I have the table like below
id timestamp speed
1 11:00:01 100
2 11:05:01 110
3 11:10:01 90
4 11:15 :01 80
I need to calculate moving average like below
id timestamp speed average
1 11:00:01 100 100
2 11:05:01 110 105
3 11:10:01 90 100
4 11:15:01 80 95
What I tried
SELECT
*,
(select avg(speed) from tbl t where tbl.timestamp<=t.timestamp) as avg
FROM
tbl
At first it looks quite easy but when the data on the table swell, it is too slow
Any faster approach?
Your query is one way to do a running average:
SELECT t.*,
(select avg(speed) from tbl tt where tt.timestamp <= t.timestamp) as avg
FROM tbl t;
The alternative is to use variables:
select t.*, (sum_speed / cnt) as running_avg_speed
from (select t.*, (#rn := #rn + 1) as cnt, (#s := #s + speed) as sum_speed
from tbl t cross join
(select #rn := 0, #s := 0) params
order by timestamp
) t;
An index on tbl(timestamp) should further improve performance.
Does MySQL support windowing functions?
select
id, timestamp, speed,
avg (speed) over (order by timestamp) as average
from tbl
If it doesn't this might work, although I doubt it's efficient:
select
min (t1.id) as id, t1.timestamp, min (t1.speed) as speed,
avg (t2.speed)
from
tbl t1
join tbl t2 on
t2.id <= t1.id
group by
t1.timestamp
order by
t1.timestamp
Or slotting neatly between GL's two answers (performancewise anyway)...
SELECT x.*, AVG(y.speed) avg
FROM my_table x
JOIN my_table y
ON y.id <= x.id
GROUP
BY x.id;
What about a simple concurrent solution?
SET #summ=0; SET #counter=0;SELECT *,(#counter := #counter +1) as cnt, (#summ := #summ+speed) as spd, (#summ/#counter) AS avg FROM tbl;
Related
I have the table like below
id timestamp speed
1 11:00:01 100
2 11:05:01 110
3 11:10:01 90
4 11:15 :01 80
I need to calculate moving average like below
id timestamp speed average
1 11:00:01 100 100
2 11:05:01 110 105
3 11:10:01 90 100
4 11:15:01 80 95
What I tried
SELECT
*,
(select avg(speed) from tbl t where tbl.timestamp<=t.timestamp) as avg
FROM
tbl
At first it looks quite easy but when the data on the table swell, it is too slow
Any faster approach?
Your query is one way to do a running average:
SELECT t.*,
(select avg(speed) from tbl tt where tt.timestamp <= t.timestamp) as avg
FROM tbl t;
The alternative is to use variables:
select t.*, (sum_speed / cnt) as running_avg_speed
from (select t.*, (#rn := #rn + 1) as cnt, (#s := #s + speed) as sum_speed
from tbl t cross join
(select #rn := 0, #s := 0) params
order by timestamp
) t;
An index on tbl(timestamp) should further improve performance.
Does MySQL support windowing functions?
select
id, timestamp, speed,
avg (speed) over (order by timestamp) as average
from tbl
If it doesn't this might work, although I doubt it's efficient:
select
min (t1.id) as id, t1.timestamp, min (t1.speed) as speed,
avg (t2.speed)
from
tbl t1
join tbl t2 on
t2.id <= t1.id
group by
t1.timestamp
order by
t1.timestamp
Or slotting neatly between GL's two answers (performancewise anyway)...
SELECT x.*, AVG(y.speed) avg
FROM my_table x
JOIN my_table y
ON y.id <= x.id
GROUP
BY x.id;
What about a simple concurrent solution?
SET #summ=0; SET #counter=0;SELECT *,(#counter := #counter +1) as cnt, (#summ := #summ+speed) as spd, (#summ/#counter) AS avg FROM tbl;
Is it possible to get specific row in query using like SUM?
Example:
id tickets
1 10 1-10 10=10
2 35 11-45 10+35=45
3 45 46-90 10+35+45=90
4 110 91-200 10+35+45+110=200
Total: 200 tickets(In SUM), I need to get row ID who have ticket with number like 23(Output would be ID: 2, because ID: 2 contains 11-45tickets in SUM)
You can do it by defining a local variable into your select query (in form clause), e.g.:
select id, #total := #total + tickets as seats
from test, (select #total := 0) t
Here is the SQL Fiddle.
You seem to want the row where "23" fits in. I think this does the trick:
select t.*
from (select t.*, (#total := #total + tickets) as running_total
from t cross join
(select #total := 0) params
order by id
) t
where 23 > running_total - tickets and 23 <= running_total;
SELECT
d.id
,d.tickets
,CONCAT(
TRIM(CAST(d.RunningTotal - d.tickets + 1 AS CHAR(10)))
,'-'
,TRIM(CAST(d.RunningTotal AS CHAR(10)))
) as TicketRange
,d.RunningTotal
FROM
(
SELECT
id
,tickets
,#total := #total + tickets as RunningTotal
FROM
test
CROSS JOIN (select #total := 0) var
ORDER BY
id
) d
This is similar to Darshan's answer but there are a few key differences:
You shouldn't use implicit join syntax, explicit join has more functionality in the long run and has been a standard for more than 20 years
ORDER BY will make a huge difference on your running total when calculated with a variable! if you change the order it will calculate differently so you need to consider how you want to do the running total, by date? by id? by??? and make sure you put it in the query.
finally I actually calculated the range as well.
And here is how you can do it without using variables:
SELECT
d.id
,d.tickets
,CONCAT(
TRIM(d.LowRange)
,'-'
,TRIM(
CAST(RunningTotal AS CHAR(10))
)
) as TicketRange
,d.RunningTotal
FROM
(
SELECT
t.id
,t.tickets
,CAST(COALESCE(SUM(t2.tickets),0) + 1 AS CHAR(10)) as LowRange
,t.tickets + COALESCE(SUM(t2.tickets),0) as RunningTotal
FROM
test t
LEFT JOIN test t2
ON t.id > t2. id
GROUP BY
t.id
,t.tickets
) d
I have the following structure in my user table:
id(INT) registered(DATETIME)
1 2016-04-01 23:23:01
2 2016-04-02 03:23:02
3 2016-04-02 05:23:03
4 2016-04-03 04:04:04
I want to get the total (accumulated) user count per day, for all days in DB
So result should be something like
day total
2016-04-01 1
2016-04-02 3
2016-04-03 4
I tried some sub querying, but somehow i have now idea how to achieve this with possibly 1 SQL statement. Of course if could group by per day count and add them programmatically, but i don't want to do that if possible.
You can use a GROUP BY that does all the counts, without the need of doing anything programmatically, please have a look at this query:
select
d.dt,
count(*) as total
from
(select distinct date(registered) dt from table1) d inner join
table1 r on d.dt>=date(r.registered)
group by
d.dt
order by
d.dt
the first subquery returns all distinct dates, then we can join all dates with all previous registrations, and do the counts, all in one query.
An alternative join condition that can give some improvements in performance is:
on d.dt + interval 1 day > r.registered
Not sure why not just use GROUP BY, without it this thing will be more complicated, anyway, try this;)
select
date_format(main.registered, '%Y-%m-%d') as `day`,
main.total
from (
select
table1.*,
#cnt := #cnt + 1 as total
from table1
cross join (select #cnt := 0) t
) main
inner join (
select
a.*,
if(#param = date_format(registered, '%Y-%m-%d'), #rowno := #rowno + 1 ,#rowno := 1) as rowno,
#param := date_format(registered, '%Y-%m-%d')
from (select * from table1 order by registered desc) a
cross join (select #param := null, #rowno := 0) tmp
having rowno = 1
) sub on main.id = sub.id
SQLFiddle DEMO
I have table structure as shown in below
Temp
Customer_id | sum
Now I have to create view with extra column customer_type and assign value 1 if customer lies in top 10% customer (with descending order of sum,also total number of customer may vary) and 2 if customer lies between 10%-20%, 3 if customer lies between 20%-60% and 4 if customer lies between 60%-100%. How can I do this?
I just able to extract top 10% and between 10% - 20% data but couldn't able to assign value as (source)
SELECT * FROM temp WHERE sum >= (SELECT sum FROM temp t1
WHERE(SELECT count(*) FROM temp t2 WHERE t2.sum >= t1.sum) <=
(SELECT 0.1 * count(*) FROM temp));
and (not efficient just enhance above code)
select * from temp t1
where (select count(*) from temp t2 where t2.sum>=t2.sum)
>= (select 0.1 * count(*) from temp) and (select count(*) from temp t2 where t2.sum>=t1.sum)
<= (select 0.2 * count(*) from temp);
Sample data are available at sqlfiddle.com
This should help you. You need to get row number for sum and total number of rows. I'm sure you can figure out the rest easily.
SELECT
*,
#curRow := #curRow + 1 AS row_number,
(#curRow2 := #curRow2 + 1) / c as pct_row
FROM
temp t
JOIN (SELECT #curRow := 0) r
JOIN (SELECT #curRow2 := 0) r2
join (select count(*) c from temp) s
order by
sum desc
This is based on this answer
I had solve this as like this. Thanks for #twn08 for his answer which guide me upto this.
select customer_id,sum,case
when pct_row<=0.10 then 1
when pct_row>0.10 and pct_row<=0.20 then 2
when pct_row>0.20 and pct_row<=0.60 then 3
when pct_row>0.60 then 4
end as customer_label from (
select customer_id,sum,(#curRow := #curRow+1)/c as pct_row
from temp t
jOIN (SELECT #curRow := 0) r
JOIN (SELECT #curRow2 := 0) r2
join (select count(*) c from temp) s
order by sum desc) p;
I don't know whether this is efficient method or not but work fine for small data set.
I have this table:
Date_on deposited withdrawal in_bank
2012-09-1 3000 2000 50000
2012-09-2/t 4000/t 0 54000
2013-09-3/t 3000 2000 55000
Now I want to execute a query to add the deposited amounts and subtract the withdrawals from the previous days entry in_bank. How can I do that? Can any one help me on that? This is my query:
select date_on, in_bank,((in_bank+deposited)-withdrawal)
from tablename where date_on > '2012-09-01' order by date_on
SELECT today.withdrawal, today.deposited, today.date_on,
(IFNULL(today.deposited, 0) - IFNULL(today.withdrawal, 0)) + IFNULL((SELECT in_bank FROM tablename AS yesterday WHERE yesterday.date_on = DATE_SUB(today.date_on, INTERVAL 1 DAY)), 0)
FROM tablename AS today
WHERE date_on BETWEEN '2012-09-01' AND '2012-09-02'
ORDER BY date_on ASC
The query will assume the balance was zero if no previous days date can be found.
Added a fiddle
http://sqlfiddle.com/#!2/d8261/14
Try this:
SELECT
t1.date_on,
t1.in_bank,
(
SELECT in_bank
FROM
(
SELECT *, (#rownum2 := #rownum2 +1) rank
FROM Table1,(SELECT #rownum2 :=0 ) t
ORDER BY date_on
) t2 WHERE t1.rank - t2.rank = 1
) - t1.withdrawal - deposited "total"
FROM
(
SELECT *, (#rownum := #rownum +1) rank
FROM Table1,(SELECT #rownum :=0 ) t
ORDER BY date_on
) t1;
SQL Fiddle Demo
SELECT
date_on,
in_bank,
(( ISNULL(in_bank, 0) + ISNULL(in_deposited,0)) - ISNULL(withdrawal,0))
FROM tablename
WHERE date_on > '2012-09-01'
ORDER BY date_on
I am not very familiar with mysql but this might help
select account.*,D.expectable from account left join
(
select deposited - withdrawal + in_bank as expectable,DATE_ADD(Date_on,INTERVAL 1 DAY) as nDate from account
)
D on account.Date_on = D.nDate