Thank you for improving this optimization !
Le mardi 21 février 2023, 04:14:02 CET David Rowley a écrit :
> I still need to look to see if there's some small amount of data that
> can be loaded into the table to help coax the planner into producing
> the ordered scan for this one. It works fine as-is for ORDER BY a,b
> and ORDER BY a; so I've put tests in for that.
I haven't looked too deeply into it, but it seems reasonable that the whole
sort would cost cheaper than individual sorts on partitions + incremental
sorts, except when the the whole sort would spill to disk much more than the
incremental ones. I find it quite difficult to reason about what that threshold
should be, but I managed to find a case which could fit in a test:
create table range_parted (a int, b int, c int) partition by range(a, b);
create table range_parted1 partition of range_parted for values from (0,0) to
(10,10);
create table range_parted2 partition of range_parted for values from (10,10)
to (20,20);
insert into range_parted(a, b, c) select i, j, k from generate_series(1, 19)
i, generate_series(1, 19) j, generate_series(1, 5) k;
analyze range_parted;
set random_page_cost = 10;
set work_mem = '64kB';
explain (costs off) select * from range_parted order by a,b,c;
It's quite convoluted, because it needs the following:
- estimate the individual partition sorts to fit into work_mem (even if that's
not the case here at runtime)
- estimate the whole table sort to not fit into work_mem
- the difference between the two should be big enough to compensate the
incremental sort penalty (hence raising random_page_cost).
This is completely tangential to the subject at hand, but maybe we have
improvements to do with the way we estimate what type of sort will be
performed ? It seems to underestimate the memory amount needed. I'm not sure
it makes a real difference in real use cases though.
Regards,
--
Ronan Dunklau
