Message-ID: <4A4D7F9B.2020107@free.fr>
Date: Fri, 03 Jul 2009 05:48:43 +0200
From: nha <lyondif02@free.fr>
Organization: Lyon, Rhone, France
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To: Jasmin Dizdarevic <Jasmin.Dizdarevic@gmail.com>
CC: pgsql-sql@postgresql.org
Subject: Re: Create custom aggregate function and custom sfunc
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	<407d949e0907021421qb1df8b1o9468f343298f08dd@mail.gmail.com>
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Hello,

Le 2/07/09 23:21, Greg Stark a écrit :
> On Thu, Jul 2, 2009 at 3:48 PM, Jasmin
> Dizdarevic<jasmin.dizdarevic@gmail.com>  wrote:
>> customer ; seg
>> 111 ; L1
>> 111 ; L2
>> 111 ; L1
>> 222 ; L3
>> 222 ; L3
>> 222 ; L2
>>
>> the result should look like this:
>>
>> 111: L1 - because L1 is higher than L2 and ratio of L1 : L2 is 2 : 1
>> 222: L3 - because L3 is higher than L2 and ratio of L3 : L2 is 2 : 1
>
> Are you just looking for the most frequent seg for each customer?
>
> select distinct on (customer) customer,seg
>     from (select customer, seg, count(*) as n from tab group by seg)
>    order by customer, n desc
>
> That doesn't give the ratios though.
>

Some errors would occur with the above query from within the subquery 
because of a projection on customer column that is not a grouped column 
(neither an uniquely identified column by seg--seg is assumed not to be 
a primary key here according to the given examples of value). By the 
way, the ordered column n is not a member of the projected columns.

Nevertheless, in the same direction as Greg Stark, the following query 
would approach the target result:

SELECT T5A.customer, T5A.seg
FROM (
	SELECT T2.customer, T2.seg, COUNT(*) AS nb
	FROM cst T2
	GROUP BY T2.customer, T2.seg
) T5A INNER JOIN (
	SELECT T4.customer, MAX(T4.nb) AS maxNb
	FROM (
		SELECT T2B.customer, T2B.seg, COUNT(*) AS nb
		FROM cst T2B
		GROUP BY T2B.customer, T2B.seg
	) T4
	GROUP BY T4.customer
) T5B
ON T5A.customer = T5B.customer
AND T5A.nb = T5B.maxNb

where 'cst' is the reference table including 'customer' and 'seg' 
columns. This query considers a join between two tables:
- the first table, aliased T5A, counts the multiplicity for each couple 
of 'customer' and 'seg';
- the secund table, aliased T5B, reveals the highest multiplicity for 
each 'customer' with regard to each 'seg' attached to customer.
Then the join only retains the couples of 'customer' and 'seg' whom 
multiplicity equals the higher for the 'customer' of the current couple.

Each multiplicity may be easily added to the resulting records by 
spanning the projected columns with T5A.nb column. A sort of ratio may 
also be added; one aggregated column may first be added to table T4 as 
SUM(T4.nb) -- that is, sum of multiplicity for each customer; then the 
resulting join may compute the ratio between the highest multiplicity 
and the sum of multiplicity for each customer, in addition to the seg 
(or list of seg) for which the multiplicity is the highest.

Here is the modified query:

SELECT T5A.customer, T5A.seg, T5A.nb/T5B.sumNb
FROM (
	SELECT T2.customer, T2.seg, COUNT(*) AS nb
	FROM cst T2
	GROUP BY T2.customer, T2.seg
) T5A INNER JOIN (
	SELECT T4.customer, MAX(T4.nb) AS maxNb, SUM(T4.nb) AS sumNb
	FROM (
		SELECT T2B.customer, T2B.seg, COUNT(*) AS nb
		FROM cst T2B
		GROUP BY T2B.customer, T2B.seg
	) T4
	GROUP BY T4.customer
) T5B
ON T5A.customer = T5B.customer
AND T5A.nb = T5B.maxNb

As is, this query may result to multiple couples of customer and seg 
with the same customer value when many seg relie as many times as many 
others for the current customer. This "edge effect" may be avoided in 
many ways depending on the original purpose.

Regards.

--
nha / Lyon / France.