Re: Difference from average
От | Richard Huxton |
---|---|
Тема | Re: Difference from average |
Дата | |
Msg-id | 434BCB1F.7040403@archonet.com обсуждение исходный текст |
Ответ на | Difference from average (Neil Saunders <n.j.saunders@gmail.com>) |
Список | pgsql-sql |
Neil Saunders wrote: > Hi all, > > I'm developing a property rental database. One of the tables tracks > the price per week for different properties: > > CREATE TABLE "public"."prices" ( > "id" SERIAL, > "property_id" INTEGER, > "start_date" TIMESTAMP WITHOUT TIME ZONE, > "end_date" TIMESTAMP WITHOUT TIME ZONE, > "price" DOUBLE PRECISION NOT NULL > ) WITH OIDS; > > CREATE INDEX "prices_idx" ON "public"."prices" > USING btree ("property_id"); > > I'd like to display the prices per property in a table, with each row > coloured different shades; darker shades representing the more > expensive periods for that property. To do this, I propose to > calculate the percentage difference of each rows price from the > average for that property, so if for example I have two rows, one for > price=200 and one for price=300, i'd like to retrieve both records > along with the calculated field indicating that the rows are -20%, > +20% from the average, respectively. > > I've started with the following query, but since I'm still learning > how PostgreSQL works, I'm confused as to the efficiency of the > following statement: > > SELECT *, (price - (SELECT avg(price) from prices)) as diff FROM prices; I'd personally write it something like: SELECT prices.property_id, prices.price AS actual_price, averages.avg_price, (averages.avg_price - prices.price) AS price_diff ((averages.avg_price - prices.price)/averages.avg_price) AS pc_diff FROM prices, (SELECT property_id, avg(price) as avg_price FROM prices) AS averages WHERE prices.property_id = averages.property_id ; That's as much to do with how I think about the problem as to any testing though. -- Richard Huxton Archonet Ltd
В списке pgsql-sql по дате отправления: