olympics ranking query

See http://www.athens2004.com/en/OlympicMedals/medals?noc=MGL .

create table countrymedal (
countryid CHAR(3) PRIMARY KEY,
gold INT NOT NULL,
silver INT NOT NULL,
bronze INT NOT NULL);

COPY countrymedal (countryid, gold, silver, bronze) FROM stdin;
ITA 5 6 3
FRA 5 3 5
UKR 5 1 1
RUS 4 8 10
GER 4 4 7
TUR 3 0 1
KOR 2 7 3
NED 2 5 4
HUN 2 3 1
SVK 2 2 1
ROM 2 0 2
GRE 2 0 1
POL 1 2 1
BLR 1 1 2
SUI 1 0 1
UAE 1 0 0
GBR 0 4 4
AUT 0 3 0
PRK 0 2 1
ESP 0 2 0
CUB 0 1 5
CZE 0 1 2
ZIM 0 1 1
USA 15 11 10
CHN 15 9 8
JPN 9 4 2
AUS 7 5 8
GEO 1 1 0
RSA 1 1 0
BUL 1 0 2
THA 1 0 2
IND 0 1 0
INA 0 1 0
KAZ 0 1 0
POR 0 1 0
SCG 0 1 0
AZE 0 0 2
BEL 0 0 2
BRA 0 0 2
DEN 0 0 2
ARG 0 0 1
CAN 0 0 1
COL 0 0 1
CRO 0 0 1
ISR 0 0 1
MGL 0 0 1
SLO 0 0 1
TRI 0 0 1
\.

create sequence seq1;
create sequence seq2;

-- query #1: list of ranks by gold
select setval('seq1', 1);
select setval('seq2', 1);
select
setval('seq1', currval('seq1')+setval('seq2',count(*)))-count(*) as rank,
count(*) as numranker,
gold, silver, bronze
from countrymedal
group by gold, silver, bronze
order by gold desc, silver desc, bronze desc;

-- result of query #1
rank | numranker | gold | silver | bronze
------+-----------+------+--------+--------
1 | 1 | 15 | 11 | 10
2 | 1 | 15 | 9 | 8
3 | 1 | 9 | 4 | 2
4 | 1 | 7 | 5 | 8
5 | 1 | 5 | 6 | 3
6 | 1 | 5 | 3 | 5
7 | 1 | 5 | 1 | 1
8 | 1 | 4 | 8 | 10
9 | 1 | 4 | 4 | 7
10 | 1 | 3 | 0 | 1
11 | 1 | 2 | 7 | 3
12 | 1 | 2 | 5 | 4
13 | 1 | 2 | 3 | 1
14 | 1 | 2 | 2 | 1
15 | 1 | 2 | 0 | 2
16 | 1 | 2 | 0 | 1
17 | 1 | 1 | 2 | 1
18 | 1 | 1 | 1 | 2
19 | 2 | 1 | 1 | 0
21 | 2 | 1 | 0 | 2
23 | 1 | 1 | 0 | 1
24 | 1 | 1 | 0 | 0
25 | 1 | 0 | 4 | 4
26 | 1 | 0 | 3 | 0
27 | 1 | 0 | 2 | 1
28 | 1 | 0 | 2 | 0
29 | 1 | 0 | 1 | 5
30 | 1 | 0 | 1 | 2
31 | 1 | 0 | 1 | 1
32 | 5 | 0 | 1 | 0
37 | 4 | 0 | 0 | 2
41 | 8 | 0 | 0 | 1
(32 rows)

-- query #2: list of countries ordered by their ranks
select setval('seq1', 1);
select setval('seq2', 1);
select
(case numranker when 1 then '' else '=' end) || rank as rank,
countryid,
cm.gold, cm.silver, cm.bronze
from countrymedal cm
left join
(select
setval('seq1',
currval('seq1')+setval('seq2',count(*))
)-count(*) as rank,
count(*) as numranker,
gold, silver, bronze
from countrymedal
group by gold, silver, bronze
order by gold desc, silver desc, bronze desc
) t1 on cm.gold=t1.gold and cm.silver=t1.silver and cm.bronze=t1.bronze
order by t1.rank;

-- result of query #2
rank | countryid | gold | silver | bronze
------+-----------+------+--------+--------
1 | USA | 15 | 11 | 10
2 | CHN | 15 | 9 | 8
3 | JPN | 9 | 4 | 2
4 | AUS | 7 | 5 | 8
5 | ITA | 5 | 6 | 3
6 | FRA | 5 | 3 | 5
7 | UKR | 5 | 1 | 1
8 | RUS | 4 | 8 | 10
9 | GER | 4 | 4 | 7
10 | TUR | 3 | 0 | 1
11 | KOR | 2 | 7 | 3
12 | NED | 2 | 5 | 4
13 | HUN | 2 | 3 | 1
14 | SVK | 2 | 2 | 1
15 | ROM | 2 | 0 | 2
16 | GRE | 2 | 0 | 1
17 | POL | 1 | 2 | 1
18 | BLR | 1 | 1 | 2
=19 | GEO | 1 | 1 | 0
=19 | RSA | 1 | 1 | 0
=21 | BUL | 1 | 0 | 2
=21 | THA | 1 | 0 | 2
23 | SUI | 1 | 0 | 1
24 | UAE | 1 | 0 | 0
25 | GBR | 0 | 4 | 4
26 | AUT | 0 | 3 | 0
27 | PRK | 0 | 2 | 1
28 | ESP | 0 | 2 | 0
29 | CUB | 0 | 1 | 5
30 | CZE | 0 | 1 | 2
31 | ZIM | 0 | 1 | 1
=32 | IND | 0 | 1 | 0
=32 | INA | 0 | 1 | 0
=32 | KAZ | 0 | 1 | 0
=32 | POR | 0 | 1 | 0
=32 | SCG | 0 | 1 | 0
=37 | AZE | 0 | 0 | 2
=37 | BEL | 0 | 0 | 2
=37 | BRA | 0 | 0 | 2
=37 | DEN | 0 | 0 | 2
=41 | ARG | 0 | 0 | 1
=41 | CAN | 0 | 0 | 1
=41 | COL | 0 | 0 | 1
=41 | CRO | 0 | 0 | 1
=41 | ISR | 0 | 0 | 1
=41 | MGL | 0 | 0 | 1
=41 | SLO | 0 | 0 | 1
=41 | TRI | 0 | 0 | 1


Challenge question: is there a simpler way to do query #1 (without any
PL, and if possible without sequences too?

--
dave


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Re: olympics ranking query

David Garamond writes:
> Challenge question: is there a simpler way to do query #1 (without any
> PL, and if possible without sequences too?

Can't without sequences AFAIK, but you certainly can do it simpler:

select setval('seq1', 0);

select nextval('seq1'), * from
(select count(*) as numranker,
gold, silver, bronze
from countrymedal
group by gold, silver, bronze
order by gold desc, silver desc, bronze desc) ss;


regards, tom lane

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Re: olympics ranking query

On Fri, Aug 20, 2004 at 23:40:08 +0700,
David Garamond wrote:
>
> Challenge question: is there a simpler way to do query #1 (without any
> PL, and if possible without sequences too?

You could use a subselect to count how many countries had a lower
medal ranking and add 1 to get the rank. This should be a lot more standard
than using sequences. It will probably be a little slower, but for tables
of that size it shouldn't be a big deal.

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Re: olympics ranking query

Tom Lane wrote:
>>Challenge question: is there a simpler way to do query #1 (without any
>>PL, and if possible without sequences too?
>
> Can't without sequences AFAIK, but you certainly can do it simpler:
>
> select setval('seq1', 0);
>
> select nextval('seq1'), * from
> (select count(*) as numranker,
> gold, silver, bronze
> from countrymedal
> group by gold, silver, bronze
> order by gold desc, silver desc, bronze desc) ss;

This is not quite the same. The ranks are sequential, but they skip, so
as to match the number of participating countries.

--
dave

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Re: olympics ranking query

David Garamond writes:
> This is not quite the same. The ranks are sequential, but they skip, so
> as to match the number of participating countries.

Oh, I missed that bit.

What you really want here is a "running sum" function, that is

SELECT running_sum(numranker) as rank, * FROM
(same subselect as before) ss;

There is no such thing in standard SQL, because it's fundamentally
dependent on the assumption of the input data coming in a particular
order, which is Evil Incarnate according to the relational worldview.
But it's not too hard to do in PLs that allow persistent state.
I recall Elein having exhibited one in plpython(?) not too long ago
--- you might find it on techdocs or varlena.com.

You could brute-force it with a subselect (essentially "SUM(numranker)
over all rows that should precede this one") but that would involve
recomputing the whole subselect for each output row, which doesn't seem
very attractive.

regards, tom lane

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Re: olympics ranking query

Bruno Wolff III wrote:
> On Fri, Aug 20, 2004 at 23:40:08 +0700,
> David Garamond wrote:
>
>>Challenge question: is there a simpler way to do query #1 (without any
>>PL, and if possible without sequences too?
>
> You could use a subselect to count how many countries had a lower
> medal ranking and add 1 to get the rank. This should be a lot more standard
> than using sequences. It will probably be a little slower, but for tables
> of that size it shouldn't be a big deal.

Thanks for the tip. This is what I came up with:

select
(select count(*) from countrymedal c1 where
c1.gold>c2.gold or
(c1.gold=c2.gold and (c1.silver>c2.silver or
(c1.silver=c2.silver and c1.bronze>c2.bronze))))+1 as rank,
count(*) as numranker,
gold, silver, bronze
from countrymedal c2
group by gold, silver, bronze
order by rank;

--
dave

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Re: olympics ranking query

That 'running aggregate' notion comes up quite regularily,
and it has a number of nonintuitive holes, to do with
what happens when your ordering allows for ties.
ASTRID had it, RedBrick SQL had it, the latest MSSQL
has it ... not necessarily a recommendation.

Tom Lane wrote:
> David Garamond writes:
>
>>This is not quite the same. The ranks are sequential, but they skip, so
>>as to match the number of participating countries.
>
>
> Oh, I missed that bit.
>
> What you really want here is a "running sum" function, that is
>
> SELECT running_sum(numranker) as rank, * FROM
> (same subselect as before) ss;
>
> There is no such thing in standard SQL, because it's fundamentally
> dependent on the assumption of the input data coming in a particular
> order, which is Evil Incarnate according to the relational worldview.
> But it's not too hard to do in PLs that allow persistent state.
> I recall Elein having exhibited one in plpython(?) not too long ago
> --- you might find it on techdocs or varlena.com.
>
> You could brute-force it with a subselect (essentially "SUM(numranker)
> over all rows that should precede this one") but that would involve
> recomputing the whole subselect for each output row, which doesn't seem
> very attractive.
>
> regards, tom lane
>
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>