Re: grouping subsets
| От | Rainer Stengele |
|---|---|
| Тема | Re: grouping subsets |
| Дата | |
| Msg-id | 4C529CF6.5040408@diplan.de обсуждение исходный текст |
| Ответ на | Re: grouping subsets ("Oliveiros d'Azevedo Cristina" <oliveiros.cristina@marktest.pt>) |
| Список | 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 oneset 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 yourselfa 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 untilthe 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 >>>>> >>>>> >>> > >
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