I´m trying to do some analysis in the following data
WeekDay Date Count
5 06/09/2018 20
6 07/09/2018 Null
7 08/09/2018 19
1 09/09/2018 16
2 10/09/2018 17
3 11/09/2018 24
4 12/09/2018 25
5 13/09/2018 24
6 14/09/2018 23
7 15/09/2018 23
1 16/09/2018 9
2 17/09/2018 23
3 18/09/2018 33
4 19/09/2018 22
5 20/09/2018 31
6 21/09/2018 17
7 22/09/2018 10
1 23/09/2018 12
2 24/09/2018 26
3 25/09/2018 29
4 26/09/2018 27
5 27/09/2018 24
6 28/09/2018 29
7 29/09/2018 27
1 30/09/2018 19
2 01/10/2018 26
3 02/10/2018 39
4 03/10/2018 32
5 04/10/2018 37
6 05/10/2018 Null
7 06/10/2018 26
1 07/10/2018 11
2 08/10/2018 32
3 09/10/2018 41
4 10/10/2018 37
5 11/10/2018 25
6 12/10/2018 20
The problem that I want to solve is: I want to create a table with the average of the 3 last same weekdays related to the day. But, when there is a NULL in the weekday, I want to ignore and do the average only with the remain numbers, not count NULL as an 0. I will give you an example here:
The date in this table is day/month/year :)
Ex: On day 12/10/2018, I need the average from
the days 05/10/2018; 28/09/2018; 21/09/2018. These are the last 3 same weekday(six) as 12/10/2018.
. Their values are Null; 29; 17. Then the result of this average must be 23, because I need to ignore the NULL, and not be 15,333.
How can I do this?
The count() function ignores nulls (i.e. does NOT increment if it encounters null) so I suggest you simply count the values then may contain the nulls you wish to ignore.
dow datecol value
6 21/09/2018 17
6 28/09/2018 29
6 05/10/2018 Null
e.g. sum(value) above = 46, and the count(value) = 2 so the average is 23.0 (and avg(value) will also return 23.0 as it also ignores nulls)
select
weekday
, `date`
, `count`
, (select (sum(`count`) * 1.0) / (count(`count`) * 1.0)
from atable as t2
where t2.weekday = t1.weekday
and t2.`date` < t1.`date
order by t2.`date` DESC
limit 3
) as average
from atable as t1
You could just use avg(count) in the query above, and get the same result.
ps. I do hope you do NOT use count as a column name! I also would suggest you do NOT use date as a column name either. i.e. Avoid using SQL terms as names.
SELECT WeekDay, AVG(Count)
FROM myTable
WHERE Count IS NOT NULL
GROUP BY WeekDay
Use IsNULL(Count,0) in your Select
SELECT WeekDay, AVG(IsNULL(Count,0))
FROM myTable
GROUP BY WeekDay
First off, you need to get the number of instances of that weekday in the data since you just need the last 3 same week days
create table table2
as
select
row_number() over(partition by weekday order by date desc) as rn
,weekday
,date
,count
from table
From here, you can get what you want. With you explanation, you don't need to filter out the NULL values for count. Just doing the avg() aggregation will simply ignore it.
select
weekday
,avg(count)
from table2
where rn in (1,2,3)
group by weekday
Related
I have table 1:
historial_id
timestamp
address
value
insertion_time
1
2022-01-29
1
84
2022-01-31
2
2022-01-29
2
40
2022-01-31
3
2022-01-30
1
84
2022-01-31
4
2022-01-30
2
41
2022-01-31
5
2022-01-30
2
41
2022-01-31
(sometimes it has repeated rows)
...
I need a Query to get:
timestamp
value(address 1)
value(address 2)
2022-01-29
84
40
2022-01-30
84
41
......
I tried with:
SELECT timestamp, ( SELECT value
FROM historical
WHERE register_type=11
AND address=2
AND timestamp=t1.timestamp
GROUP BY value
) AS CORRIENTE_mA,
( SELECT value
FROM historical
WHERE register_type=11
AND address=1
AND timestamp=t1.timestamp
GROUP BY value ) AS Q_M3pH
FROM historical AS t1
GROUP BY timestamp;
But it's too slow, it even stops because of exceeded time.
I tried with distinct too instead of group by
I think you need dynamic pivot.
Please try and avoid MySQL reserved words like timestamp.
Below query return only the max value for address 1 and 2 grouping by timestamp.
This is a simplified version of your query :
select
`timestamp`
, max(case when address=1 then value end) as value_address1
, max(case when address=2 then value end) as value_address2
from historical
group by `timestamp`;
Result:
timestamp value_address1 value_address2
2022-01-29 84 40
2022-01-30 84 41
Demo
I have a simple query where I would like a drop down box to show the results in numeric (integer) order. Below is my query.
Select fv
FROM fvqc
ORDER BY Value ASC
The results I get are like this -
1
13
14
18
2
23
27
3
30
31
What I would like to see is this -
1
2
3
13
14
18
23
27
30
31
You need to cast your values as numeric:
Select fv
FROM fvqc
ORDER BY CAST(Value AS INT) ASC
I have table following :
ClientNUM PIECES DID
NEWAGENC 10 5
NEWAGENC 25 5
WAY 30 4
CHCAH 20 2
AVC 21 2
i want the Result that sum the value for each client as below
CleintNUM Pieces DID
NEWAGENC 35 5
WAY 30 4
CHCAH 20 2
AVC 21 2
My query
SELECT
CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID, PIECES
So how can i do the sum for each CLIENTNUM in my query Means DISTINCT For each client Pieces like NEWAGENC has value 10 and in second row 25 so the pieces will be 10+ 25 = 35
Don't group by PIECES if you want to aggregate it
SELECT CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID
I want a query that can give result with sum of last 7 day look back.
I want output date and sum of last 7 day look back impressions for each date
e.g. I have a table tblFactImps with below data:
dateFact impressions id
2015-07-01 4022 30
2015-07-02 4021 33
2015-07-03 4011 34
2015-07-04 4029 35
2015-07-05 1023 39
2015-07-06 3023 92
2015-07-07 8027 66
2015-07-08 2024 89
I need output with 2 columns:
dateFact impressions_last_7
query I got:
select dateFact, sum(if(datediff(curdate(), dateFact)<=7, impressions,0)) impressions_last_7 from tblFactImps group by dateFact;
Thanks!
If your fact table is not too big, then a correlated subquery is a simple way to do what you want:
select i.dateFact,
(select sum(i2.impressions)
from tblFactImps i2
where i2.dateFact >= i.dateFact - interval 6 day
) as impressions_last_7
from tblFactImps i;
You can achieve this by LEFT OUTER JOINing the table with itself on a date range, and summing the impressions grouped by date, as follows:
SELECT
t1.dateFact,
SUM(t2.impressions) AS impressions_last_7
FROM
tblFactImps t1
LEFT OUTER JOIN
tblFactImps t2
ON
t2.dateFact BETWEEN
DATE_SUB(t1.dateFact, INTERVAL 6 DAY)
AND t1.dateFact
GROUP BY
t1.dateFact;
This should give you a sliding 7-day sum for each date in your table.
Assuming your dateFact column is indexed, this query should also be relatively fast.
Fixture Table
uid home_uid away_uid winner date season_division_uid
1 26 6 6 2013-07-30 18
2 8 21 8 2013-06-30 18
3 6 8 8 2013-06-29 18
4 21 26 21 2013-05-20 18
5 6 26 6 2013-04-19 18
This table contains hundreds of rows.
Currently I have a query to select all the teams in a division, i.e.
SELECT team_uid
FROM Season_Division_Team
WHERE season_division_uid='18'
which lists the rows of team uid's i.e. [6,26,8,21,26].
Now for each of the unique team ids, I would like to return the last 3 winner values, ordered by the date column, that they were involved in (they could be an away_uid or home_uid).
So the returned value example would be:
team_id winner date
6 6 2013-07-30
6 8 2013-06-29
6 26 2013-04-19
26 6 2013-07-30
26 21 2013-05-20
26 6 2013-04-19
Any ideas? Thank you
Im not sure how to get it direct, a query like
select * from Season_division_Team where
`date >= (select min(`date`) from
(select `date` from season_division_team order by date desc limit 3))
and (home_uid = 6 or away_uid = 6)
Thats not going to be a good query. But only way i can think of currently
Its hard to get the 3rd largest value from SQL Example
the sub query is trying to get the date where the last win occured, and then getting all dates after that where the team played.
EDIT:
SELECT * FROM Season_Division_Team WHERE winner = 6 ORDER BY `date` DESC LIMIT 3
that sounds more like your latter comment