From: Neil Saunders <n.j.saunders@gmail.com>
To: pgsql-sql@postgresql.org
Subject: Difference from average
Date: Tue, 11 Oct 2005 14:20:16 +0100
Message-ID: <ddcd549e0510110620v601d5d5es6c88e87e609d1a73@mail.gmail.com> (raw)
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;
EXPLAIN reveals (albeit not a real test, as only the two rows above)
Seq Scan on prices (cost=1.03..2.05 rows=2 width=32)
InitPlan
-> Aggregate (cost=1.03..1.03 rows=1 width=8)
-> Seq Scan on prices (cost=0.00..1.02 rows=2 width=8)
Does this mean that I'll be performing a nested table scan every time
I run this query? Also, I haven't yet calculated the percentage
difference for this, which in my eyes means another instance of
"SELECT avg(price) from prices". Is this the best way of doing this?
Can I optimize this away by re-writing this as a function and storing
"SELECT avg(price) from prices)" in a variable?
All opinions gratefully received.
Kind Regards,
Neil
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Cc: n.j.saunders@gmail.com
Subject: Re: Difference from average
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