From: Craig Ringer <craig@postnewspapers.com.au>
Cc: Allan Kamau <allank@sanbi.ac.za>
Cc: pgsql-sql@postgresql.org
Subject: Re: Removing redundant itemsets
Date: Mon, 31 Mar 2008 19:48:34 +0800
Message-ID: <47F0CF92.5080900@postnewspapers.com.au> (raw)
In-Reply-To: <47F0CB12.5010506@postnewspapers.com.au>
References: <47F09DD1.1080905@sanbi.ac.za>
<47F0C2A8.9010008@postnewspapers.com.au>
<47F0CB12.5010506@postnewspapers.com.au>
> -- Find any `a' for which `item_from_a_is_in_b' is
> -- true for all items in `a'
> SELECT a_tid AS is_redundant, b_tid AS contained_by
> FROM (
> -- For every item in every pair of purchases,
> -- determine whether the item in purchase `a'
> -- was also in purchase `b'.
> SELECT
> a.tid AS a_tid,
> b.tid AS b_tid,
> a.item AS item,
> EXISTS(
> -- Was this item from `a' also in the `b' purchase?
> SELECT 1 FROM togo x WHERE x.tid = b.tid AND x.item = a.item
> ) AS item_from_a_is_in_b
> FROM togo a INNER JOIN togo b ON (a.tid <> b.tid)
> GROUP BY a.tid, b.tid, a.item) AS item_containment
> GROUP BY a_tid, b_tid
> HAVING every(item_from_a_is_in_b);
That really should've been written as:
SELECT
a.tid AS is_redundant,
b.tid AS contained_by
FROM togo a INNER JOIN togo b ON (a.tid <> b.tid)
GROUP BY a.tid, b.tid
HAVING
EVERY(EXISTS(
SELECT 1 FROM togo x WHERE x.tid = b.tid AND x.item = a.item
));
... but I'm a bit of an idiot, and couldn't figure out why the
EVERY(EXISTS(subq)) wasn't working when testing it before.
Sorry for all the noise.
--
Craig Ringer
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To: pgsql-sql@postgresql.org
Cc: craig@postnewspapers.com.au, allank@sanbi.ac.za
Subject: Re: Removing redundant itemsets
In-Reply-To: <47F0CF92.5080900@postnewspapers.com.au>
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