Обсуждение: A counter argument about DISTINCT and GROUP BY in PostgreSQL

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A counter argument about DISTINCT and GROUP BY in PostgreSQL

От
dterrors@hotmail.com
Дата:
I've just spent a few hours searching and reading about the postgres
way of selecting distinct records.  I understand the points made about
the ORDER BY limitation of DISTINCT ON, and the abiguity of GROUP BY,
but I think there's a (simple, common) case that have been missed in
the discussion. Here is my sitation:

table "projects":
id  title     more stuff (pretend there's 20 more columns.)
-----------------------------------------------------------
1   buildrome     moredata       inothercolumns
2   housework   evenmoredata letssay20columns

table "todos":
id projectid name     duedate
-----------------------------------------
1  1         conquer    1pm
2  1         laybricks  10pm
3  2         dolaundry  5pm


In english, I want to "select projects and order them by the ones that
have todos due the soonest."  Does that sound like a reasonable
request?

This won't work in postgresql 8.2.4:

select distinct on(a.id) a.* from projects a inner join todos b on
b.projectid = a.id order by b.duedate offset 10 limit 20;
ERROR: SELECT DISTINCT ON expressions must match initial ORDER BY
expressions

What to do?  I could:

Option A. select distinct on(b.duedate, a.id) a.* from projects a
inner join todos b on b.projectid = a.id order by b.duedate, a.id
offset 10 limit 20;

But that's not equivalent because the duedates will break uniqueness.
The results will have two records where the a.id is 1.  So I guess
that's a non-option.

Option B. I could try group by:
select a.* from projects a inner join todos b on b.projectid = a.id
group by a.id order by b.duedate offset 10 limit 20;
Query failed: ERROR: column "a.title" must appear in the GROUP BY
clause or be used in an aggregate function

(And presumably every other column of projects would too).

Option C. I could list out each and every one of the 20+ columns of
the project table in MAX() functions so that they'd be "in an
aggregate function":
select MAX(a.title) as title, MAX(a.column2) as column2,
MAX(a.column3) as column3, MAX(a.column4) as column4,  MAX(a.column5)
as column5,  MAX(a.column6) as column6,  MAX(a.column7) as column7,
MAX(a.column8) as column8, MAX(a.column9) as column9, MAX(a.column8)
as ihatepostgresql, MAX(a.column9) as imgoingbacktomysql,  from
projects a inner join todos b on b.projectid = a.id group by a.id
order by b.duedate offset 10 limit 20;

http://groups.google.com/group/comp.databases.postgresql.general/msg/923aed5b8ea5faa1

Option D: select distinct a.* from projects a inner join todos b on
b.projectid = a.id order by b.duedate offset 10 limit 20;
Query failed: ERROR: for SELECT DISTINCT, ORDER BY expressions must
appear in select list

But I don't want b.duedate in my results. And if I add it, it will,
again, break the desired uniqueness (see option A).  (And, actually
even if that one worked, it would suck. Instead of making postgres
compare only one column to determine uniqueness, it now has to compare
all of them.  (In other words, the whole advantage of DISTINCT ON is
out the window).)

Option E: I could use a subselect.  But notice my offset, limit.  If I
use a subselect, then postgresql would have to build ALL of the
results in memory (to create the subselect virtual table), before I
apply the offset and limit on the subselect.

Any suggestion would be appreciated.

BTW for those of you who are curious, in mysql (that other db), this
would be:

select a.* from projects a inner join todos b on b.projectid = a.id
group by a.id order by b.duedate limit 10,20;


Re: A counter argument about DISTINCT and GROUP BY in PostgreSQL

От
Gregory Stark
Дата:
<dterrors@hotmail.com> writes:

> I've just spent a few hours searching and reading about the postgres
> way of selecting distinct records.  I understand the points made about
> the ORDER BY limitation of DISTINCT ON, and the abiguity of GROUP BY,
> but I think there's a (simple, common) case that have been missed in
> the discussion. Here is my sitation:
>
> table "projects":
> id  title     more stuff (pretend there's 20 more columns.)
> -----------------------------------------------------------
> 1   buildrome     moredata       inothercolumns
> 2   housework   evenmoredata letssay20columns
>
> table "todos":
> id projectid name     duedate
> -----------------------------------------
> 1  1         conquer    1pm
> 2  1         laybricks  10pm
> 3  2         dolaundry  5pm
>
>
> In english, I want to "select projects and order them by the ones that
> have todos due the soonest."  Does that sound like a reasonable
> request?

SELECT *
  FROM (
        SELECT DISTINCT ON (projects.id) projects.*
          FROM projects
          JOIN todos ON (todos.projectid = projects.id)
         ORDER BY projects.id, projects.duedate ASC
        )
 ORDER BY duedate ASC
OFFSET 10
 LIMIT 20

> Option E: I could use a subselect.  But notice my offset, limit.  If I
> use a subselect, then postgresql would have to build ALL of the
> results in memory (to create the subselect virtual table), before I
> apply the offset and limit on the subselect.

Don't assume Postgres has to do things a particular way just because there's a
subselect involved. In this case however I expect Postgres would have to build
the results in memory, but not because of the subselect, just because that's
the only way to do what you're asking.

You're asking for it to pick out distinct values according to one sort key
then return the results sorted according to another key. Even if you had an
index for the first key or Postgres used a hash to perform the distinct, the
ORDER BY will require a sort.

Another way to do it would be:

SELECT *,
       (select min(duedate) from todos where projectid = projects.id) as duedate
  FROM projects
 ORDER BY duedate
OFFSET 10
 LIMIT 20


> Any suggestion would be appreciated.
>
> BTW for those of you who are curious, in mysql (that other db), this
> would be:
>
> select a.* from projects a inner join todos b on b.projectid = a.id
> group by a.id order by b.duedate limit 10,20;

And what does the plan look like?


--
  Gregory Stark
  EnterpriseDB          http://www.enterprisedb.com
  Ask me about EnterpriseDB's PostGIS support!

Re: A counter argument about DISTINCT and GROUP BY in PostgreSQL

От
Alvaro Herrera
Дата:
dterrors@hotmail.com wrote:
>
> I've just spent a few hours searching and reading about the postgres
> way of selecting distinct records.  I understand the points made about
> the ORDER BY limitation of DISTINCT ON, and the abiguity of GROUP BY,
> but I think there's a (simple, common) case that have been missed in
> the discussion. Here is my sitation:

Did you try putting the DISTINCT ON in a subselect, and the ORDER BY in
an outer select?

--
Alvaro Herrera                                http://www.CommandPrompt.com/
The PostgreSQL Company - Command Prompt, Inc.

Re: A counter argument about DISTINCT and GROUP BY in PostgreSQL

От
dterrors@hotmail.com
Дата:
On Jan 4, 11:48 am, st...@enterprisedb.com (Gregory Stark) wrote:
> <dterr...@hotmail.com> writes:
> > I've just spent a few hours searching and reading about the postgres
> > way of selecting distinct records.  I understand the points made about
> > the ORDER BY limitation of DISTINCT ON, and the abiguity of GROUP BY,
> > but I think there's a (simple, common) case that have been missed in
> > the discussion. Here is my sitation:
>
> > table "projects":
> > id  title     more stuff (pretend there's 20 more columns.)
> > -----------------------------------------------------------
> > 1   buildrome     moredata       inothercolumns
> > 2   housework   evenmoredata letssay20columns
>
> > table "todos":
> > id projectid name     duedate
> > -----------------------------------------
> > 1  1         conquer    1pm
> > 2  1         laybricks  10pm
> > 3  2         dolaundry  5pm
>
> > In english, I want to "select projects and order them by the ones that
> > have todos due the soonest."  Does that sound like a reasonable
> > request?
>
> SELECT *
>   FROM (
>         SELECT DISTINCT ON (projects.id) projects.*
>           FROM projects
>           JOIN todos ON (todos.projectid = projects.id)
>          ORDER BY projects.id, projects.duedate ASC
>         )
>  ORDER BY duedate ASC
> OFFSET 10
>  LIMIT 20
>
> > Option E: I could use a subselect.  But notice my offset, limit.  If I
> > use a subselect, then postgresql would have to build ALL of the
> > results in memory (to create the subselect virtual table), before I
> > apply the offset and limit on the subselect.
>
> Don't assume Postgres has to do things a particular way just because there's a
> subselect involved. In this case however I expect Postgres would have to build
> the results in memory, but not because of the subselect, just because that's
> the only way to do what you're asking.

When you say it would build the results in memory, do you mean the
entire subselected table?  The subselect in your example doesn't do
any offset, limit. (And, do you think what I'm asking for is odd or
unusual? I can think of a hundred examples besides a todo list.)

> You're asking for it to pick out distinct values according to one sort key
> then return the results sorted according to another key. Even if you had an
> index for the first key or Postgres used a hash to perform the distinct, the
> ORDER BY will require a sort.

I'm not trying to avoid doing a sort, actually.

> > Any suggestion would be appreciated.
>
> > BTW for those of you who are curious, in mysql (that other db), this
> > would be:
>
> > select a.* from projects a inner join todos b on b.projectid = a.id
> > group by a.id order by b.duedate limit 10,20;
>
> And what does the plan look like?

It looks great in mysql!   Works perfectly- that was from my old mysql
code before I switched, or well tried to switch to postgres. I get:

id  title     more stuff....
 -----------------------------------------------------------
1   buildrome     moredata       inothercolumns
2   housework   evenmoredata letssay20columns


Re: A counter argument about DISTINCT and GROUP BY in PostgreSQL

От
Gregory Stark
Дата:
<dterrors@hotmail.com> writes:

> On Jan 4, 11:48 am, st...@enterprisedb.com (Gregory Stark) wrote:
>
>> And what does the plan look like?
>
> It looks great in mysql!

Like what?

--
  Gregory Stark
  EnterpriseDB          http://www.enterprisedb.com
  Ask me about EnterpriseDB's Slony Replication support!