Обсуждение: Underestimated number of output rows with an aggregate function
Hi all, Working on the emaj extension (for the curious ones, https://emaj.readthedocs.io/en/latest/ and https://github.com/dalibo/emaj), I recently faced a performance problem when querying and aggregating data changes. A query with 3 CTE has a O^2 behavior (https://explain.dalibo.com/plan/1ded242d4ebf3gch#plan). I have found a workaround by setting enable_nestloop to FALSE. But this has drawbacks. So I want to better understand the issue. 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. A simple table is created, filled and analyzed. The simplified statement is: WITH keys AS ( SELECT c1, min(seq) AS seq FROM perf GROUP BY c1 ) SELECT tbl.* FROM perf tbl JOIN keys ON (keys.c1 = tbl.c1 AND keys.seq = tbl.seq); Its plan is: Hash Join (cost=958.00..1569.00 rows=1 width=262) (actual time=18.516..30.702 rows=10000 loops=1) Output: tbl.c1, tbl.seq, tbl.c2 Inner Unique: true Hash Cond: ((tbl.c1 = perf.c1) AND (tbl.seq = (min(perf.seq)))) Buffers: shared hit=856 -> Seq Scan on public.perf tbl (cost=0.00..548.00 rows=12000 width=262) (actual time=0.007..2.323 rows=12000 loops=1) Output: tbl.c1, tbl.seq, tbl.c2 Buffers: shared hit=428 -> Hash (cost=808.00..808.00 rows=10000 width=8) (actual time=18.480..18.484 rows=10000 loops=1) Output: perf.c1, (min(perf.seq)) Buckets: 16384 Batches: 1 Memory Usage: 519kB Buffers: shared hit=428 -> HashAggregate (cost=608.00..708.00 rows=10000 width=8) (actual time=10.688..14.321 rows=10000 loops=1) Output: perf.c1, min(perf.seq) Group Key: perf.c1 Batches: 1 Memory Usage: 1425kB Buffers: shared hit=428 -> Seq Scan on public.perf (cost=0.00..548.00 rows=12000 width=8) (actual time=0.002..2.330 rows=12000 loops=1) Output: perf.c1, perf.seq, perf.c2 Buffers: shared hit=428 It globally looks good to me, with 2 sequential scans and a hash join. But the number of returned rows estimate is always 1, while it actually depends on the data content (here 10000). For the hash join node, the plan shows a "Inner Unique: true" property. I wonder if this is normal. It look likes the optimizer doesn't take into account the presence of the GROUP BY clause in its estimate. I reproduce the case with all supported postgres versions. Thanks by advance for any explanation. Philippe.
Вложения
Philippe BEAUDOIN <phb.emaj@free.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). > 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. regards, tom lane
Le 15/10/2023 à 18:37, Tom Lane a écrit :
Philippe BEAUDOIN <phb.emaj@free.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).
OK, I see.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.
regards, tom lane