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* grouping subsets
@ 2010-07-22 08:09 Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 10:02 ` Re: grouping subsets A. Kretschmer <andreas.kretschmer@schollglas.com>
2010-07-22 13:18 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
0 siblings, 3 replies; 16+ messages in thread
From: Rainer Stengele @ 2010-07-22 08:09 UTC (permalink / raw)
To: pgsql-sql
Hi,
having a table similar to
| 1 | B | [2010-07-15 Do] |
| 1 | B | [2010-07-16 Fr] |
|---+---+-----------------|
| 2 | C | [2010-07-17 Sa] |
| 2 | C | [2010-07-18 So] |
|---+---+-----------------|
| 1 | B | [2010-07-19 Mo] |
| 1 | B | [2010-07-20 Di] |
| 1 | B | [2010-07-21 Mi] |
| 1 | B | [2010-07-22 Do] |
|---+---+-----------------|
| 3 | D | [2010-07-23 Fr] |
a simple group by gives me:
| 6 | B |
| 4 | C |
| 3 | D |
What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
Is there a way to get
| 2 | B |
| 4 | C |
| 4 | B |
| 3 | D |
by SQL only?
- Rainer
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-22 10:02 ` A. Kretschmer <andreas.kretschmer@schollglas.com>
2010-07-22 10:53 ` Re: grouping subsets Richard Huxton <dev@archonet.com>
2 siblings, 1 reply; 16+ messages in thread
From: A. Kretschmer @ 2010-07-22 10:02 UTC (permalink / raw)
To: pgsql-sql
In response to Rainer Stengele :
> Hi,
>
> having a table similar to
>
> | 1 | B | [2010-07-15 Do] |
> | 1 | B | [2010-07-16 Fr] |
> |---+---+-----------------|
> | 2 | C | [2010-07-17 Sa] |
> | 2 | C | [2010-07-18 So] |
> |---+---+-----------------|
> | 1 | B | [2010-07-19 Mo] |
> | 1 | B | [2010-07-20 Di] |
> | 1 | B | [2010-07-21 Mi] |
> | 1 | B | [2010-07-22 Do] |
> |---+---+-----------------|
> | 3 | D | [2010-07-23 Fr] |
>
> a simple group by gives me:
>
> | 6 | B |
> | 4 | C |
> | 3 | D |
>
>
> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
> Is there a way to get
>
> | 2 | B |
> | 4 | C |
> | 4 | B |
> | 3 | D |
>
> by SQL only?
I think, the problem is that there are 2 identical groups. I think, you
can write a pl/pgsql-proc, selecting all ordered by the date-field and
walking through the result to do the grouping, checking if the 2nd
column is different from the previous.
With plain SQL it's maybe possible too, but i don't know how ...
Regards, Andreas
--
Andreas Kretschmer
Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header)
GnuPG: 0x31720C99, 1006 CCB4 A326 1D42 6431 2EB0 389D 1DC2 3172 0C99
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 10:02 ` Re: grouping subsets A. Kretschmer <andreas.kretschmer@schollglas.com>
@ 2010-07-22 10:53 ` Richard Huxton <dev@archonet.com>
2010-07-22 11:31 ` Re: grouping subsets Tim Landscheidt <tim@tim-landscheidt.de>
0 siblings, 1 reply; 16+ messages in thread
From: Richard Huxton @ 2010-07-22 10:53 UTC (permalink / raw)
To: A. Kretschmer <andreas.kretschmer@schollglas.com>; +Cc: pgsql-sql
On 22/07/10 11:02, A. Kretschmer wrote:
> In response to Rainer Stengele :
>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>> Is there a way to get
>>
>> | 2 | B |
>> | 4 | C |
>> | 4 | B |
>> | 3 | D |
>>
>> by SQL only?
>
> I think, the problem is that there are 2 identical groups. I think, you
> can write a pl/pgsql-proc, selecting all ordered by the date-field and
> walking through the result to do the grouping, checking if the 2nd
> column is different from the previous.
>
> With plain SQL it's maybe possible too, but i don't know how ...
It should be do-able in 8.4 onwards, look into windowing functions. In
particular the lag() function:
SELECT
mycode,
mydate,
lag(mycode) OVER (ORDER BY mydate) AS prev_code
FROM
mytable
ORDER BY mydate;
It should be possible to use that as a subquery with an outer query that
compares mycode=prev_code to get a run length.
--
Richard Huxton
Archonet Ltd
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 10:02 ` Re: grouping subsets A. Kretschmer <andreas.kretschmer@schollglas.com>
2010-07-22 10:53 ` Re: grouping subsets Richard Huxton <dev@archonet.com>
@ 2010-07-22 11:31 ` Tim Landscheidt <tim@tim-landscheidt.de>
2010-07-30 15:58 ` Re: grouping subsets Joshua Tolley <eggyknap@gmail.com>
0 siblings, 1 reply; 16+ messages in thread
From: Tim Landscheidt @ 2010-07-22 11:31 UTC (permalink / raw)
To: pgsql-sql
Richard Huxton <dev@archonet.com> wrote:
>>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>>> Is there a way to get
>>> | 2 | B |
>>> | 4 | C |
>>> | 4 | B |
>>> | 3 | D |
>>> by SQL only?
>> I think, the problem is that there are 2 identical groups. I think, you
>> can write a pl/pgsql-proc, selecting all ordered by the date-field and
>> walking through the result to do the grouping, checking if the 2nd
>> column is different from the previous.
>> With plain SQL it's maybe possible too, but i don't know how ...
> It should be do-able in 8.4 onwards, look into windowing
> functions. In particular the lag() function:
> SELECT
> mycode,
> mydate,
> lag(mycode) OVER (ORDER BY mydate) AS prev_code
> FROM
> mytable
> ORDER BY mydate;
> It should be possible to use that as a subquery with an
> outer query that compares mycode=prev_code to get a run
> length.
Hmmm. Can the outer query be done without using "WITH
RECURSIVE"?
Tim
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 10:02 ` Re: grouping subsets A. Kretschmer <andreas.kretschmer@schollglas.com>
2010-07-22 10:53 ` Re: grouping subsets Richard Huxton <dev@archonet.com>
2010-07-22 11:31 ` Re: grouping subsets Tim Landscheidt <tim@tim-landscheidt.de>
@ 2010-07-30 15:58 ` Joshua Tolley <eggyknap@gmail.com>
0 siblings, 0 replies; 16+ messages in thread
From: Joshua Tolley @ 2010-07-30 15:58 UTC (permalink / raw)
To: pgsql-sql
On Thu, Jul 22, 2010 at 11:31:23AM +0000, Tim Landscheidt wrote:
> Richard Huxton <dev@archonet.com> wrote:
>
> >>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
> >>> Is there a way to get
>
> >>> | 2 | B |
> >>> | 4 | C |
> >>> | 4 | B |
> >>> | 3 | D |
>
> >>> by SQL only?
>
> >> I think, the problem is that there are 2 identical groups. I think, you
> >> can write a pl/pgsql-proc, selecting all ordered by the date-field and
> >> walking through the result to do the grouping, checking if the 2nd
> >> column is different from the previous.
>
> >> With plain SQL it's maybe possible too, but i don't know how ...
>
> > It should be do-able in 8.4 onwards, look into windowing
> > functions. In particular the lag() function:
>
> > SELECT
> > mycode,
> > mydate,
> > lag(mycode) OVER (ORDER BY mydate) AS prev_code
> > FROM
> > mytable
> > ORDER BY mydate;
>
> > It should be possible to use that as a subquery with an
> > outer query that compares mycode=prev_code to get a run
> > length.
>
> Hmmm. Can the outer query be done without using "WITH
> RECURSIVE"?
How about this:
select
a, b, c, d, sum
from (
select
a, b, c, d, new_partition,
sum(e) over (partition by partition_num)
from (
select
a, b, c, d, e,
case when
lag(a, 1, null) over (order by d) is null or
lag(a, 1, null) over (order by d) != a or
lag(b, 1, null) over (order by d) is null or
lag(b, 1, null) over (order by d) != b or
lag(c, 1, null) over (order by d) is null or
lag(c, 1, null) over (order by d) != c
then nextval('a')
else currval('a')
end as partition_num,
case when
lag(a, 1, null) over (order by d) is null or
lag(a, 1, null) over (order by d) != a or
lag(b, 1, null) over (order by d) is null or
lag(b, 1, null) over (order by d) != b or
lag(c, 1, null) over (order by d) is null or
lag(c, 1, null) over (order by d) != c
then 'T'::boolean
else 'f'::boolean
end as new_partition
from foo
) bar
) baz
where
new_partition;
Here's my test table:
5432 josh@postgres# select * from foo;
a | b | c | d | e
---+---+---+------------------------+---
1 | 9 | 1 | 2007-01-01 00:00:05-07 | 8
9 | 2 | 1 | 2007-01-01 00:00:10-07 | 4
9 | 2 | 1 | 2007-01-01 00:00:15-07 | 6
9 | 2 | 1 | 2007-01-01 00:00:20-07 | 2
6 | 5 | 7 | 2007-01-01 00:00:25-07 | 3
4 | 9 | 0 | 2007-01-01 00:00:30-07 | 0
4 | 9 | 0 | 2007-01-01 00:00:35-07 | 7
5 | 2 | 7 | 2007-01-01 00:01:25-07 | 7
5 | 2 | 7 | 2007-01-01 00:01:30-07 | 7
5 | 2 | 7 | 2007-01-01 00:01:35-07 | 9
5 | 2 | 7 | 2007-01-01 00:01:40-07 | 2
5 | 2 | 7 | 2007-01-01 00:01:45-07 | 5
5 | 2 | 7 | 2007-01-01 00:01:50-07 | 8
5 | 2 | 7 | 2007-01-01 00:01:55-07 | 5
5 | 2 | 7 | 2007-01-01 00:02:00-07 | 9
7 | 8 | 8 | 2007-01-01 00:02:05-07 | 7
7 | 8 | 8 | 2007-01-01 00:02:10-07 | 8
9 | 3 | 0 | 2007-01-01 00:02:15-07 | 0
9 | 3 | 0 | 2007-01-01 00:02:20-07 | 8
9 | 2 | 1 | 2007-01-01 00:02:25-07 | 3
9 | 2 | 1 | 2007-01-01 00:02:30-07 | 3
(21 rows)
...and these results...
a | b | c | d | sum
---+---+---+------------------------+-----
1 | 9 | 1 | 2007-01-01 00:00:05-07 | 8
9 | 2 | 1 | 2007-01-01 00:00:10-07 | 12
6 | 5 | 7 | 2007-01-01 00:00:25-07 | 3
4 | 9 | 0 | 2007-01-01 00:00:30-07 | 7
5 | 2 | 7 | 2007-01-01 00:01:25-07 | 52
7 | 8 | 8 | 2007-01-01 00:02:05-07 | 15
9 | 3 | 0 | 2007-01-01 00:02:15-07 | 8
9 | 2 | 1 | 2007-01-01 00:02:25-07 | 6
(8 rows)
--
Joshua Tolley / eggyknap
End Point Corporation
http://www.endpoint.com
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^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-22 13:18 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-22 13:25 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2 siblings, 1 reply; 16+ messages in thread
From: Oliveiros d'Azevedo Cristina @ 2010-07-22 13:18 UTC (permalink / raw)
To: pgsql-sql; Rainer Stengele <rainer.stengele@diplan.de>
Howdy, Rainer.
Please advice me,
The dates always follow that sequential pattern?
Or can be holes on the dates sequence?
Best,
Oliveiros
----- Original Message -----
From: "Rainer Stengele" <rainer.stengele@diplan.de>
To: <pgsql-sql@postgresql.org>
Sent: Thursday, July 22, 2010 9:09 AM
Subject: [SQL] grouping subsets
> Hi,
>
> having a table similar to
>
> | 1 | B | [2010-07-15 Do] |
> | 1 | B | [2010-07-16 Fr] |
> |---+---+-----------------|
> | 2 | C | [2010-07-17 Sa] |
> | 2 | C | [2010-07-18 So] |
> |---+---+-----------------|
> | 1 | B | [2010-07-19 Mo] |
> | 1 | B | [2010-07-20 Di] |
> | 1 | B | [2010-07-21 Mi] |
> | 1 | B | [2010-07-22 Do] |
> |---+---+-----------------|
> | 3 | D | [2010-07-23 Fr] |
>
> a simple group by gives me:
>
> | 6 | B |
> | 4 | C |
> | 3 | D |
>
>
> What I want to get is the values grouped by "subset", where a subset is a
> set of rows with identical column until the colum changes.
> Is there a way to get
>
> | 2 | B |
> | 4 | C |
> | 4 | B |
> | 3 | D |
>
> by SQL only?
>
> - Rainer
>
>
>
>
>
> --
> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
> To make changes to your subscription:
> http://www.postgresql.org/mailpref/pgsql-sql
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 13:18 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
@ 2010-07-22 13:25 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 14:18 ` Re: grouping subsets Tim Landscheidt <tim@tim-landscheidt.de>
0 siblings, 1 reply; 16+ messages in thread
From: Rainer Stengele @ 2010-07-22 13:25 UTC (permalink / raw)
To: Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>; +Cc: pgsql-sql
Hi Oliveiros,
yes, the date is always incremented - but anyway the date column is not really the point!
Actually the first tow columns are relevant.
I want them gouped together as indicated, adding up column 1 in the blocks with identical second column, but not adding up over all the rows.
Hope I could express it!
Mit freundlichen Grüßen / Best Regards
Rainer Stengele
__|___
| Dipl. Inf. (Univ.) Rainer Stengele
| Technical Control - System Administration
|
| email : Rainer.Stengele@diplan.de
| voice/fax: ++49-9131-7778-85/88
| WWW : http://www.diplan.de
|
| diplan GmbH
| Wetterkreuz 27
| 91058 Erlangen, Germany
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Am 22.07.2010 15:18, schrieb Oliveiros d'Azevedo Cristina:
> Howdy, Rainer.
>
> Please advice me,
>
> The dates always follow that sequential pattern?
>
> Or can be holes on the dates sequence?
>
> Best,
> Oliveiros
>
> ----- Original Message ----- From: "Rainer Stengele" <rainer.stengele@diplan.de>
> To: <pgsql-sql@postgresql.org>
> Sent: Thursday, July 22, 2010 9:09 AM
> Subject: [SQL] grouping subsets
>
>
>> Hi,
>>
>> having a table similar to
>>
>> | 1 | B | [2010-07-15 Do] |
>> | 1 | B | [2010-07-16 Fr] |
>> |---+---+-----------------|
>> | 2 | C | [2010-07-17 Sa] |
>> | 2 | C | [2010-07-18 So] |
>> |---+---+-----------------|
>> | 1 | B | [2010-07-19 Mo] |
>> | 1 | B | [2010-07-20 Di] |
>> | 1 | B | [2010-07-21 Mi] |
>> | 1 | B | [2010-07-22 Do] |
>> |---+---+-----------------|
>> | 3 | D | [2010-07-23 Fr] |
>>
>> a simple group by gives me:
>>
>> | 6 | B |
>> | 4 | C |
>> | 3 | D |
>>
>>
>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>> Is there a way to get
>>
>> | 2 | B |
>> | 4 | C |
>> | 4 | B |
>> | 3 | D |
>>
>> by SQL only?
>>
>> - Rainer
>>
>>
>>
>>
>>
>> --
>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>> To make changes to your subscription:
>> http://www.postgresql.org/mailpref/pgsql-sql
>
>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 13:18 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-22 13:25 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-22 14:18 ` Tim Landscheidt <tim@tim-landscheidt.de>
0 siblings, 0 replies; 16+ messages in thread
From: Tim Landscheidt @ 2010-07-22 14:18 UTC (permalink / raw)
To: pgsql-sql
Rainer Stengele <rainer.stengele@diplan.de> wrote:
> yes, the date is always incremented - but anyway the date
> column is not really the point! Actually the first tow
> columns are relevant. I want them gouped together as
> indicated, adding up column 1 in the blocks with identical
> second column, but not adding up over all the rows.
> [...]
If the date column wasn't relevant, how would you group the
first two columns?
Tim
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-27 10:37 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2 siblings, 1 reply; 16+ messages in thread
From: Oliveiros d'Azevedo Cristina @ 2010-07-27 10:37 UTC (permalink / raw)
To: pgsql-sql; Rainer Stengele <rainer.stengele@diplan.de>
Howdy, Rainer.
It's been a while, so I don't know if you are still interested in this
problem or if you, in the meantime, found yourself a solution,
but I've tried this on a local copy of the example you provided and it
seems to work.
The problem is that I suspect that if you have several thousands of records
on your table it will become slow...
Best,
Oliveiros
SELECT SUM(tudo.parcela),tudo.a
FROM
(
SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
FROM
(
SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
FROM
yourTable se
LEFT JOIN
(
SELECT a.*
FROM yourTable a
JOIN yourTable b
ON (b.b <> a.b)
AND ((age(a.c,b.c) = '1 day'::interval)
)
) pr
ON pr.b = se.b
AND se.c >= pr.c
GROUP BY se.a,se.b,se.c
) fo
LEFT JOIN
(
SELECT a.*
FROM yourTable a
JOIN yourTable b
ON (b.b <> a.b)
AND ((age(a.c,b.c) = '-1 day'::interval)
)
) th
ON fo.a = th.b
AND fo.b <= th.c
GROUP BY fo.parcela,fo.a,fo.b,fo.c
) tudo
GROUP BY tudo.a,tudo.c,tudo.d
To: <pgsql-sql@postgresql.org>
Sent: Thursday, July 22, 2010 9:09 AM
Subject: [SQL] grouping subsets
> Hi,
>
> having a table similar to
>
> | 1 | B | [2010-07-15 Do] |
> | 1 | B | [2010-07-16 Fr] |
> |---+---+-----------------|
> | 2 | C | [2010-07-17 Sa] |
> | 2 | C | [2010-07-18 So] |
> |---+---+-----------------|
> | 1 | B | [2010-07-19 Mo] |
> | 1 | B | [2010-07-20 Di] |
> | 1 | B | [2010-07-21 Mi] |
> | 1 | B | [2010-07-22 Do] |
> |---+---+-----------------|
> | 3 | D | [2010-07-23 Fr] |
>
> a simple group by gives me:
>
> | 6 | B |
> | 4 | C |
> | 3 | D |
>
>
> What I want to get is the values grouped by "subset", where a subset is a
> set of rows with identical column until the colum changes.
> Is there a way to get
>
> | 2 | B |
> | 4 | C |
> | 4 | B |
> | 3 | D |
>
> by SQL only?
>
> - Rainer
>
>
>
>
>
> --
> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
> To make changes to your subscription:
> http://www.postgresql.org/mailpref/pgsql-sql
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
@ 2010-07-29 09:41 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
0 siblings, 1 reply; 16+ messages in thread
From: Rainer Stengele @ 2010-07-29 09:41 UTC (permalink / raw)
To: Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>; +Cc: pgsql-sql
Howdy Cristina,
unfortunately things are more complicated. I have inserted an excerpt of the real data here:
================================================================================
TableID MasterID dtBegin dtEnd idR idL idB consumption
4057312 295530 2010-07-01 00:59:21.077 2010-07-01 01:32:59.670 1726 3212 1428 279
4061043 295574 2010-07-01 01:59:31.137 2010-07-01 02:32:09.373 1726 3212 1428 183
4083397 295838 2010-07-01 07:57:51.327 2010-07-01 08:28:28.117 318 1846 1012 30
4090858 295920 2010-07-01 09:52:33.777 2010-07-01 10:31:34.393 318 1846 1012 487
4094589 295961 2010-07-01 10:47:59.370 2010-07-01 11:32:20.903 318 1846 1012 472
4098330 296013 2010-07-01 11:58:53.890 2010-07-01 12:31:35.730 318 1846 1012 195
4102069 296058 2010-07-01 12:36:19.170 2010-07-01 13:32:13.950 318 1846 1012 338
4105809 296102 2010-07-01 13:58:53.170 2010-07-01 14:02:57.710 318 1846 1012 105
4109555 296150 2010-07-01 14:59:11.663 2010-07-01 15:32:33.810 318 1846 1012 187
4113305 296194 2010-07-01 15:59:01.797 2010-07-01 16:02:27.260 318 1846 1012 108
4117048 296238 2010-07-01 16:20:47.997 2010-07-01 17:32:49.367 318 1846 1012 179
4120791 296282 2010-07-01 17:58:27.657 2010-07-01 18:29:01.733 318 1846 1012 256
4128291 296370 2010-07-01 19:54:17.687 2010-07-01 20:32:53.850 318 1846 1012 239
4132044 296413 2010-07-01 20:31:37.653 2010-07-01 21:29:13.497 318 1846 1012 39
4135797 296458 2010-07-01 21:59:13.983 2010-07-01 22:32:46.503 318 1846 1012 157
4139572 296506 2010-07-01 22:58:49.530 2010-07-01 23:32:22.543 318 1846 1012 218
4142941 296554 2010-07-01 23:59:13.857 2010-07-02 00:32:30.390 318 1846 1012 248
4146289 296598 2010-07-02 00:58:55.763 2010-07-02 01:32:41.983 318 1846 1012 204
4149616 296642 2010-07-02 01:46:57.357 2010-07-02 02:32:56.983 318 1846 1012 42
4152952 296686 2010-07-02 02:55:19.653 2010-07-02 03:32:28.013 318 1846 1012 135
4156289 296730 2010-07-02 03:43:52.777 2010-07-02 04:32:55.250 318 1846 1012 743
4159624 296774 2010-07-02 04:43:15.310 2010-07-02 05:32:44.547 318 1846 1012 277
4162961 296817 2010-07-02 05:58:59.483 2010-07-02 06:32:37.340 318 1846 1012 121
4166303 296862 2010-07-02 06:58:50.733 2010-07-02 07:32:39.113 318 1846 1012 239
4172981 296950 2010-07-02 07:28:55.293 2010-07-02 09:33:01.200 318 1846 1012 512
4176322 296993 2010-07-02 09:59:04.607 2010-07-02 10:33:01.903 318 1846 1012 139
4179667 297038 2010-07-02 10:55:27.760 2010-07-02 11:32:56.560 318 1846 1012 722
4183012 297082 2010-07-02 11:59:33.650 2010-07-02 12:32:14.700 318 1846 1012 163
4186351 297126 2010-07-02 12:23:45.997 2010-07-02 13:32:59.500 318 1846 1012 284
4189689 297169 2010-07-02 13:44:21.253 2010-07-02 14:18:05.080 318 1846 1012 254
4196371 297258 2010-07-02 16:16:19.123 2010-07-02 16:32:53.437 1706 3541 1511 161
4199720 297301 2010-07-02 16:59:35.127 2010-07-02 17:32:57.950 1706 3541 1511 250
4203068 297346 2010-07-02 17:59:34.027 2010-07-02 18:32:54.337 1706 3541 1511 302
4206413 297389 2010-07-02 18:59:28.730 2010-07-02 19:32:37.950 1706 3541 1511 276
4209758 297434 2010-07-02 19:54:00.243 2010-07-02 20:32:57.433 1706 3541 1511 209
4213102 297473 2010-07-02 20:49:10.963 2010-07-02 21:30:44.540 1706 3541 1511 76
4216447 297511 2010-07-02 21:59:34.810 2010-07-02 22:33:00.603 1706 3541 1511 287
4219818 297569 2010-07-02 22:56:52.750 2010-07-02 23:59:31.607 1706 3541 1511 1877
4219819 297570 2010-07-02 23:59:21.577 2010-07-03 00:54:40.153 1706 3541 1511 1798
4219821 297572 2010-07-03 00:48:03.310 2010-07-03 01:59:37.920 1706 3541 1511 1125
4219823 297574 2010-07-03 01:51:01.057 2010-07-03 02:59:45.433 1706 3541 1511 1629
4219820 297571 2010-07-03 02:59:29.393 2010-07-03 03:59:54.920 1706 3541 1511 2462
4219822 297573 2010-07-03 03:59:18.663 2010-07-03 04:01:48.810 1706 3541 1511 70
4225738 297656 2010-07-03 06:13:34.980 2010-07-03 06:28:09.697 1726 3212 1428 46
4228694 297695 2010-07-03 06:59:15.560 2010-07-03 07:32:45.653 1726 3212 1428 251
4231649 297733 2010-07-03 07:59:11.937 2010-07-03 08:32:57.217 1726 3212 1428 284
4234604 297771 2010-07-03 08:57:00.357 2010-07-03 09:32:47.903 1726 3212 1428 227
4237559 297809 2010-07-03 09:59:19.813 2010-07-03 10:33:02.063 1726 3212 1428 285
4261156 298596 2010-07-04 22:59:09.863 2010-07-04 23:33:45.530 1726 3212 1428 1286
4264114 298646 2010-07-04 23:59:16.967 2010-07-05 00:33:08.107 1726 3212 1428 297
4267067 298690 2010-07-05 00:59:15.187 2010-07-05 01:32:48.300 1726 3212 1428 333
4270023 298734 2010-07-05 01:59:02.497 2010-07-05 02:32:48.780 1726 3212 1428 270
4272977 298778 2010-07-05 02:41:43.737 2010-07-05 03:32:56.043 1726 3212 1428 317
4275927 298822 2010-07-05 03:59:17.027 2010-07-05 04:33:14.947 1726 3212 1428 1623
================================================================================
Description:
1. Column: some ID
2. Column: reference to another table
3. and 4. column: timestamp from/to of the item
5. Column: ID R
6. Column: ID L
7. Column: ID B
8. Column: Sum of components
Requirement:
Sum over all components (from column 8) for each combination of ID R, ID L, ID B, but (!)
rows with same keys (R,L,B) should be summed up only until the keys change.
Do not sum up the components for identical keys, if there are other keys between them.
Example result:
idR idL idB SUM
1726 3212 1428 462
318 1846 1012 ...
1706 3541 1511 ...
1726 3212 1428 ...
Note that the first and last entry here has the same keys
Maybe you find a similar monster SQL solving such a requirement.
Thanks for considering!
Rainer
Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
> Howdy, Rainer.
>
> It's been a while, so I don't know if you are still interested in this problem or if you, in the meantime, found yourself a solution,
> but I've tried this on a local copy of the example you provided and it seems to work.
>
> The problem is that I suspect that if you have several thousands of records on your table it will become slow...
>
> Best,
> Oliveiros
>
> SELECT SUM(tudo.parcela),tudo.a
> FROM
> (
> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
> FROM
> (
> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
> FROM
> yourTable se
> LEFT JOIN
> (
> SELECT a.*
> FROM yourTable a
> JOIN yourTable b
> ON (b.b <> a.b)
> AND ((age(a.c,b.c) = '1 day'::interval)
>
> )
> ) pr
> ON pr.b = se.b
> AND se.c >= pr.c
> GROUP BY se.a,se.b,se.c
> ) fo
> LEFT JOIN
> (
> SELECT a.*
> FROM yourTable a
> JOIN yourTable b
> ON (b.b <> a.b)
> AND ((age(a.c,b.c) = '-1 day'::interval)
> )
> ) th
> ON fo.a = th.b
> AND fo.b <= th.c
> GROUP BY fo.parcela,fo.a,fo.b,fo.c
> ) tudo
> GROUP BY tudo.a,tudo.c,tudo.d
>
>
>
>
> To: <pgsql-sql@postgresql.org>
> Sent: Thursday, July 22, 2010 9:09 AM
> Subject: [SQL] grouping subsets
>
>
>> Hi,
>>
>> having a table similar to
>>
>> | 1 | B | [2010-07-15 Do] |
>> | 1 | B | [2010-07-16 Fr] |
>> |---+---+-----------------|
>> | 2 | C | [2010-07-17 Sa] |
>> | 2 | C | [2010-07-18 So] |
>> |---+---+-----------------|
>> | 1 | B | [2010-07-19 Mo] |
>> | 1 | B | [2010-07-20 Di] |
>> | 1 | B | [2010-07-21 Mi] |
>> | 1 | B | [2010-07-22 Do] |
>> |---+---+-----------------|
>> | 3 | D | [2010-07-23 Fr] |
>>
>> a simple group by gives me:
>>
>> | 6 | B |
>> | 4 | C |
>> | 3 | D |
>>
>>
>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>> Is there a way to get
>>
>> | 2 | B |
>> | 4 | C |
>> | 4 | B |
>> | 3 | D |
>>
>> by SQL only?
>>
>> - Rainer
>>
>>
>>
>>
>>
>> --
>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>> To make changes to your subscription:
>> http://www.postgresql.org/mailpref/pgsql-sql
>
>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-29 11:47 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
0 siblings, 1 reply; 16+ messages in thread
From: Oliveiros d'Azevedo Cristina @ 2010-07-29 11:47 UTC (permalink / raw)
To: Rainer Stengele <rainer.stengele@diplan.de>; +Cc: pgsql-sql
Yes. This is somewhat more complicated because it has more constraints.
I've noticed that a given combination doesn't appear with holes on a certain
day.
For ex, on a daily basis, we have every three key combinations together.
We dont have things like
2010-7-01 1726 3212 1428
2010-7-01 1726 3212 1428
... 318 1846 1012
2010-7-01 1726 3212 1428
Can I assume that, for a certain day , the records for the same three
combination are all together? There is just one set per day for a given
combination?
Or is it possible to have the same combination on one day with several sets?
Best,
Oliveiros
----- Original Message -----
From: "Rainer Stengele" <rainer.stengele@diplan.de>
Newsgroups: gmane.comp.db.postgresql.sql
To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
Cc: <>
Sent: Thursday, July 29, 2010 10:41 AM
Subject: Re: grouping subsets
> Howdy Cristina,
>
> unfortunately things are more complicated. I have inserted an excerpt of
> the real data here:
>
> ================================================================================
> TableID MasterID dtBegin dtEnd
> idR idL idB consumption
> 4057312 295530 2010-07-01 00:59:21.077 2010-07-01
> 01:32:59.670 1726 3212 1428 279
> 4061043 295574 2010-07-01 01:59:31.137 2010-07-01
> 02:32:09.373 1726 3212 1428 183
> 4083397 295838 2010-07-01 07:57:51.327 2010-07-01
> 08:28:28.117 318 1846 1012 30
> 4090858 295920 2010-07-01 09:52:33.777 2010-07-01
> 10:31:34.393 318 1846 1012 487
> 4094589 295961 2010-07-01 10:47:59.370 2010-07-01
> 11:32:20.903 318 1846 1012 472
> 4098330 296013 2010-07-01 11:58:53.890 2010-07-01
> 12:31:35.730 318 1846 1012 195
> 4102069 296058 2010-07-01 12:36:19.170 2010-07-01
> 13:32:13.950 318 1846 1012 338
> 4105809 296102 2010-07-01 13:58:53.170 2010-07-01
> 14:02:57.710 318 1846 1012 105
> 4109555 296150 2010-07-01 14:59:11.663 2010-07-01
> 15:32:33.810 318 1846 1012 187
> 4113305 296194 2010-07-01 15:59:01.797 2010-07-01
> 16:02:27.260 318 1846 1012 108
> 4117048 296238 2010-07-01 16:20:47.997 2010-07-01
> 17:32:49.367 318 1846 1012 179
> 4120791 296282 2010-07-01 17:58:27.657 2010-07-01
> 18:29:01.733 318 1846 1012 256
> 4128291 296370 2010-07-01 19:54:17.687 2010-07-01
> 20:32:53.850 318 1846 1012 239
> 4132044 296413 2010-07-01 20:31:37.653 2010-07-01
> 21:29:13.497 318 1846 1012 39
> 4135797 296458 2010-07-01 21:59:13.983 2010-07-01
> 22:32:46.503 318 1846 1012 157
> 4139572 296506 2010-07-01 22:58:49.530 2010-07-01
> 23:32:22.543 318 1846 1012 218
> 4142941 296554 2010-07-01 23:59:13.857 2010-07-02
> 00:32:30.390 318 1846 1012 248
> 4146289 296598 2010-07-02 00:58:55.763 2010-07-02
> 01:32:41.983 318 1846 1012 204
> 4149616 296642 2010-07-02 01:46:57.357 2010-07-02
> 02:32:56.983 318 1846 1012 42
> 4152952 296686 2010-07-02 02:55:19.653 2010-07-02
> 03:32:28.013 318 1846 1012 135
> 4156289 296730 2010-07-02 03:43:52.777 2010-07-02
> 04:32:55.250 318 1846 1012 743
> 4159624 296774 2010-07-02 04:43:15.310 2010-07-02
> 05:32:44.547 318 1846 1012 277
> 4162961 296817 2010-07-02 05:58:59.483 2010-07-02
> 06:32:37.340 318 1846 1012 121
> 4166303 296862 2010-07-02 06:58:50.733 2010-07-02
> 07:32:39.113 318 1846 1012 239
> 4172981 296950 2010-07-02 07:28:55.293 2010-07-02
> 09:33:01.200 318 1846 1012 512
> 4176322 296993 2010-07-02 09:59:04.607 2010-07-02
> 10:33:01.903 318 1846 1012 139
> 4179667 297038 2010-07-02 10:55:27.760 2010-07-02
> 11:32:56.560 318 1846 1012 722
> 4183012 297082 2010-07-02 11:59:33.650 2010-07-02
> 12:32:14.700 318 1846 1012 163
> 4186351 297126 2010-07-02 12:23:45.997 2010-07-02
> 13:32:59.500 318 1846 1012 284
> 4189689 297169 2010-07-02 13:44:21.253 2010-07-02
> 14:18:05.080 318 1846 1012 254
> 4196371 297258 2010-07-02 16:16:19.123 2010-07-02
> 16:32:53.437 1706 3541 1511 161
> 4199720 297301 2010-07-02 16:59:35.127 2010-07-02
> 17:32:57.950 1706 3541 1511 250
> 4203068 297346 2010-07-02 17:59:34.027 2010-07-02
> 18:32:54.337 1706 3541 1511 302
> 4206413 297389 2010-07-02 18:59:28.730 2010-07-02
> 19:32:37.950 1706 3541 1511 276
> 4209758 297434 2010-07-02 19:54:00.243 2010-07-02
> 20:32:57.433 1706 3541 1511 209
> 4213102 297473 2010-07-02 20:49:10.963 2010-07-02
> 21:30:44.540 1706 3541 1511 76
> 4216447 297511 2010-07-02 21:59:34.810 2010-07-02
> 22:33:00.603 1706 3541 1511 287
> 4219818 297569 2010-07-02 22:56:52.750 2010-07-02
> 23:59:31.607 1706 3541 1511 1877
> 4219819 297570 2010-07-02 23:59:21.577 2010-07-03
> 00:54:40.153 1706 3541 1511 1798
> 4219821 297572 2010-07-03 00:48:03.310 2010-07-03
> 01:59:37.920 1706 3541 1511 1125
> 4219823 297574 2010-07-03 01:51:01.057 2010-07-03
> 02:59:45.433 1706 3541 1511 1629
> 4219820 297571 2010-07-03 02:59:29.393 2010-07-03
> 03:59:54.920 1706 3541 1511 2462
> 4219822 297573 2010-07-03 03:59:18.663 2010-07-03
> 04:01:48.810 1706 3541 1511 70
> 4225738 297656 2010-07-03 06:13:34.980 2010-07-03
> 06:28:09.697 1726 3212 1428 46
> 4228694 297695 2010-07-03 06:59:15.560 2010-07-03
> 07:32:45.653 1726 3212 1428 251
> 4231649 297733 2010-07-03 07:59:11.937 2010-07-03
> 08:32:57.217 1726 3212 1428 284
> 4234604 297771 2010-07-03 08:57:00.357 2010-07-03
> 09:32:47.903 1726 3212 1428 227
> 4237559 297809 2010-07-03 09:59:19.813 2010-07-03
> 10:33:02.063 1726 3212 1428 285
> 4261156 298596 2010-07-04 22:59:09.863 2010-07-04
> 23:33:45.530 1726 3212 1428 1286
> 4264114 298646 2010-07-04 23:59:16.967 2010-07-05
> 00:33:08.107 1726 3212 1428 297
> 4267067 298690 2010-07-05 00:59:15.187 2010-07-05
> 01:32:48.300 1726 3212 1428 333
> 4270023 298734 2010-07-05 01:59:02.497 2010-07-05
> 02:32:48.780 1726 3212 1428 270
> 4272977 298778 2010-07-05 02:41:43.737 2010-07-05
> 03:32:56.043 1726 3212 1428 317
> 4275927 298822 2010-07-05 03:59:17.027 2010-07-05
> 04:33:14.947 1726 3212 1428 1623
> ================================================================================
>
> Description:
> 1. Column: some ID
> 2. Column: reference to another table
> 3. and 4. column: timestamp from/to of the item
> 5. Column: ID R
> 6. Column: ID L
> 7. Column: ID B
> 8. Column: Sum of components
>
> Requirement:
> Sum over all components (from column 8) for each combination of ID R, ID
> L, ID B, but (!)
> rows with same keys (R,L,B) should be summed up only until the keys
> change.
> Do not sum up the components for identical keys, if there are other keys
> between them.
>
> Example result:
>
> idR idL idB SUM
> 1726 3212 1428 462
> 318 1846 1012 ...
> 1706 3541 1511 ...
> 1726 3212 1428 ...
>
>
> Note that the first and last entry here has the same keys
>
> Maybe you find a similar monster SQL solving such a requirement.
> Thanks for considering!
>
> Rainer
>
>
>
> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>> Howdy, Rainer.
>>
>> It's been a while, so I don't know if you are still interested in this
>> problem or if you, in the meantime, found yourself a solution,
>> but I've tried this on a local copy of the example you provided and it
>> seems to work.
>>
>> The problem is that I suspect that if you have several thousands of
>> records on your table it will become slow...
>>
>> Best,
>> Oliveiros
>>
>> SELECT SUM(tudo.parcela),tudo.a
>> FROM
>> (
>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>> FROM
>> (
>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>> FROM
>> yourTable se
>> LEFT JOIN
>> (
>> SELECT a.*
>> FROM yourTable a
>> JOIN yourTable b
>> ON (b.b <> a.b)
>> AND ((age(a.c,b.c) = '1 day'::interval)
>>
>> )
>> ) pr
>> ON pr.b = se.b
>> AND se.c >= pr.c
>> GROUP BY se.a,se.b,se.c
>> ) fo
>> LEFT JOIN
>> (
>> SELECT a.*
>> FROM yourTable a
>> JOIN yourTable b
>> ON (b.b <> a.b)
>> AND ((age(a.c,b.c) = '-1 day'::interval)
>> )
>> ) th
>> ON fo.a = th.b
>> AND fo.b <= th.c
>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>> ) tudo
>> GROUP BY tudo.a,tudo.c,tudo.d
>>
>>
>>
>>
>> To: <pgsql-sql@postgresql.org>
>> Sent: Thursday, July 22, 2010 9:09 AM
>> Subject: [SQL] grouping subsets
>>
>>
>>> Hi,
>>>
>>> having a table similar to
>>>
>>> | 1 | B | [2010-07-15 Do] |
>>> | 1 | B | [2010-07-16 Fr] |
>>> |---+---+-----------------|
>>> | 2 | C | [2010-07-17 Sa] |
>>> | 2 | C | [2010-07-18 So] |
>>> |---+---+-----------------|
>>> | 1 | B | [2010-07-19 Mo] |
>>> | 1 | B | [2010-07-20 Di] |
>>> | 1 | B | [2010-07-21 Mi] |
>>> | 1 | B | [2010-07-22 Do] |
>>> |---+---+-----------------|
>>> | 3 | D | [2010-07-23 Fr] |
>>>
>>> a simple group by gives me:
>>>
>>> | 6 | B |
>>> | 4 | C |
>>> | 3 | D |
>>>
>>>
>>> What I want to get is the values grouped by "subset", where a subset is
>>> a set of rows with identical column until the colum changes.
>>> Is there a way to get
>>>
>>> | 2 | B |
>>> | 4 | C |
>>> | 4 | B |
>>> | 3 | D |
>>>
>>> by SQL only?
>>>
>>> - Rainer
>>>
>>>
>>>
>>>
>>>
>>> --
>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>> To make changes to your subscription:
>>> http://www.postgresql.org/mailpref/pgsql-sql
>>
>>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
@ 2010-07-29 12:10 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 13:31 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
0 siblings, 1 reply; 16+ messages in thread
From: Rainer Stengele @ 2010-07-29 12:10 UTC (permalink / raw)
To: Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>; +Cc: pgsql-sql
No. This is by accident.
We have to assume that the combinations do change anytime, and many times per day.
So
"Or is it possible to have the same combination on one day with several sets?"
YES!
Rainer
Am 29.07.2010 13:47, schrieb Oliveiros d'Azevedo Cristina:
> Yes. This is somewhat more complicated because it has more constraints.
> I've noticed that a given combination doesn't appear with holes on a certain day.
>
> For ex, on a daily basis, we have every three key combinations together.
>
> We dont have things like
> 2010-7-01 1726 3212 1428
> 2010-7-01 1726 3212 1428
> ... 318 1846 1012
> 2010-7-01 1726 3212 1428
>
> Can I assume that, for a certain day , the records for the same three combination are all together? There is just one set per day for a given combination?
>
> Or is it possible to have the same combination on one day with several sets?
>
> Best,
> Oliveiros
>
>
> ----- Original Message ----- From: "Rainer Stengele" <rainer.stengele@diplan.de>
> Newsgroups: gmane.comp.db.postgresql.sql
> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
> Cc: <>
> Sent: Thursday, July 29, 2010 10:41 AM
> Subject: Re: grouping subsets
>
>
>> Howdy Cristina,
>>
>> unfortunately things are more complicated. I have inserted an excerpt of the real data here:
>>
>> ================================================================================
>> TableID MasterID dtBegin dtEnd idR idL idB consumption
>> 4057312 295530 2010-07-01 00:59:21.077 2010-07-01 01:32:59.670 1726 3212 1428 279
>> 4061043 295574 2010-07-01 01:59:31.137 2010-07-01 02:32:09.373 1726 3212 1428 183
>> 4083397 295838 2010-07-01 07:57:51.327 2010-07-01 08:28:28.117 318 1846 1012 30
>> 4090858 295920 2010-07-01 09:52:33.777 2010-07-01 10:31:34.393 318 1846 1012 487
>> 4094589 295961 2010-07-01 10:47:59.370 2010-07-01 11:32:20.903 318 1846 1012 472
>> 4098330 296013 2010-07-01 11:58:53.890 2010-07-01 12:31:35.730 318 1846 1012 195
>> 4102069 296058 2010-07-01 12:36:19.170 2010-07-01 13:32:13.950 318 1846 1012 338
>> 4105809 296102 2010-07-01 13:58:53.170 2010-07-01 14:02:57.710 318 1846 1012 105
>> 4109555 296150 2010-07-01 14:59:11.663 2010-07-01 15:32:33.810 318 1846 1012 187
>> 4113305 296194 2010-07-01 15:59:01.797 2010-07-01 16:02:27.260 318 1846 1012 108
>> 4117048 296238 2010-07-01 16:20:47.997 2010-07-01 17:32:49.367 318 1846 1012 179
>> 4120791 296282 2010-07-01 17:58:27.657 2010-07-01 18:29:01.733 318 1846 1012 256
>> 4128291 296370 2010-07-01 19:54:17.687 2010-07-01 20:32:53.850 318 1846 1012 239
>> 4132044 296413 2010-07-01 20:31:37.653 2010-07-01 21:29:13.497 318 1846 1012 39
>> 4135797 296458 2010-07-01 21:59:13.983 2010-07-01 22:32:46.503 318 1846 1012 157
>> 4139572 296506 2010-07-01 22:58:49.530 2010-07-01 23:32:22.543 318 1846 1012 218
>> 4142941 296554 2010-07-01 23:59:13.857 2010-07-02 00:32:30.390 318 1846 1012 248
>> 4146289 296598 2010-07-02 00:58:55.763 2010-07-02 01:32:41.983 318 1846 1012 204
>> 4149616 296642 2010-07-02 01:46:57.357 2010-07-02 02:32:56.983 318 1846 1012 42
>> 4152952 296686 2010-07-02 02:55:19.653 2010-07-02 03:32:28.013 318 1846 1012 135
>> 4156289 296730 2010-07-02 03:43:52.777 2010-07-02 04:32:55.250 318 1846 1012 743
>> 4159624 296774 2010-07-02 04:43:15.310 2010-07-02 05:32:44.547 318 1846 1012 277
>> 4162961 296817 2010-07-02 05:58:59.483 2010-07-02 06:32:37.340 318 1846 1012 121
>> 4166303 296862 2010-07-02 06:58:50.733 2010-07-02 07:32:39.113 318 1846 1012 239
>> 4172981 296950 2010-07-02 07:28:55.293 2010-07-02 09:33:01.200 318 1846 1012 512
>> 4176322 296993 2010-07-02 09:59:04.607 2010-07-02 10:33:01.903 318 1846 1012 139
>> 4179667 297038 2010-07-02 10:55:27.760 2010-07-02 11:32:56.560 318 1846 1012 722
>> 4183012 297082 2010-07-02 11:59:33.650 2010-07-02 12:32:14.700 318 1846 1012 163
>> 4186351 297126 2010-07-02 12:23:45.997 2010-07-02 13:32:59.500 318 1846 1012 284
>> 4189689 297169 2010-07-02 13:44:21.253 2010-07-02 14:18:05.080 318 1846 1012 254
>> 4196371 297258 2010-07-02 16:16:19.123 2010-07-02 16:32:53.437 1706 3541 1511 161
>> 4199720 297301 2010-07-02 16:59:35.127 2010-07-02 17:32:57.950 1706 3541 1511 250
>> 4203068 297346 2010-07-02 17:59:34.027 2010-07-02 18:32:54.337 1706 3541 1511 302
>> 4206413 297389 2010-07-02 18:59:28.730 2010-07-02 19:32:37.950 1706 3541 1511 276
>> 4209758 297434 2010-07-02 19:54:00.243 2010-07-02 20:32:57.433 1706 3541 1511 209
>> 4213102 297473 2010-07-02 20:49:10.963 2010-07-02 21:30:44.540 1706 3541 1511 76
>> 4216447 297511 2010-07-02 21:59:34.810 2010-07-02 22:33:00.603 1706 3541 1511 287
>> 4219818 297569 2010-07-02 22:56:52.750 2010-07-02 23:59:31.607 1706 3541 1511 1877
>> 4219819 297570 2010-07-02 23:59:21.577 2010-07-03 00:54:40.153 1706 3541 1511 1798
>> 4219821 297572 2010-07-03 00:48:03.310 2010-07-03 01:59:37.920 1706 3541 1511 1125
>> 4219823 297574 2010-07-03 01:51:01.057 2010-07-03 02:59:45.433 1706 3541 1511 1629
>> 4219820 297571 2010-07-03 02:59:29.393 2010-07-03 03:59:54.920 1706 3541 1511 2462
>> 4219822 297573 2010-07-03 03:59:18.663 2010-07-03 04:01:48.810 1706 3541 1511 70
>> 4225738 297656 2010-07-03 06:13:34.980 2010-07-03 06:28:09.697 1726 3212 1428 46
>> 4228694 297695 2010-07-03 06:59:15.560 2010-07-03 07:32:45.653 1726 3212 1428 251
>> 4231649 297733 2010-07-03 07:59:11.937 2010-07-03 08:32:57.217 1726 3212 1428 284
>> 4234604 297771 2010-07-03 08:57:00.357 2010-07-03 09:32:47.903 1726 3212 1428 227
>> 4237559 297809 2010-07-03 09:59:19.813 2010-07-03 10:33:02.063 1726 3212 1428 285
>> 4261156 298596 2010-07-04 22:59:09.863 2010-07-04 23:33:45.530 1726 3212 1428 1286
>> 4264114 298646 2010-07-04 23:59:16.967 2010-07-05 00:33:08.107 1726 3212 1428 297
>> 4267067 298690 2010-07-05 00:59:15.187 2010-07-05 01:32:48.300 1726 3212 1428 333
>> 4270023 298734 2010-07-05 01:59:02.497 2010-07-05 02:32:48.780 1726 3212 1428 270
>> 4272977 298778 2010-07-05 02:41:43.737 2010-07-05 03:32:56.043 1726 3212 1428 317
>> 4275927 298822 2010-07-05 03:59:17.027 2010-07-05 04:33:14.947 1726 3212 1428 1623
>> ================================================================================
>>
>> Description:
>> 1. Column: some ID
>> 2. Column: reference to another table
>> 3. and 4. column: timestamp from/to of the item
>> 5. Column: ID R
>> 6. Column: ID L
>> 7. Column: ID B
>> 8. Column: Sum of components
>>
>> Requirement:
>> Sum over all components (from column 8) for each combination of ID R, ID L, ID B, but (!)
>> rows with same keys (R,L,B) should be summed up only until the keys change.
>> Do not sum up the components for identical keys, if there are other keys between them.
>>
>> Example result:
>>
>> idR idL idB SUM
>> 1726 3212 1428 462
>> 318 1846 1012 ...
>> 1706 3541 1511 ...
>> 1726 3212 1428 ...
>>
>>
>> Note that the first and last entry here has the same keys
>>
>> Maybe you find a similar monster SQL solving such a requirement.
>> Thanks for considering!
>>
>> Rainer
>>
>>
>>
>> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>>> Howdy, Rainer.
>>>
>>> It's been a while, so I don't know if you are still interested in this problem or if you, in the meantime, found yourself a solution,
>>> but I've tried this on a local copy of the example you provided and it seems to work.
>>>
>>> The problem is that I suspect that if you have several thousands of records on your table it will become slow...
>>>
>>> Best,
>>> Oliveiros
>>>
>>> SELECT SUM(tudo.parcela),tudo.a
>>> FROM
>>> (
>>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>>> FROM
>>> (
>>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>>> FROM
>>> yourTable se
>>> LEFT JOIN
>>> (
>>> SELECT a.*
>>> FROM yourTable a
>>> JOIN yourTable b
>>> ON (b.b <> a.b)
>>> AND ((age(a.c,b.c) = '1 day'::interval)
>>>
>>> )
>>> ) pr
>>> ON pr.b = se.b
>>> AND se.c >= pr.c
>>> GROUP BY se.a,se.b,se.c
>>> ) fo
>>> LEFT JOIN
>>> (
>>> SELECT a.*
>>> FROM yourTable a
>>> JOIN yourTable b
>>> ON (b.b <> a.b)
>>> AND ((age(a.c,b.c) = '-1 day'::interval)
>>> )
>>> ) th
>>> ON fo.a = th.b
>>> AND fo.b <= th.c
>>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>>> ) tudo
>>> GROUP BY tudo.a,tudo.c,tudo.d
>>>
>>>
>>>
>>>
>>> To: <pgsql-sql@postgresql.org>
>>> Sent: Thursday, July 22, 2010 9:09 AM
>>> Subject: [SQL] grouping subsets
>>>
>>>
>>>> Hi,
>>>>
>>>> having a table similar to
>>>>
>>>> | 1 | B | [2010-07-15 Do] |
>>>> | 1 | B | [2010-07-16 Fr] |
>>>> |---+---+-----------------|
>>>> | 2 | C | [2010-07-17 Sa] |
>>>> | 2 | C | [2010-07-18 So] |
>>>> |---+---+-----------------|
>>>> | 1 | B | [2010-07-19 Mo] |
>>>> | 1 | B | [2010-07-20 Di] |
>>>> | 1 | B | [2010-07-21 Mi] |
>>>> | 1 | B | [2010-07-22 Do] |
>>>> |---+---+-----------------|
>>>> | 3 | D | [2010-07-23 Fr] |
>>>>
>>>> a simple group by gives me:
>>>>
>>>> | 6 | B |
>>>> | 4 | C |
>>>> | 3 | D |
>>>>
>>>>
>>>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>>>> Is there a way to get
>>>>
>>>> | 2 | B |
>>>> | 4 | C |
>>>> | 4 | B |
>>>> | 3 | D |
>>>>
>>>> by SQL only?
>>>>
>>>> - Rainer
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>>> To make changes to your subscription:
>>>> http://www.postgresql.org/mailpref/pgsql-sql
>>>
>>>
>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-29 13:31 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-30 09:35 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
0 siblings, 1 reply; 16+ messages in thread
From: Oliveiros d'Azevedo Cristina @ 2010-07-29 13:31 UTC (permalink / raw)
To: Rainer Stengele <rainer.stengele@diplan.de>; +Cc: pgsql-sql
Fine.
Please advice me,
How long can
your table be? Thousands? Millions of records?
Do you really need it in pure SQL
?
It seems to me that it might be possible, I'm just affraid that the query
would become too complex and thus slow...
Best,
Oliveiros
----- Original Message -----
From: "Rainer Stengele" <rainer.stengele@diplan.de>
To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
Cc: <pgsql-sql@postgresql.org>
Sent: Thursday, July 29, 2010 1:10 PM
Subject: Re: grouping subsets
> No. This is by accident.
> We have to assume that the combinations do change anytime, and many times
> per day.
>
> So
>
> "Or is it possible to have the same combination on one day with several
> sets?"
>
> YES!
>
> Rainer
>
>
>
> Am 29.07.2010 13:47, schrieb Oliveiros d'Azevedo Cristina:
>> Yes. This is somewhat more complicated because it has more constraints.
>> I've noticed that a given combination doesn't appear with holes on a
>> certain day.
>>
>> For ex, on a daily basis, we have every three key combinations together.
>>
>> We dont have things like
>> 2010-7-01 1726 3212 1428
>> 2010-7-01 1726 3212 1428
>> ... 318 1846 1012
>> 2010-7-01 1726 3212 1428
>>
>> Can I assume that, for a certain day , the records for the same three
>> combination are all together? There is just one set per day for a given
>> combination?
>>
>> Or is it possible to have the same combination on one day with several
>> sets?
>>
>> Best,
>> Oliveiros
>>
>>
>> ----- Original Message ----- From: "Rainer Stengele"
>> <rainer.stengele@diplan.de>
>> Newsgroups: gmane.comp.db.postgresql.sql
>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>> Cc: <>
>> Sent: Thursday, July 29, 2010 10:41 AM
>> Subject: Re: grouping subsets
>>
>>
>>> Howdy Cristina,
>>>
>>> unfortunately things are more complicated. I have inserted an excerpt of
>>> the real data here:
>>>
>>> ================================================================================
>>> TableID MasterID dtBegin dtEnd idR
>>> idL idB consumption
>>> 4057312 295530 2010-07-01 00:59:21.077 2010-07-01
>>> 01:32:59.670 1726 3212 1428 279
>>> 4061043 295574 2010-07-01 01:59:31.137 2010-07-01
>>> 02:32:09.373 1726 3212 1428 183
>>> 4083397 295838 2010-07-01 07:57:51.327 2010-07-01
>>> 08:28:28.117 318 1846 1012 30
>>> 4090858 295920 2010-07-01 09:52:33.777 2010-07-01
>>> 10:31:34.393 318 1846 1012 487
>>> 4094589 295961 2010-07-01 10:47:59.370 2010-07-01
>>> 11:32:20.903 318 1846 1012 472
>>> 4098330 296013 2010-07-01 11:58:53.890 2010-07-01
>>> 12:31:35.730 318 1846 1012 195
>>> 4102069 296058 2010-07-01 12:36:19.170 2010-07-01
>>> 13:32:13.950 318 1846 1012 338
>>> 4105809 296102 2010-07-01 13:58:53.170 2010-07-01
>>> 14:02:57.710 318 1846 1012 105
>>> 4109555 296150 2010-07-01 14:59:11.663 2010-07-01
>>> 15:32:33.810 318 1846 1012 187
>>> 4113305 296194 2010-07-01 15:59:01.797 2010-07-01
>>> 16:02:27.260 318 1846 1012 108
>>> 4117048 296238 2010-07-01 16:20:47.997 2010-07-01
>>> 17:32:49.367 318 1846 1012 179
>>> 4120791 296282 2010-07-01 17:58:27.657 2010-07-01
>>> 18:29:01.733 318 1846 1012 256
>>> 4128291 296370 2010-07-01 19:54:17.687 2010-07-01
>>> 20:32:53.850 318 1846 1012 239
>>> 4132044 296413 2010-07-01 20:31:37.653 2010-07-01
>>> 21:29:13.497 318 1846 1012 39
>>> 4135797 296458 2010-07-01 21:59:13.983 2010-07-01
>>> 22:32:46.503 318 1846 1012 157
>>> 4139572 296506 2010-07-01 22:58:49.530 2010-07-01
>>> 23:32:22.543 318 1846 1012 218
>>> 4142941 296554 2010-07-01 23:59:13.857 2010-07-02
>>> 00:32:30.390 318 1846 1012 248
>>> 4146289 296598 2010-07-02 00:58:55.763 2010-07-02
>>> 01:32:41.983 318 1846 1012 204
>>> 4149616 296642 2010-07-02 01:46:57.357 2010-07-02
>>> 02:32:56.983 318 1846 1012 42
>>> 4152952 296686 2010-07-02 02:55:19.653 2010-07-02
>>> 03:32:28.013 318 1846 1012 135
>>> 4156289 296730 2010-07-02 03:43:52.777 2010-07-02
>>> 04:32:55.250 318 1846 1012 743
>>> 4159624 296774 2010-07-02 04:43:15.310 2010-07-02
>>> 05:32:44.547 318 1846 1012 277
>>> 4162961 296817 2010-07-02 05:58:59.483 2010-07-02
>>> 06:32:37.340 318 1846 1012 121
>>> 4166303 296862 2010-07-02 06:58:50.733 2010-07-02
>>> 07:32:39.113 318 1846 1012 239
>>> 4172981 296950 2010-07-02 07:28:55.293 2010-07-02
>>> 09:33:01.200 318 1846 1012 512
>>> 4176322 296993 2010-07-02 09:59:04.607 2010-07-02
>>> 10:33:01.903 318 1846 1012 139
>>> 4179667 297038 2010-07-02 10:55:27.760 2010-07-02
>>> 11:32:56.560 318 1846 1012 722
>>> 4183012 297082 2010-07-02 11:59:33.650 2010-07-02
>>> 12:32:14.700 318 1846 1012 163
>>> 4186351 297126 2010-07-02 12:23:45.997 2010-07-02
>>> 13:32:59.500 318 1846 1012 284
>>> 4189689 297169 2010-07-02 13:44:21.253 2010-07-02
>>> 14:18:05.080 318 1846 1012 254
>>> 4196371 297258 2010-07-02 16:16:19.123 2010-07-02
>>> 16:32:53.437 1706 3541 1511 161
>>> 4199720 297301 2010-07-02 16:59:35.127 2010-07-02
>>> 17:32:57.950 1706 3541 1511 250
>>> 4203068 297346 2010-07-02 17:59:34.027 2010-07-02
>>> 18:32:54.337 1706 3541 1511 302
>>> 4206413 297389 2010-07-02 18:59:28.730 2010-07-02
>>> 19:32:37.950 1706 3541 1511 276
>>> 4209758 297434 2010-07-02 19:54:00.243 2010-07-02
>>> 20:32:57.433 1706 3541 1511 209
>>> 4213102 297473 2010-07-02 20:49:10.963 2010-07-02
>>> 21:30:44.540 1706 3541 1511 76
>>> 4216447 297511 2010-07-02 21:59:34.810 2010-07-02
>>> 22:33:00.603 1706 3541 1511 287
>>> 4219818 297569 2010-07-02 22:56:52.750 2010-07-02
>>> 23:59:31.607 1706 3541 1511 1877
>>> 4219819 297570 2010-07-02 23:59:21.577 2010-07-03
>>> 00:54:40.153 1706 3541 1511 1798
>>> 4219821 297572 2010-07-03 00:48:03.310 2010-07-03
>>> 01:59:37.920 1706 3541 1511 1125
>>> 4219823 297574 2010-07-03 01:51:01.057 2010-07-03
>>> 02:59:45.433 1706 3541 1511 1629
>>> 4219820 297571 2010-07-03 02:59:29.393 2010-07-03
>>> 03:59:54.920 1706 3541 1511 2462
>>> 4219822 297573 2010-07-03 03:59:18.663 2010-07-03
>>> 04:01:48.810 1706 3541 1511 70
>>> 4225738 297656 2010-07-03 06:13:34.980 2010-07-03
>>> 06:28:09.697 1726 3212 1428 46
>>> 4228694 297695 2010-07-03 06:59:15.560 2010-07-03
>>> 07:32:45.653 1726 3212 1428 251
>>> 4231649 297733 2010-07-03 07:59:11.937 2010-07-03
>>> 08:32:57.217 1726 3212 1428 284
>>> 4234604 297771 2010-07-03 08:57:00.357 2010-07-03
>>> 09:32:47.903 1726 3212 1428 227
>>> 4237559 297809 2010-07-03 09:59:19.813 2010-07-03
>>> 10:33:02.063 1726 3212 1428 285
>>> 4261156 298596 2010-07-04 22:59:09.863 2010-07-04
>>> 23:33:45.530 1726 3212 1428 1286
>>> 4264114 298646 2010-07-04 23:59:16.967 2010-07-05
>>> 00:33:08.107 1726 3212 1428 297
>>> 4267067 298690 2010-07-05 00:59:15.187 2010-07-05
>>> 01:32:48.300 1726 3212 1428 333
>>> 4270023 298734 2010-07-05 01:59:02.497 2010-07-05
>>> 02:32:48.780 1726 3212 1428 270
>>> 4272977 298778 2010-07-05 02:41:43.737 2010-07-05
>>> 03:32:56.043 1726 3212 1428 317
>>> 4275927 298822 2010-07-05 03:59:17.027 2010-07-05
>>> 04:33:14.947 1726 3212 1428 1623
>>> ================================================================================
>>>
>>> Description:
>>> 1. Column: some ID
>>> 2. Column: reference to another table
>>> 3. and 4. column: timestamp from/to of the item
>>> 5. Column: ID R
>>> 6. Column: ID L
>>> 7. Column: ID B
>>> 8. Column: Sum of components
>>>
>>> Requirement:
>>> Sum over all components (from column 8) for each combination of ID R, ID
>>> L, ID B, but (!)
>>> rows with same keys (R,L,B) should be summed up only until the keys
>>> change.
>>> Do not sum up the components for identical keys, if there are other keys
>>> between them.
>>>
>>> Example result:
>>>
>>> idR idL idB SUM
>>> 1726 3212 1428 462
>>> 318 1846 1012 ...
>>> 1706 3541 1511 ...
>>> 1726 3212 1428 ...
>>>
>>>
>>> Note that the first and last entry here has the same keys
>>>
>>> Maybe you find a similar monster SQL solving such a requirement.
>>> Thanks for considering!
>>>
>>> Rainer
>>>
>>>
>>>
>>> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>>>> Howdy, Rainer.
>>>>
>>>> It's been a while, so I don't know if you are still interested in this
>>>> problem or if you, in the meantime, found yourself a solution,
>>>> but I've tried this on a local copy of the example you provided and
>>>> it seems to work.
>>>>
>>>> The problem is that I suspect that if you have several thousands of
>>>> records on your table it will become slow...
>>>>
>>>> Best,
>>>> Oliveiros
>>>>
>>>> SELECT SUM(tudo.parcela),tudo.a
>>>> FROM
>>>> (
>>>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>>>> FROM
>>>> (
>>>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>>>> FROM
>>>> yourTable se
>>>> LEFT JOIN
>>>> (
>>>> SELECT a.*
>>>> FROM yourTable a
>>>> JOIN yourTable b
>>>> ON (b.b <> a.b)
>>>> AND ((age(a.c,b.c) = '1 day'::interval)
>>>>
>>>> )
>>>> ) pr
>>>> ON pr.b = se.b
>>>> AND se.c >= pr.c
>>>> GROUP BY se.a,se.b,se.c
>>>> ) fo
>>>> LEFT JOIN
>>>> (
>>>> SELECT a.*
>>>> FROM yourTable a
>>>> JOIN yourTable b
>>>> ON (b.b <> a.b)
>>>> AND ((age(a.c,b.c) = '-1 day'::interval)
>>>> )
>>>> ) th
>>>> ON fo.a = th.b
>>>> AND fo.b <= th.c
>>>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>>>> ) tudo
>>>> GROUP BY tudo.a,tudo.c,tudo.d
>>>>
>>>>
>>>>
>>>>
>>>> To: <pgsql-sql@postgresql.org>
>>>> Sent: Thursday, July 22, 2010 9:09 AM
>>>> Subject: [SQL] grouping subsets
>>>>
>>>>
>>>>> Hi,
>>>>>
>>>>> having a table similar to
>>>>>
>>>>> | 1 | B | [2010-07-15 Do] |
>>>>> | 1 | B | [2010-07-16 Fr] |
>>>>> |---+---+-----------------|
>>>>> | 2 | C | [2010-07-17 Sa] |
>>>>> | 2 | C | [2010-07-18 So] |
>>>>> |---+---+-----------------|
>>>>> | 1 | B | [2010-07-19 Mo] |
>>>>> | 1 | B | [2010-07-20 Di] |
>>>>> | 1 | B | [2010-07-21 Mi] |
>>>>> | 1 | B | [2010-07-22 Do] |
>>>>> |---+---+-----------------|
>>>>> | 3 | D | [2010-07-23 Fr] |
>>>>>
>>>>> a simple group by gives me:
>>>>>
>>>>> | 6 | B |
>>>>> | 4 | C |
>>>>> | 3 | D |
>>>>>
>>>>>
>>>>> What I want to get is the values grouped by "subset", where a subset
>>>>> is a set of rows with identical column until the colum changes.
>>>>> Is there a way to get
>>>>>
>>>>> | 2 | B |
>>>>> | 4 | C |
>>>>> | 4 | B |
>>>>> | 3 | D |
>>>>>
>>>>> by SQL only?
>>>>>
>>>>> - Rainer
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>>>> To make changes to your subscription:
>>>>> http://www.postgresql.org/mailpref/pgsql-sql
>>>>
>>>>
>>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 13:31 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
@ 2010-07-30 09:35 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-30 15:07 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
0 siblings, 1 reply; 16+ messages in thread
From: Rainer Stengele @ 2010-07-30 09:35 UTC (permalink / raw)
To: Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>; +Cc: pgsql-sql
the table may include up to maybe 30 entries per day, average maybe 10-15
After a year this makes about 10.000 entries - maximum, average about 5000 entries.
For the problem described I have to use a Microsoft SQL database and would like to use pure SQL.
As I use postgres on my Linux servers I found this newsgroup and thought I ask here.
Thanks!
Rainer
Am 29.07.2010 15:31, schrieb Oliveiros d'Azevedo Cristina:
> Fine.
>
> Please advice me,
>
> How long can
> your table be? Thousands? Millions of records?
>
> Do you really need it in pure SQL
> ?
>
> It seems to me that it might be possible, I'm just affraid that the query would become too complex and thus slow...
>
> Best,
> Oliveiros
>
> ----- Original Message ----- From: "Rainer Stengele" <rainer.stengele@diplan.de>
> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
> Cc: <pgsql-sql@postgresql.org>
> Sent: Thursday, July 29, 2010 1:10 PM
> Subject: Re: grouping subsets
>
>
>> No. This is by accident.
>> We have to assume that the combinations do change anytime, and many times per day.
>>
>> So
>>
>> "Or is it possible to have the same combination on one day with several sets?"
>>
>> YES!
>>
>> Rainer
>>
>>
>>
>> Am 29.07.2010 13:47, schrieb Oliveiros d'Azevedo Cristina:
>>> Yes. This is somewhat more complicated because it has more constraints.
>>> I've noticed that a given combination doesn't appear with holes on a certain day.
>>>
>>> For ex, on a daily basis, we have every three key combinations together.
>>>
>>> We dont have things like
>>> 2010-7-01 1726 3212 1428
>>> 2010-7-01 1726 3212 1428
>>> ... 318 1846 1012
>>> 2010-7-01 1726 3212 1428
>>>
>>> Can I assume that, for a certain day , the records for the same three combination are all together? There is just one set per day for a given combination?
>>>
>>> Or is it possible to have the same combination on one day with several sets?
>>>
>>> Best,
>>> Oliveiros
>>>
>>>
>>> ----- Original Message ----- From: "Rainer Stengele" <rainer.stengele@diplan.de>
>>> Newsgroups: gmane.comp.db.postgresql.sql
>>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>>> Cc: <>
>>> Sent: Thursday, July 29, 2010 10:41 AM
>>> Subject: Re: grouping subsets
>>>
>>>
>>>> Howdy Cristina,
>>>>
>>>> unfortunately things are more complicated. I have inserted an excerpt of the real data here:
>>>>
>>>> ================================================================================
>>>> TableID MasterID dtBegin dtEnd idR idL idB consumption
>>>> 4057312 295530 2010-07-01 00:59:21.077 2010-07-01 01:32:59.670 1726 3212 1428 279
>>>> 4061043 295574 2010-07-01 01:59:31.137 2010-07-01 02:32:09.373 1726 3212 1428 183
>>>> 4083397 295838 2010-07-01 07:57:51.327 2010-07-01 08:28:28.117 318 1846 1012 30
>>>> 4090858 295920 2010-07-01 09:52:33.777 2010-07-01 10:31:34.393 318 1846 1012 487
>>>> 4094589 295961 2010-07-01 10:47:59.370 2010-07-01 11:32:20.903 318 1846 1012 472
>>>> 4098330 296013 2010-07-01 11:58:53.890 2010-07-01 12:31:35.730 318 1846 1012 195
>>>> 4102069 296058 2010-07-01 12:36:19.170 2010-07-01 13:32:13.950 318 1846 1012 338
>>>> 4105809 296102 2010-07-01 13:58:53.170 2010-07-01 14:02:57.710 318 1846 1012 105
>>>> 4109555 296150 2010-07-01 14:59:11.663 2010-07-01 15:32:33.810 318 1846 1012 187
>>>> 4113305 296194 2010-07-01 15:59:01.797 2010-07-01 16:02:27.260 318 1846 1012 108
>>>> 4117048 296238 2010-07-01 16:20:47.997 2010-07-01 17:32:49.367 318 1846 1012 179
>>>> 4120791 296282 2010-07-01 17:58:27.657 2010-07-01 18:29:01.733 318 1846 1012 256
>>>> 4128291 296370 2010-07-01 19:54:17.687 2010-07-01 20:32:53.850 318 1846 1012 239
>>>> 4132044 296413 2010-07-01 20:31:37.653 2010-07-01 21:29:13.497 318 1846 1012 39
>>>> 4135797 296458 2010-07-01 21:59:13.983 2010-07-01 22:32:46.503 318 1846 1012 157
>>>> 4139572 296506 2010-07-01 22:58:49.530 2010-07-01 23:32:22.543 318 1846 1012 218
>>>> 4142941 296554 2010-07-01 23:59:13.857 2010-07-02 00:32:30.390 318 1846 1012 248
>>>> 4146289 296598 2010-07-02 00:58:55.763 2010-07-02 01:32:41.983 318 1846 1012 204
>>>> 4149616 296642 2010-07-02 01:46:57.357 2010-07-02 02:32:56.983 318 1846 1012 42
>>>> 4152952 296686 2010-07-02 02:55:19.653 2010-07-02 03:32:28.013 318 1846 1012 135
>>>> 4156289 296730 2010-07-02 03:43:52.777 2010-07-02 04:32:55.250 318 1846 1012 743
>>>> 4159624 296774 2010-07-02 04:43:15.310 2010-07-02 05:32:44.547 318 1846 1012 277
>>>> 4162961 296817 2010-07-02 05:58:59.483 2010-07-02 06:32:37.340 318 1846 1012 121
>>>> 4166303 296862 2010-07-02 06:58:50.733 2010-07-02 07:32:39.113 318 1846 1012 239
>>>> 4172981 296950 2010-07-02 07:28:55.293 2010-07-02 09:33:01.200 318 1846 1012 512
>>>> 4176322 296993 2010-07-02 09:59:04.607 2010-07-02 10:33:01.903 318 1846 1012 139
>>>> 4179667 297038 2010-07-02 10:55:27.760 2010-07-02 11:32:56.560 318 1846 1012 722
>>>> 4183012 297082 2010-07-02 11:59:33.650 2010-07-02 12:32:14.700 318 1846 1012 163
>>>> 4186351 297126 2010-07-02 12:23:45.997 2010-07-02 13:32:59.500 318 1846 1012 284
>>>> 4189689 297169 2010-07-02 13:44:21.253 2010-07-02 14:18:05.080 318 1846 1012 254
>>>> 4196371 297258 2010-07-02 16:16:19.123 2010-07-02 16:32:53.437 1706 3541 1511 161
>>>> 4199720 297301 2010-07-02 16:59:35.127 2010-07-02 17:32:57.950 1706 3541 1511 250
>>>> 4203068 297346 2010-07-02 17:59:34.027 2010-07-02 18:32:54.337 1706 3541 1511 302
>>>> 4206413 297389 2010-07-02 18:59:28.730 2010-07-02 19:32:37.950 1706 3541 1511 276
>>>> 4209758 297434 2010-07-02 19:54:00.243 2010-07-02 20:32:57.433 1706 3541 1511 209
>>>> 4213102 297473 2010-07-02 20:49:10.963 2010-07-02 21:30:44.540 1706 3541 1511 76
>>>> 4216447 297511 2010-07-02 21:59:34.810 2010-07-02 22:33:00.603 1706 3541 1511 287
>>>> 4219818 297569 2010-07-02 22:56:52.750 2010-07-02 23:59:31.607 1706 3541 1511 1877
>>>> 4219819 297570 2010-07-02 23:59:21.577 2010-07-03 00:54:40.153 1706 3541 1511 1798
>>>> 4219821 297572 2010-07-03 00:48:03.310 2010-07-03 01:59:37.920 1706 3541 1511 1125
>>>> 4219823 297574 2010-07-03 01:51:01.057 2010-07-03 02:59:45.433 1706 3541 1511 1629
>>>> 4219820 297571 2010-07-03 02:59:29.393 2010-07-03 03:59:54.920 1706 3541 1511 2462
>>>> 4219822 297573 2010-07-03 03:59:18.663 2010-07-03 04:01:48.810 1706 3541 1511 70
>>>> 4225738 297656 2010-07-03 06:13:34.980 2010-07-03 06:28:09.697 1726 3212 1428 46
>>>> 4228694 297695 2010-07-03 06:59:15.560 2010-07-03 07:32:45.653 1726 3212 1428 251
>>>> 4231649 297733 2010-07-03 07:59:11.937 2010-07-03 08:32:57.217 1726 3212 1428 284
>>>> 4234604 297771 2010-07-03 08:57:00.357 2010-07-03 09:32:47.903 1726 3212 1428 227
>>>> 4237559 297809 2010-07-03 09:59:19.813 2010-07-03 10:33:02.063 1726 3212 1428 285
>>>> 4261156 298596 2010-07-04 22:59:09.863 2010-07-04 23:33:45.530 1726 3212 1428 1286
>>>> 4264114 298646 2010-07-04 23:59:16.967 2010-07-05 00:33:08.107 1726 3212 1428 297
>>>> 4267067 298690 2010-07-05 00:59:15.187 2010-07-05 01:32:48.300 1726 3212 1428 333
>>>> 4270023 298734 2010-07-05 01:59:02.497 2010-07-05 02:32:48.780 1726 3212 1428 270
>>>> 4272977 298778 2010-07-05 02:41:43.737 2010-07-05 03:32:56.043 1726 3212 1428 317
>>>> 4275927 298822 2010-07-05 03:59:17.027 2010-07-05 04:33:14.947 1726 3212 1428 1623
>>>> ================================================================================
>>>>
>>>> Description:
>>>> 1. Column: some ID
>>>> 2. Column: reference to another table
>>>> 3. and 4. column: timestamp from/to of the item
>>>> 5. Column: ID R
>>>> 6. Column: ID L
>>>> 7. Column: ID B
>>>> 8. Column: Sum of components
>>>>
>>>> Requirement:
>>>> Sum over all components (from column 8) for each combination of ID R, ID L, ID B, but (!)
>>>> rows with same keys (R,L,B) should be summed up only until the keys change.
>>>> Do not sum up the components for identical keys, if there are other keys between them.
>>>>
>>>> Example result:
>>>>
>>>> idR idL idB SUM
>>>> 1726 3212 1428 462
>>>> 318 1846 1012 ...
>>>> 1706 3541 1511 ...
>>>> 1726 3212 1428 ...
>>>>
>>>>
>>>> Note that the first and last entry here has the same keys
>>>>
>>>> Maybe you find a similar monster SQL solving such a requirement.
>>>> Thanks for considering!
>>>>
>>>> Rainer
>>>>
>>>>
>>>>
>>>> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>>>>> Howdy, Rainer.
>>>>>
>>>>> It's been a while, so I don't know if you are still interested in this problem or if you, in the meantime, found yourself a solution,
>>>>> but I've tried this on a local copy of the example you provided and it seems to work.
>>>>>
>>>>> The problem is that I suspect that if you have several thousands of records on your table it will become slow...
>>>>>
>>>>> Best,
>>>>> Oliveiros
>>>>>
>>>>> SELECT SUM(tudo.parcela),tudo.a
>>>>> FROM
>>>>> (
>>>>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>>>>> FROM
>>>>> (
>>>>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>>>>> FROM
>>>>> yourTable se
>>>>> LEFT JOIN
>>>>> (
>>>>> SELECT a.*
>>>>> FROM yourTable a
>>>>> JOIN yourTable b
>>>>> ON (b.b <> a.b)
>>>>> AND ((age(a.c,b.c) = '1 day'::interval)
>>>>>
>>>>> )
>>>>> ) pr
>>>>> ON pr.b = se.b
>>>>> AND se.c >= pr.c
>>>>> GROUP BY se.a,se.b,se.c
>>>>> ) fo
>>>>> LEFT JOIN
>>>>> (
>>>>> SELECT a.*
>>>>> FROM yourTable a
>>>>> JOIN yourTable b
>>>>> ON (b.b <> a.b)
>>>>> AND ((age(a.c,b.c) = '-1 day'::interval)
>>>>> )
>>>>> ) th
>>>>> ON fo.a = th.b
>>>>> AND fo.b <= th.c
>>>>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>>>>> ) tudo
>>>>> GROUP BY tudo.a,tudo.c,tudo.d
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> To: <pgsql-sql@postgresql.org>
>>>>> Sent: Thursday, July 22, 2010 9:09 AM
>>>>> Subject: [SQL] grouping subsets
>>>>>
>>>>>
>>>>>> Hi,
>>>>>>
>>>>>> having a table similar to
>>>>>>
>>>>>> | 1 | B | [2010-07-15 Do] |
>>>>>> | 1 | B | [2010-07-16 Fr] |
>>>>>> |---+---+-----------------|
>>>>>> | 2 | C | [2010-07-17 Sa] |
>>>>>> | 2 | C | [2010-07-18 So] |
>>>>>> |---+---+-----------------|
>>>>>> | 1 | B | [2010-07-19 Mo] |
>>>>>> | 1 | B | [2010-07-20 Di] |
>>>>>> | 1 | B | [2010-07-21 Mi] |
>>>>>> | 1 | B | [2010-07-22 Do] |
>>>>>> |---+---+-----------------|
>>>>>> | 3 | D | [2010-07-23 Fr] |
>>>>>>
>>>>>> a simple group by gives me:
>>>>>>
>>>>>> | 6 | B |
>>>>>> | 4 | C |
>>>>>> | 3 | D |
>>>>>>
>>>>>>
>>>>>> What I want to get is the values grouped by "subset", where a subset is a set of rows with identical column until the colum changes.
>>>>>> Is there a way to get
>>>>>>
>>>>>> | 2 | B |
>>>>>> | 4 | C |
>>>>>> | 4 | B |
>>>>>> | 3 | D |
>>>>>>
>>>>>> by SQL only?
>>>>>>
>>>>>> - Rainer
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>>>>> To make changes to your subscription:
>>>>>> http://www.postgresql.org/mailpref/pgsql-sql
>>>>>
>>>>>
>>>
>
>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 13:31 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-30 09:35 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
@ 2010-07-30 15:07 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-31 19:48 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
0 siblings, 1 reply; 16+ messages in thread
From: Oliveiros d'Azevedo Cristina @ 2010-07-30 15:07 UTC (permalink / raw)
To: Rainer Stengele <rainer.stengele@diplan.de>; +Cc: pgsql-sql
I See.
And the analysis you need to do, the sum of the rows with the same keys
(until they change) will have to be done over all
table?
Or just over some predefined interval ?
Best,
Oliveiros
----- Original Message -----
From: "Rainer Stengele" <rainer.stengele@diplan.de>
Newsgroups: gmane.comp.db.postgresql.sql
To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
Cc: <pgsql-sql@postgresql.org>
Sent: Friday, July 30, 2010 10:35 AM
Subject: Re: grouping subsets
> the table may include up to maybe 30 entries per day, average maybe 10-15
> After a year this makes about 10.000 entries - maximum, average about 5000
> entries.
>
> For the problem described I have to use a Microsoft SQL database and would
> like to use pure SQL.
> As I use postgres on my Linux servers I found this newsgroup and thought I
> ask here.
>
> Thanks!
> Rainer
>
> Am 29.07.2010 15:31, schrieb Oliveiros d'Azevedo Cristina:
>> Fine.
>>
>> Please advice me,
>>
>> How long can
>> your table be? Thousands? Millions of records?
>>
>> Do you really need it in pure SQL
>> ?
>>
>> It seems to me that it might be possible, I'm just affraid that the query
>> would become too complex and thus slow...
>>
>> Best,
>> Oliveiros
>>
>> ----- Original Message ----- From: "Rainer Stengele"
>> <rainer.stengele@diplan.de>
>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>> Cc: <pgsql-sql@postgresql.org>
>> Sent: Thursday, July 29, 2010 1:10 PM
>> Subject: Re: grouping subsets
>>
>>
>>> No. This is by accident.
>>> We have to assume that the combinations do change anytime, and many
>>> times per day.
>>>
>>> So
>>>
>>> "Or is it possible to have the same combination on one day with several
>>> sets?"
>>>
>>> YES!
>>>
>>> Rainer
>>>
>>>
>>>
>>> Am 29.07.2010 13:47, schrieb Oliveiros d'Azevedo Cristina:
>>>> Yes. This is somewhat more complicated because it has more constraints.
>>>> I've noticed that a given combination doesn't appear with holes on a
>>>> certain day.
>>>>
>>>> For ex, on a daily basis, we have every three key combinations
>>>> together.
>>>>
>>>> We dont have things like
>>>> 2010-7-01 1726 3212 1428
>>>> 2010-7-01 1726 3212 1428
>>>> ... 318 1846 1012
>>>> 2010-7-01 1726 3212 1428
>>>>
>>>> Can I assume that, for a certain day , the records for the same three
>>>> combination are all together? There is just one set per day for a given
>>>> combination?
>>>>
>>>> Or is it possible to have the same combination on one day with several
>>>> sets?
>>>>
>>>> Best,
>>>> Oliveiros
>>>>
>>>>
>>>> ----- Original Message ----- From: "Rainer Stengele"
>>>> <rainer.stengele@diplan.de>
>>>> Newsgroups: gmane.comp.db.postgresql.sql
>>>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>>>> Cc: <>
>>>> Sent: Thursday, July 29, 2010 10:41 AM
>>>> Subject: Re: grouping subsets
>>>>
>>>>
>>>>> Howdy Cristina,
>>>>>
>>>>> unfortunately things are more complicated. I have inserted an excerpt
>>>>> of the real data here:
>>>>>
>>>>> ================================================================================
>>>>> TableID MasterID dtBegin dtEnd idR
>>>>> idL idB consumption
>>>>> 4057312 295530 2010-07-01 00:59:21.077 2010-07-01
>>>>> 01:32:59.670 1726 3212 1428 279
>>>>> 4061043 295574 2010-07-01 01:59:31.137 2010-07-01
>>>>> 02:32:09.373 1726 3212 1428 183
>>>>> 4083397 295838 2010-07-01 07:57:51.327 2010-07-01
>>>>> 08:28:28.117 318 1846 1012 30
>>>>> 4090858 295920 2010-07-01 09:52:33.777 2010-07-01
>>>>> 10:31:34.393 318 1846 1012 487
>>>>> 4094589 295961 2010-07-01 10:47:59.370 2010-07-01
>>>>> 11:32:20.903 318 1846 1012 472
>>>>> 4098330 296013 2010-07-01 11:58:53.890 2010-07-01
>>>>> 12:31:35.730 318 1846 1012 195
>>>>> 4102069 296058 2010-07-01 12:36:19.170 2010-07-01
>>>>> 13:32:13.950 318 1846 1012 338
>>>>> 4105809 296102 2010-07-01 13:58:53.170 2010-07-01
>>>>> 14:02:57.710 318 1846 1012 105
>>>>> 4109555 296150 2010-07-01 14:59:11.663 2010-07-01
>>>>> 15:32:33.810 318 1846 1012 187
>>>>> 4113305 296194 2010-07-01 15:59:01.797 2010-07-01
>>>>> 16:02:27.260 318 1846 1012 108
>>>>> 4117048 296238 2010-07-01 16:20:47.997 2010-07-01
>>>>> 17:32:49.367 318 1846 1012 179
>>>>> 4120791 296282 2010-07-01 17:58:27.657 2010-07-01
>>>>> 18:29:01.733 318 1846 1012 256
>>>>> 4128291 296370 2010-07-01 19:54:17.687 2010-07-01
>>>>> 20:32:53.850 318 1846 1012 239
>>>>> 4132044 296413 2010-07-01 20:31:37.653 2010-07-01
>>>>> 21:29:13.497 318 1846 1012 39
>>>>> 4135797 296458 2010-07-01 21:59:13.983 2010-07-01
>>>>> 22:32:46.503 318 1846 1012 157
>>>>> 4139572 296506 2010-07-01 22:58:49.530 2010-07-01
>>>>> 23:32:22.543 318 1846 1012 218
>>>>> 4142941 296554 2010-07-01 23:59:13.857 2010-07-02
>>>>> 00:32:30.390 318 1846 1012 248
>>>>> 4146289 296598 2010-07-02 00:58:55.763 2010-07-02
>>>>> 01:32:41.983 318 1846 1012 204
>>>>> 4149616 296642 2010-07-02 01:46:57.357 2010-07-02
>>>>> 02:32:56.983 318 1846 1012 42
>>>>> 4152952 296686 2010-07-02 02:55:19.653 2010-07-02
>>>>> 03:32:28.013 318 1846 1012 135
>>>>> 4156289 296730 2010-07-02 03:43:52.777 2010-07-02
>>>>> 04:32:55.250 318 1846 1012 743
>>>>> 4159624 296774 2010-07-02 04:43:15.310 2010-07-02
>>>>> 05:32:44.547 318 1846 1012 277
>>>>> 4162961 296817 2010-07-02 05:58:59.483 2010-07-02
>>>>> 06:32:37.340 318 1846 1012 121
>>>>> 4166303 296862 2010-07-02 06:58:50.733 2010-07-02
>>>>> 07:32:39.113 318 1846 1012 239
>>>>> 4172981 296950 2010-07-02 07:28:55.293 2010-07-02
>>>>> 09:33:01.200 318 1846 1012 512
>>>>> 4176322 296993 2010-07-02 09:59:04.607 2010-07-02
>>>>> 10:33:01.903 318 1846 1012 139
>>>>> 4179667 297038 2010-07-02 10:55:27.760 2010-07-02
>>>>> 11:32:56.560 318 1846 1012 722
>>>>> 4183012 297082 2010-07-02 11:59:33.650 2010-07-02
>>>>> 12:32:14.700 318 1846 1012 163
>>>>> 4186351 297126 2010-07-02 12:23:45.997 2010-07-02
>>>>> 13:32:59.500 318 1846 1012 284
>>>>> 4189689 297169 2010-07-02 13:44:21.253 2010-07-02
>>>>> 14:18:05.080 318 1846 1012 254
>>>>> 4196371 297258 2010-07-02 16:16:19.123 2010-07-02
>>>>> 16:32:53.437 1706 3541 1511 161
>>>>> 4199720 297301 2010-07-02 16:59:35.127 2010-07-02
>>>>> 17:32:57.950 1706 3541 1511 250
>>>>> 4203068 297346 2010-07-02 17:59:34.027 2010-07-02
>>>>> 18:32:54.337 1706 3541 1511 302
>>>>> 4206413 297389 2010-07-02 18:59:28.730 2010-07-02
>>>>> 19:32:37.950 1706 3541 1511 276
>>>>> 4209758 297434 2010-07-02 19:54:00.243 2010-07-02
>>>>> 20:32:57.433 1706 3541 1511 209
>>>>> 4213102 297473 2010-07-02 20:49:10.963 2010-07-02
>>>>> 21:30:44.540 1706 3541 1511 76
>>>>> 4216447 297511 2010-07-02 21:59:34.810 2010-07-02
>>>>> 22:33:00.603 1706 3541 1511 287
>>>>> 4219818 297569 2010-07-02 22:56:52.750 2010-07-02
>>>>> 23:59:31.607 1706 3541 1511 1877
>>>>> 4219819 297570 2010-07-02 23:59:21.577 2010-07-03
>>>>> 00:54:40.153 1706 3541 1511 1798
>>>>> 4219821 297572 2010-07-03 00:48:03.310 2010-07-03
>>>>> 01:59:37.920 1706 3541 1511 1125
>>>>> 4219823 297574 2010-07-03 01:51:01.057 2010-07-03
>>>>> 02:59:45.433 1706 3541 1511 1629
>>>>> 4219820 297571 2010-07-03 02:59:29.393 2010-07-03
>>>>> 03:59:54.920 1706 3541 1511 2462
>>>>> 4219822 297573 2010-07-03 03:59:18.663 2010-07-03
>>>>> 04:01:48.810 1706 3541 1511 70
>>>>> 4225738 297656 2010-07-03 06:13:34.980 2010-07-03
>>>>> 06:28:09.697 1726 3212 1428 46
>>>>> 4228694 297695 2010-07-03 06:59:15.560 2010-07-03
>>>>> 07:32:45.653 1726 3212 1428 251
>>>>> 4231649 297733 2010-07-03 07:59:11.937 2010-07-03
>>>>> 08:32:57.217 1726 3212 1428 284
>>>>> 4234604 297771 2010-07-03 08:57:00.357 2010-07-03
>>>>> 09:32:47.903 1726 3212 1428 227
>>>>> 4237559 297809 2010-07-03 09:59:19.813 2010-07-03
>>>>> 10:33:02.063 1726 3212 1428 285
>>>>> 4261156 298596 2010-07-04 22:59:09.863 2010-07-04
>>>>> 23:33:45.530 1726 3212 1428 1286
>>>>> 4264114 298646 2010-07-04 23:59:16.967 2010-07-05
>>>>> 00:33:08.107 1726 3212 1428 297
>>>>> 4267067 298690 2010-07-05 00:59:15.187 2010-07-05
>>>>> 01:32:48.300 1726 3212 1428 333
>>>>> 4270023 298734 2010-07-05 01:59:02.497 2010-07-05
>>>>> 02:32:48.780 1726 3212 1428 270
>>>>> 4272977 298778 2010-07-05 02:41:43.737 2010-07-05
>>>>> 03:32:56.043 1726 3212 1428 317
>>>>> 4275927 298822 2010-07-05 03:59:17.027 2010-07-05
>>>>> 04:33:14.947 1726 3212 1428 1623
>>>>> ================================================================================
>>>>>
>>>>> Description:
>>>>> 1. Column: some ID
>>>>> 2. Column: reference to another table
>>>>> 3. and 4. column: timestamp from/to of the item
>>>>> 5. Column: ID R
>>>>> 6. Column: ID L
>>>>> 7. Column: ID B
>>>>> 8. Column: Sum of components
>>>>>
>>>>> Requirement:
>>>>> Sum over all components (from column 8) for each combination of ID R,
>>>>> ID L, ID B, but (!)
>>>>> rows with same keys (R,L,B) should be summed up only until the keys
>>>>> change.
>>>>> Do not sum up the components for identical keys, if there are other
>>>>> keys between them.
>>>>>
>>>>> Example result:
>>>>>
>>>>> idR idL idB SUM
>>>>> 1726 3212 1428 462
>>>>> 318 1846 1012 ...
>>>>> 1706 3541 1511 ...
>>>>> 1726 3212 1428 ...
>>>>>
>>>>>
>>>>> Note that the first and last entry here has the same keys
>>>>>
>>>>> Maybe you find a similar monster SQL solving such a requirement.
>>>>> Thanks for considering!
>>>>>
>>>>> Rainer
>>>>>
>>>>>
>>>>>
>>>>> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>>>>>> Howdy, Rainer.
>>>>>>
>>>>>> It's been a while, so I don't know if you are still interested in
>>>>>> this problem or if you, in the meantime, found yourself a solution,
>>>>>> but I've tried this on a local copy of the example you provided and
>>>>>> it seems to work.
>>>>>>
>>>>>> The problem is that I suspect that if you have several thousands of
>>>>>> records on your table it will become slow...
>>>>>>
>>>>>> Best,
>>>>>> Oliveiros
>>>>>>
>>>>>> SELECT SUM(tudo.parcela),tudo.a
>>>>>> FROM
>>>>>> (
>>>>>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>>>>>> FROM
>>>>>> (
>>>>>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>>>>>> FROM
>>>>>> yourTable se
>>>>>> LEFT JOIN
>>>>>> (
>>>>>> SELECT a.*
>>>>>> FROM yourTable a
>>>>>> JOIN yourTable b
>>>>>> ON (b.b <> a.b)
>>>>>> AND ((age(a.c,b.c) = '1 day'::interval)
>>>>>>
>>>>>> )
>>>>>> ) pr
>>>>>> ON pr.b = se.b
>>>>>> AND se.c >= pr.c
>>>>>> GROUP BY se.a,se.b,se.c
>>>>>> ) fo
>>>>>> LEFT JOIN
>>>>>> (
>>>>>> SELECT a.*
>>>>>> FROM yourTable a
>>>>>> JOIN yourTable b
>>>>>> ON (b.b <> a.b)
>>>>>> AND ((age(a.c,b.c) = '-1 day'::interval)
>>>>>> )
>>>>>> ) th
>>>>>> ON fo.a = th.b
>>>>>> AND fo.b <= th.c
>>>>>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>>>>>> ) tudo
>>>>>> GROUP BY tudo.a,tudo.c,tudo.d
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> To: <pgsql-sql@postgresql.org>
>>>>>> Sent: Thursday, July 22, 2010 9:09 AM
>>>>>> Subject: [SQL] grouping subsets
>>>>>>
>>>>>>
>>>>>>> Hi,
>>>>>>>
>>>>>>> having a table similar to
>>>>>>>
>>>>>>> | 1 | B | [2010-07-15 Do] |
>>>>>>> | 1 | B | [2010-07-16 Fr] |
>>>>>>> |---+---+-----------------|
>>>>>>> | 2 | C | [2010-07-17 Sa] |
>>>>>>> | 2 | C | [2010-07-18 So] |
>>>>>>> |---+---+-----------------|
>>>>>>> | 1 | B | [2010-07-19 Mo] |
>>>>>>> | 1 | B | [2010-07-20 Di] |
>>>>>>> | 1 | B | [2010-07-21 Mi] |
>>>>>>> | 1 | B | [2010-07-22 Do] |
>>>>>>> |---+---+-----------------|
>>>>>>> | 3 | D | [2010-07-23 Fr] |
>>>>>>>
>>>>>>> a simple group by gives me:
>>>>>>>
>>>>>>> | 6 | B |
>>>>>>> | 4 | C |
>>>>>>> | 3 | D |
>>>>>>>
>>>>>>>
>>>>>>> What I want to get is the values grouped by "subset", where a subset
>>>>>>> is a set of rows with identical column until the colum changes.
>>>>>>> Is there a way to get
>>>>>>>
>>>>>>> | 2 | B |
>>>>>>> | 4 | C |
>>>>>>> | 4 | B |
>>>>>>> | 3 | D |
>>>>>>>
>>>>>>> by SQL only?
>>>>>>>
>>>>>>> - Rainer
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> --
>>>>>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>>>>>> To make changes to your subscription:
>>>>>>> http://www.postgresql.org/mailpref/pgsql-sql
>>>>>>
>>>>>>
>>>>
>>
>>
^ permalink raw reply [nested|flat] 16+ messages in thread
* Re: grouping subsets
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-27 10:37 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 13:31 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-30 09:35 ` Re: grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-30 15:07 ` Re: grouping subsets Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
@ 2010-07-31 19:48 ` Rainer Stengele <rainer.stengele@diplan.de>
0 siblings, 0 replies; 16+ messages in thread
From: Rainer Stengele @ 2010-07-31 19:48 UTC (permalink / raw)
To: Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>; +Cc: pgsql-sql
The analysis will have to be done over a calendar range which the user
selects via web interface. So - yes - normally not the whole table, but
maybe he chooses one month and thats already a lot of rows in the table ...
Best, too,
Rainer
Am 30.07.2010 17:07, schrieb Oliveiros d'Azevedo Cristina:
> I See.
>
> And the analysis you need to do, the sum of the rows with the same
> keys (until they change) will have to be done over all
> table?
> Or just over some predefined interval ?
>
> Best,
> Oliveiros
> ----- Original Message ----- From: "Rainer Stengele"
> <rainer.stengele@diplan.de>
> Newsgroups: gmane.comp.db.postgresql.sql
> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
> Cc: <pgsql-sql@postgresql.org>
> Sent: Friday, July 30, 2010 10:35 AM
> Subject: Re: grouping subsets
>
>
>> the table may include up to maybe 30 entries per day, average maybe
>> 10-15
>> After a year this makes about 10.000 entries - maximum, average about
>> 5000 entries.
>>
>> For the problem described I have to use a Microsoft SQL database and
>> would like to use pure SQL.
>> As I use postgres on my Linux servers I found this newsgroup and
>> thought I ask here.
>>
>> Thanks!
>> Rainer
>>
>> Am 29.07.2010 15:31, schrieb Oliveiros d'Azevedo Cristina:
>>> Fine.
>>>
>>> Please advice me,
>>>
>>> How long can
>>> your table be? Thousands? Millions of records?
>>>
>>> Do you really need it in pure SQL
>>> ?
>>>
>>> It seems to me that it might be possible, I'm just affraid that the
>>> query would become too complex and thus slow...
>>>
>>> Best,
>>> Oliveiros
>>>
>>> ----- Original Message ----- From: "Rainer Stengele"
>>> <rainer.stengele@diplan.de>
>>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>>> Cc: <pgsql-sql@postgresql.org>
>>> Sent: Thursday, July 29, 2010 1:10 PM
>>> Subject: Re: grouping subsets
>>>
>>>
>>>> No. This is by accident.
>>>> We have to assume that the combinations do change anytime, and many
>>>> times per day.
>>>>
>>>> So
>>>>
>>>> "Or is it possible to have the same combination on one day with
>>>> several sets?"
>>>>
>>>> YES!
>>>>
>>>> Rainer
>>>>
>>>>
>>>>
>>>> Am 29.07.2010 13:47, schrieb Oliveiros d'Azevedo Cristina:
>>>>> Yes. This is somewhat more complicated because it has more
>>>>> constraints.
>>>>> I've noticed that a given combination doesn't appear with holes on
>>>>> a certain day.
>>>>>
>>>>> For ex, on a daily basis, we have every three key combinations
>>>>> together.
>>>>>
>>>>> We dont have things like
>>>>> 2010-7-01 1726 3212 1428
>>>>> 2010-7-01 1726 3212 1428
>>>>> ... 318 1846 1012
>>>>> 2010-7-01 1726 3212 1428
>>>>>
>>>>> Can I assume that, for a certain day , the records for the same
>>>>> three combination are all together? There is just one set per day
>>>>> for a given combination?
>>>>>
>>>>> Or is it possible to have the same combination on one day with
>>>>> several sets?
>>>>>
>>>>> Best,
>>>>> Oliveiros
>>>>>
>>>>>
>>>>> ----- Original Message ----- From: "Rainer Stengele"
>>>>> <rainer.stengele@diplan.de>
>>>>> Newsgroups: gmane.comp.db.postgresql.sql
>>>>> To: "Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>
>>>>> Cc: <>
>>>>> Sent: Thursday, July 29, 2010 10:41 AM
>>>>> Subject: Re: grouping subsets
>>>>>
>>>>>
>>>>>> Howdy Cristina,
>>>>>>
>>>>>> unfortunately things are more complicated. I have inserted an
>>>>>> excerpt of the real data here:
>>>>>>
>>>>>> ================================================================================
>>>>>>
>>>>>> TableID MasterID dtBegin dtEnd
>>>>>> idR idL idB consumption
>>>>>> 4057312 295530 2010-07-01 00:59:21.077
>>>>>> 2010-07-01 01:32:59.670 1726 3212 1428 279
>>>>>> 4061043 295574 2010-07-01 01:59:31.137
>>>>>> 2010-07-01 02:32:09.373 1726 3212 1428 183
>>>>>> 4083397 295838 2010-07-01 07:57:51.327
>>>>>> 2010-07-01 08:28:28.117 318 1846 1012 30
>>>>>> 4090858 295920 2010-07-01 09:52:33.777
>>>>>> 2010-07-01 10:31:34.393 318 1846 1012 487
>>>>>> 4094589 295961 2010-07-01 10:47:59.370
>>>>>> 2010-07-01 11:32:20.903 318 1846 1012 472
>>>>>> 4098330 296013 2010-07-01 11:58:53.890
>>>>>> 2010-07-01 12:31:35.730 318 1846 1012 195
>>>>>> 4102069 296058 2010-07-01 12:36:19.170
>>>>>> 2010-07-01 13:32:13.950 318 1846 1012 338
>>>>>> 4105809 296102 2010-07-01 13:58:53.170
>>>>>> 2010-07-01 14:02:57.710 318 1846 1012 105
>>>>>> 4109555 296150 2010-07-01 14:59:11.663
>>>>>> 2010-07-01 15:32:33.810 318 1846 1012 187
>>>>>> 4113305 296194 2010-07-01 15:59:01.797
>>>>>> 2010-07-01 16:02:27.260 318 1846 1012 108
>>>>>> 4117048 296238 2010-07-01 16:20:47.997
>>>>>> 2010-07-01 17:32:49.367 318 1846 1012 179
>>>>>> 4120791 296282 2010-07-01 17:58:27.657
>>>>>> 2010-07-01 18:29:01.733 318 1846 1012 256
>>>>>> 4128291 296370 2010-07-01 19:54:17.687
>>>>>> 2010-07-01 20:32:53.850 318 1846 1012 239
>>>>>> 4132044 296413 2010-07-01 20:31:37.653
>>>>>> 2010-07-01 21:29:13.497 318 1846 1012 39
>>>>>> 4135797 296458 2010-07-01 21:59:13.983
>>>>>> 2010-07-01 22:32:46.503 318 1846 1012 157
>>>>>> 4139572 296506 2010-07-01 22:58:49.530
>>>>>> 2010-07-01 23:32:22.543 318 1846 1012 218
>>>>>> 4142941 296554 2010-07-01 23:59:13.857
>>>>>> 2010-07-02 00:32:30.390 318 1846 1012 248
>>>>>> 4146289 296598 2010-07-02 00:58:55.763
>>>>>> 2010-07-02 01:32:41.983 318 1846 1012 204
>>>>>> 4149616 296642 2010-07-02 01:46:57.357
>>>>>> 2010-07-02 02:32:56.983 318 1846 1012 42
>>>>>> 4152952 296686 2010-07-02 02:55:19.653
>>>>>> 2010-07-02 03:32:28.013 318 1846 1012 135
>>>>>> 4156289 296730 2010-07-02 03:43:52.777
>>>>>> 2010-07-02 04:32:55.250 318 1846 1012 743
>>>>>> 4159624 296774 2010-07-02 04:43:15.310
>>>>>> 2010-07-02 05:32:44.547 318 1846 1012 277
>>>>>> 4162961 296817 2010-07-02 05:58:59.483
>>>>>> 2010-07-02 06:32:37.340 318 1846 1012 121
>>>>>> 4166303 296862 2010-07-02 06:58:50.733
>>>>>> 2010-07-02 07:32:39.113 318 1846 1012 239
>>>>>> 4172981 296950 2010-07-02 07:28:55.293
>>>>>> 2010-07-02 09:33:01.200 318 1846 1012 512
>>>>>> 4176322 296993 2010-07-02 09:59:04.607
>>>>>> 2010-07-02 10:33:01.903 318 1846 1012 139
>>>>>> 4179667 297038 2010-07-02 10:55:27.760
>>>>>> 2010-07-02 11:32:56.560 318 1846 1012 722
>>>>>> 4183012 297082 2010-07-02 11:59:33.650
>>>>>> 2010-07-02 12:32:14.700 318 1846 1012 163
>>>>>> 4186351 297126 2010-07-02 12:23:45.997
>>>>>> 2010-07-02 13:32:59.500 318 1846 1012 284
>>>>>> 4189689 297169 2010-07-02 13:44:21.253
>>>>>> 2010-07-02 14:18:05.080 318 1846 1012 254
>>>>>> 4196371 297258 2010-07-02 16:16:19.123
>>>>>> 2010-07-02 16:32:53.437 1706 3541 1511 161
>>>>>> 4199720 297301 2010-07-02 16:59:35.127
>>>>>> 2010-07-02 17:32:57.950 1706 3541 1511 250
>>>>>> 4203068 297346 2010-07-02 17:59:34.027
>>>>>> 2010-07-02 18:32:54.337 1706 3541 1511 302
>>>>>> 4206413 297389 2010-07-02 18:59:28.730
>>>>>> 2010-07-02 19:32:37.950 1706 3541 1511 276
>>>>>> 4209758 297434 2010-07-02 19:54:00.243
>>>>>> 2010-07-02 20:32:57.433 1706 3541 1511 209
>>>>>> 4213102 297473 2010-07-02 20:49:10.963
>>>>>> 2010-07-02 21:30:44.540 1706 3541 1511 76
>>>>>> 4216447 297511 2010-07-02 21:59:34.810
>>>>>> 2010-07-02 22:33:00.603 1706 3541 1511 287
>>>>>> 4219818 297569 2010-07-02 22:56:52.750
>>>>>> 2010-07-02 23:59:31.607 1706 3541 1511 1877
>>>>>> 4219819 297570 2010-07-02 23:59:21.577
>>>>>> 2010-07-03 00:54:40.153 1706 3541 1511 1798
>>>>>> 4219821 297572 2010-07-03 00:48:03.310
>>>>>> 2010-07-03 01:59:37.920 1706 3541 1511 1125
>>>>>> 4219823 297574 2010-07-03 01:51:01.057
>>>>>> 2010-07-03 02:59:45.433 1706 3541 1511 1629
>>>>>> 4219820 297571 2010-07-03 02:59:29.393
>>>>>> 2010-07-03 03:59:54.920 1706 3541 1511 2462
>>>>>> 4219822 297573 2010-07-03 03:59:18.663
>>>>>> 2010-07-03 04:01:48.810 1706 3541 1511 70
>>>>>> 4225738 297656 2010-07-03 06:13:34.980
>>>>>> 2010-07-03 06:28:09.697 1726 3212 1428 46
>>>>>> 4228694 297695 2010-07-03 06:59:15.560
>>>>>> 2010-07-03 07:32:45.653 1726 3212 1428 251
>>>>>> 4231649 297733 2010-07-03 07:59:11.937
>>>>>> 2010-07-03 08:32:57.217 1726 3212 1428 284
>>>>>> 4234604 297771 2010-07-03 08:57:00.357
>>>>>> 2010-07-03 09:32:47.903 1726 3212 1428 227
>>>>>> 4237559 297809 2010-07-03 09:59:19.813
>>>>>> 2010-07-03 10:33:02.063 1726 3212 1428 285
>>>>>> 4261156 298596 2010-07-04 22:59:09.863
>>>>>> 2010-07-04 23:33:45.530 1726 3212 1428 1286
>>>>>> 4264114 298646 2010-07-04 23:59:16.967
>>>>>> 2010-07-05 00:33:08.107 1726 3212 1428 297
>>>>>> 4267067 298690 2010-07-05 00:59:15.187
>>>>>> 2010-07-05 01:32:48.300 1726 3212 1428 333
>>>>>> 4270023 298734 2010-07-05 01:59:02.497
>>>>>> 2010-07-05 02:32:48.780 1726 3212 1428 270
>>>>>> 4272977 298778 2010-07-05 02:41:43.737
>>>>>> 2010-07-05 03:32:56.043 1726 3212 1428 317
>>>>>> 4275927 298822 2010-07-05 03:59:17.027
>>>>>> 2010-07-05 04:33:14.947 1726 3212 1428 1623
>>>>>> ================================================================================
>>>>>>
>>>>>>
>>>>>> Description:
>>>>>> 1. Column: some ID
>>>>>> 2. Column: reference to another table
>>>>>> 3. and 4. column: timestamp from/to of the item
>>>>>> 5. Column: ID R
>>>>>> 6. Column: ID L
>>>>>> 7. Column: ID B
>>>>>> 8. Column: Sum of components
>>>>>>
>>>>>> Requirement:
>>>>>> Sum over all components (from column 8) for each combination of
>>>>>> ID R, ID L, ID B, but (!)
>>>>>> rows with same keys (R,L,B) should be summed up only until the
>>>>>> keys change.
>>>>>> Do not sum up the components for identical keys, if there are
>>>>>> other keys between them.
>>>>>>
>>>>>> Example result:
>>>>>>
>>>>>> idR idL idB SUM
>>>>>> 1726 3212 1428 462
>>>>>> 318 1846 1012 ...
>>>>>> 1706 3541 1511 ...
>>>>>> 1726 3212 1428 ...
>>>>>>
>>>>>>
>>>>>> Note that the first and last entry here has the same keys
>>>>>>
>>>>>> Maybe you find a similar monster SQL solving such a requirement.
>>>>>> Thanks for considering!
>>>>>>
>>>>>> Rainer
>>>>>>
>>>>>>
>>>>>>
>>>>>> Am 27.07.2010 12:37, schrieb Oliveiros d'Azevedo Cristina:
>>>>>>> Howdy, Rainer.
>>>>>>>
>>>>>>> It's been a while, so I don't know if you are still interested
>>>>>>> in this problem or if you, in the meantime, found yourself a
>>>>>>> solution,
>>>>>>> but I've tried this on a local copy of the example you
>>>>>>> provided and it seems to work.
>>>>>>>
>>>>>>> The problem is that I suspect that if you have several thousands
>>>>>>> of records on your table it will become slow...
>>>>>>>
>>>>>>> Best,
>>>>>>> Oliveiros
>>>>>>>
>>>>>>> SELECT SUM(tudo.parcela),tudo.a
>>>>>>> FROM
>>>>>>> (
>>>>>>> SELECT fo.parcela,fo.a,fo.b,fo.c,MIN(th.c) as d
>>>>>>> FROM
>>>>>>> (
>>>>>>> SELECT se.a as parcela,se.b as a,se.c as b, MAX(pr.c) as c
>>>>>>> FROM
>>>>>>> yourTable se
>>>>>>> LEFT JOIN
>>>>>>> (
>>>>>>> SELECT a.*
>>>>>>> FROM yourTable a
>>>>>>> JOIN yourTable b
>>>>>>> ON (b.b <> a.b)
>>>>>>> AND ((age(a.c,b.c) = '1 day'::interval)
>>>>>>>
>>>>>>> )
>>>>>>> ) pr
>>>>>>> ON pr.b = se.b
>>>>>>> AND se.c >= pr.c
>>>>>>> GROUP BY se.a,se.b,se.c
>>>>>>> ) fo
>>>>>>> LEFT JOIN
>>>>>>> (
>>>>>>> SELECT a.*
>>>>>>> FROM yourTable a
>>>>>>> JOIN yourTable b
>>>>>>> ON (b.b <> a.b)
>>>>>>> AND ((age(a.c,b.c) = '-1 day'::interval)
>>>>>>> )
>>>>>>> ) th
>>>>>>> ON fo.a = th.b
>>>>>>> AND fo.b <= th.c
>>>>>>> GROUP BY fo.parcela,fo.a,fo.b,fo.c
>>>>>>> ) tudo
>>>>>>> GROUP BY tudo.a,tudo.c,tudo.d
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> To: <pgsql-sql@postgresql.org>
>>>>>>> Sent: Thursday, July 22, 2010 9:09 AM
>>>>>>> Subject: [SQL] grouping subsets
>>>>>>>
>>>>>>>
>>>>>>>> Hi,
>>>>>>>>
>>>>>>>> having a table similar to
>>>>>>>>
>>>>>>>> | 1 | B | [2010-07-15 Do] |
>>>>>>>> | 1 | B | [2010-07-16 Fr] |
>>>>>>>> |---+---+-----------------|
>>>>>>>> | 2 | C | [2010-07-17 Sa] |
>>>>>>>> | 2 | C | [2010-07-18 So] |
>>>>>>>> |---+---+-----------------|
>>>>>>>> | 1 | B | [2010-07-19 Mo] |
>>>>>>>> | 1 | B | [2010-07-20 Di] |
>>>>>>>> | 1 | B | [2010-07-21 Mi] |
>>>>>>>> | 1 | B | [2010-07-22 Do] |
>>>>>>>> |---+---+-----------------|
>>>>>>>> | 3 | D | [2010-07-23 Fr] |
>>>>>>>>
>>>>>>>> a simple group by gives me:
>>>>>>>>
>>>>>>>> | 6 | B |
>>>>>>>> | 4 | C |
>>>>>>>> | 3 | D |
>>>>>>>>
>>>>>>>>
>>>>>>>> What I want to get is the values grouped by "subset", where a
>>>>>>>> subset is a set of rows with identical column until the colum
>>>>>>>> changes.
>>>>>>>> Is there a way to get
>>>>>>>>
>>>>>>>> | 2 | B |
>>>>>>>> | 4 | C |
>>>>>>>> | 4 | B |
>>>>>>>> | 3 | D |
>>>>>>>>
>>>>>>>> by SQL only?
>>>>>>>>
>>>>>>>> - Rainer
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> --
>>>>>>>> Sent via pgsql-sql mailing list (pgsql-sql@postgresql.org)
>>>>>>>> To make changes to your subscription:
>>>>>>>> http://www.postgresql.org/mailpref/pgsql-sql
>>>>>>>
>>>>>>>
>>>>>
>>>
>>>
>
^ permalink raw reply [nested|flat] 16+ messages in thread
end of thread, other threads:[~2010-07-31 19:48 UTC | newest]
Thread overview: 16+ messages (download: mbox mbox.gz follow: Atom feed)
-- links below jump to the message on this page --
2010-07-22 08:09 grouping subsets Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 10:02 ` A. Kretschmer <andreas.kretschmer@schollglas.com>
2010-07-22 10:53 ` Richard Huxton <dev@archonet.com>
2010-07-22 11:31 ` Tim Landscheidt <tim@tim-landscheidt.de>
2010-07-30 15:58 ` Joshua Tolley <eggyknap@gmail.com>
2010-07-22 13:18 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-22 13:25 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-22 14:18 ` Tim Landscheidt <tim@tim-landscheidt.de>
2010-07-27 10:37 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 09:41 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 11:47 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-29 12:10 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-29 13:31 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-30 09:35 ` Rainer Stengele <rainer.stengele@diplan.de>
2010-07-30 15:07 ` Oliveiros d'Azevedo Cristina <oliveiros.cristina@marktest.pt>
2010-07-31 19:48 ` Rainer Stengele <rainer.stengele@diplan.de>
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