylist
( fyyear, fydate )
SELECT fyear, DATEADD(DAY, Number, DATEADD(DAY, -1, fy.fstart)) AS fydate
FROM Common.NUMBERS
CROSS APPLY (SELECT * FROM #fy WHERE fyear = 2013) fy
WHERE fy.fend >= DATEADD(DAY, Number, DATEADD(DAY, -1, fy.fstart));
WITH weekcalc AS
(
SELECT DISTINCT DATEPART(YEAR, fydate) yr, DATEPART(week, fydate) dt
FROM #fylist
),
ridcalc AS
(
SELECT
ROW_NUMBER() OVER (ORDER BY yr, dt) AS rid, yr, dt
FROM weekcalc
)
UPDATE #fylist
SET fyweek = rid
FROM #fylist
JOIN ridcalc
ON DATEPART(YEAR, fydate) = yr
AND DATEPART(week, fydate) = dt;
SELECT list.fyyear, list.fyweek, p.[date], COUNT(bread) AS Bread, COUNT(Milk) AS Milk
FROM products p
JOIN #fylist list
ON p.[date] = list.fydate
GROUP BY list.fyyear, list.fyweek, p.[date] WITH ROLLUP;
The Common.Numbers reference above is a simple numbers table that I use for this sort of thing (goes from 1 to 1M). You could also build that on the fly as needed.