Re: So, is COUNT(*) fast now?
От | Robert Haas |
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Тема | Re: So, is COUNT(*) fast now? |
Дата | |
Msg-id | CA+TgmoZKCULEZDNPU008rCGtsvTZHC4r0BaykAjH6UrjfTVqVg@mail.gmail.com обсуждение исходный текст |
Ответ на | Re: So, is COUNT(*) fast now? (Kääriäinen Anssi <anssi.kaariainen@thl.fi>) |
Ответы |
Re: So, is COUNT(*) fast now?
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Список | pgsql-hackers |
On Sun, Oct 30, 2011 at 8:02 AM, Kääriäinen Anssi <anssi.kaariainen@thl.fi> wrote: > Table size is around 600MB, index size is around 350MB and VM on-disk > size is 16kB with default fillfactor. With fillfactor = 10, the VM size is 104 > KB, and table size is around 6GB. The index size is the same. What I think you're probably measuring here (oprofile would tell us for sure) is that once the size of the table goes beyond about half a gigabyte, it will have more than one page in the visibility map. The index-only scan code keeps the most recently used visibility map page pinned to save on overhead, but if you're bouncing back and forth between data in the first ~500MB of the table and data in the last ~100MB, each switch will result in dropping the current pin and getting a new one, which figures to be fairly expensive. With the table is only a little over 500GB, you're probably only changing VM pages every couple of tuples, but with a 6GB table just about every tuple will switch to a new VM page. Now, maybe you're right and the CPU caches are the more significant effect. But I wouldn't like to bet on it without seeing how much the drop-and-get-new-pin operations are costing us. -- Robert Haas EnterpriseDB: http://www.enterprisedb.com The Enterprise PostgreSQL Company
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