Re: Underestimated number of output rows with an aggregate function

From: Philippe BEAUDOIN <phb(dot)emaj(at)free(dot)fr>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: "pgsql-performance(at)postgresql(dot)org" <pgsql-performance(at)postgresql(dot)org>
Subject: Re: Underestimated number of output rows with an aggregate function
Date: 2023-10-16 16:52:11
Message-ID: 84cc7977-49ba-49b2-ab36-dea0ed14ebdc@free.fr
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Le 15/10/2023 à 18:37, Tom Lane a écrit :
> Philippe BEAUDOIN<phb(dot)emaj(at)free(dot)fr> writes:
>> During my analysis, I realized that the output rows estimate of the
>> second CTE is really bad, leading to a bad plan for the next CTE.
>> I reproduced the issue in a very small test case with a simplified
>> query. Attached is a shell script and its output.
> Yeah. If you try it you'll see that the estimates for the
> "keys.c1 = tbl.c1" and "keys.seq = tbl.seq" clauses are spot-on
> individually. The problem is that the planner assumes that they
> are independent clauses, so it multiplies those selectivities together.
> In reality, because seq is already unique, the condition on c1 adds
> no additional selectivity.
>
> If seq is guaranteed unique in your real application, you could just
> drop the condition on c1. Otherwise I'm not sure about a good
> answer. In principle creating extended stats on c1 and seq should
> help, but I think we don't yet apply those for join clauses.
>
> A partial answer could be to defeat application of the table's
> statistics by writing
>
> JOIN keys ON (keys.c1 = tbl.c1+0 AND keys.seq = tbl.seq+0)
>
> For me this gives an output estimate of 3000 rows, which is still not
> great but should at least prevent choice of an insane plan at the
> next join level. However, it pessimizes the plan for this query
> itself a little bit (about doubling the runtime).

Thanks for the trick (and the quick answer). In the test case, it
effectively brings a pretty good plan.

Unfortunately, as these statements are generated and depend on the base
table structure, the issue remains for some of them (but not all). So,
for the moment at least, I keep the previous workaround (disabling
nested loops).

>> For the hash join node, the plan shows a "Inner Unique: true" property.
>> I wonder if this is normal.
> Sure. The output of the WITH is visibly unique on c1.
OK, I see.
> regards, tom lane

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