Hello Rob,
So how do I improve this query speed? Thanks, Arup Rakshit a...@zeit.io > On 14-Sep-2018, at 12:27 AM, Rob Sargent <robjsarg...@gmail.com> wrote: > > > >> On Sep 13, 2018, at 12:17 PM, Arup Rakshit <a...@zeit.io >> <mailto:a...@zeit.io>> wrote: >> >> The below query basically gives the result by maintaining the order of the >> sizes in the list. >> >> explain analyze select >> "price_levels"."name", >> "price_levels"."size" >> from >> "price_levels" >> join unnest(array['M', >> 'L', >> 'XL', >> '2XL', >> '3XL', >> '4XL', >> '5XL', >> '6XL', >> 'S']) with ordinality t(size, >> ord) >> using (size) >> order by >> t.size >> >> >> I have a Btree index on the size column. >> >> Explain output is: >> >> Merge Join (cost=4.61..5165.38 rows=60000 width=46) (actual >> time=0.157..57.872 rows=60000 loops=1) >> Merge Cond: ((price_levels.size)::text = t.size) >> -> Index Scan using price_levels_size_idx on price_levels >> (cost=0.29..4111.05 rows=60000 width=14) (actual time=0.044..25.941 >> rows=60000 loops=1) >> -> Sort (cost=4.32..4.57 rows=100 width=32) (actual time=0.108..3.946 >> rows=53289 loops=1) >> Sort Key: t.size >> Sort Method: quicksort Memory: 25kB >> -> Function Scan on unnest t (cost=0.00..1.00 rows=100 width=32) >> (actual time=0.030..0.033 rows=9 loops=1) >> Planning time: 0.667 ms >> Execution time: 62.846 ms >> >> >> >> Thanks, >> >> Arup Rakshit >> a...@zeit.io <mailto:a...@zeit.io> >> >> > There are not value of size fit it to be a worthwhile key. >> >
