SELECT syntax question - combining COUNT and DISTINCT

On 1 Oct 2003 at 17:26, Cath Lawrence wrote:

Does this help?

select count(distinct pdb_id),chainid from chain group by chain_id;

Try indexing the column in the where clause

This should give you distinct values counts:

select pdb_id,count(*)
from chain
where chain_id = 'A'
group by pdb_id;

zhix.

This should do:

select pdb_id,count(*)
from chain
where chain_id = 'A'
group by pdb_id;

rgds,
zhix

http://www.postgresql.org/docs/7.3/static/tutorial-agg.html

:)

Louise

-----Original Message-----
From: [email protected]
On Behalf Of Cath Lawrence
Sent: Wednesday, October 01, 2003 1:27 AM
To: [email protected]
Subject: [NOVICE] SELECT syntax question - combining COUNT and DISTINCT

I'm having some trouble working with aggregates. I think I'm missing
the concept of how aggregates are treated. Is there a tutorial page?

Here's my problem, no doubt very simple. It's one table only, nice and
easy...
SELECT DISTINCT pdb_id FROM chain WHERE chain_id = 'A';

But now I want to count how many records are returned and I can't work
it out. I'm sure I'm missing something obvious.
For instance "SELECT COUNT(pdb_id) FROM chain WHERE chain_id = 'A';"
works fine but of course is no longer distinct.


A secondary question (relating to my quality control problem but not
the subject heading) is why I can't do this:
"SELECT DISTINCT chain_id FROM chain WHERE pdb_id IN (SELECT pdb_code
FROM pdb_entry WHERE chain_count=1);"
This looks OK but takes apparently forever (or until I lose patience)
on my Mac G4. There are about 10,000 values in that subquery, so I
didn't expect it to be lightning fast, but it seems ridiculous.


Relevant snippets of table:
pdb_entry - 19,587 entries
---------
pdb_code is KEY, CHAR(4), unique, indexed
chain_count is INTEGER

chain - 40,844 entries
-----
pdb_id is CHAR(4) Foreign key, indexed
chain_id is CHAR(1)



Cath Lawrence, [email protected]
Senior Scientific Programmer, Centre for Bioinformation Science,
John Curtin School of Medical Research (room 4088)
Australian National University, Canberra ACT 0200
ph: (02) 61257959 mobile: 0421-902694 fax: (02) 61252595

SELECT DISTINCT c.chain_id
FROM chain AS c, pdb_entry AS p
WHERE c.pdb_id = p.pdb_code
AND p.chain_count = 1;

Servus
Manfred