Thread: Difference in columns
Hi All,<br /><br />I have a view that generates output similar to this.<br /><br />select * from foo.view;<br /><br /><preclass="PROGRAMLISTING"> ts | size<br />-------------------+-----<br /> 2002-03-16 | 11 <br />2002-03-17 | 16 <br />2002-03-18 | 18<br /> 2002-03-19 | 12 <br /></pre>I am trying to find the difference between the size column. So the desired output would be<br /><br /><pre class="PROGRAMLISTING"> ts | size| Diff<br />-------------------+-----+------<br /> 2002-03-16 | 11 | 0<br />2002-03-17 | 15 | 4<br /> 2002-03-18 | 18 | 3<br /> 2002-03-19 | 12 | -6<br /></pre><br/>I need the first column to be 0, since it will be 11-11. The second colum is 15-11. The third column is 18-15.The fourth column is 12-18. <br /><br />Any thoughts about this?<br /><br />TIA<br />
--- Mag Gam <magawake@gmail.com> wrote: > Hi All, > > I have a view that generates output similar to this. > > select * from foo.view; > > ts | size > -------------------+----- > 2002-03-16 | 11 > 2002-03-17 | 16 > 2002-03-18 | 18 > 2002-03-19 | 12 > > I am trying to find the difference between the size column. So the > desired > output would be > > ts | size| Diff > -------------------+-----+------ > 2002-03-16 | 11 | 0 > 2002-03-17 | 15 | 4 > 2002-03-18 | 18 | 3 > 2002-03-19 | 12 | -6 > > > I need the first column to be 0, since it will be 11-11. The second > colum is > 15-11. The third column is 18-15. The fourth column is 12-18. > > Any thoughts about this? > select cur.ts, cur.size, cur.size - coalesce( (select size from view next where next.ts = cur.ts - '1 day'::interval), cur.size ) as diff from view cur; alternately: select cur.ts, cur.size, case when cur.ts = '2002-03-16' then 0 else cur.size - coalesce( (select size from view next where next.ts = cur.ts - '1 day'::interval), cur.size ) end as diff from view; ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ
On Sun, May 11, 2008 at 11:07 PM, Mag Gam <magawake@gmail.com> wrote:
Try this:
select ts,
size,
t1.size - (select t2.size
from foo.view as t2
where t2.ts < t1.ts
order by ts desc
limit 1) as diff
from foo.view as t1
order by ts asc;
HTH,
--
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Hi All,
I have a view that generates output similar to this.
select * from foo.view;ts | sizeI am trying to find the difference between the size column. So the desired output would be
-------------------+-----
2002-03-16 | 11
2002-03-17 | 16
2002-03-18 | 18
2002-03-19 | 12ts | size| Diff
-------------------+-----+------
2002-03-16 | 11 | 0
2002-03-17 | 15 | 4
2002-03-18 | 18 | 3
2002-03-19 | 12 | -6
I need the first column to be 0, since it will be 11-11. The second colum is 15-11. The third column is 18-15. The fourth column is 12-18.
Any thoughts about this?
Try this:
select ts,
size,
t1.size - (select t2.size
from foo.view as t2
where t2.ts < t1.ts
order by ts desc
limit 1) as diff
from foo.view as t1
order by ts asc;
HTH,
--
gurjeet[.singh]@EnterpriseDB.com
singh.gurjeet@{ gmail | hotmail | indiatimes | yahoo }.com
EnterpriseDB http://www.enterprisedb.com
Mail sent from my BlackLaptop device
On Sun, May 11, 2008 at 01:37:52PM -0400, Mag Gam wrote: > Any thoughts about this? 1. will there be any gaps in between dates? if yes, what should be diff be then? 2. can't you calculate it in client application? 3. is usage of pl/pgsql acceptable (i think this will be the fastest way to do it in postgresql itself). depesz -- quicksil1er: "postgres is excellent, but like any DB it requires a highly paid DBA. here's my CV!" :) http://www.depesz.com/ - blog dla ciebie (i moje CV)
Mag Gam wrote: > > ts | size| Diff > -------------------+-----+------ > 2002-03-16 | 11 | 0 > > 2002-03-17 | 15 | 4 > 2002-03-18 | 18 | 3 > 2002-03-19 | 12 | -6 > > > I need the first column to be 0, since it will be 11-11. The second > colum is 15-11. The third column is 18-15. The fourth column is 12-18. > > Any thoughts about this? Without making any comments on the advisability of the structure you're trying to use, here are a few ideas. The easy way is to use PL/PgSQL and FOR EACH .. SELECT . It's probably going to be rather fast too as it can use a single sequential scan. Otherwise (all examples use the following code): CREATE TABLE x (ts timestamp, size int); INSERT INTO x (ts, size) VALUES ('2002-03-16',11), ('2002-03-17',15), ('2002-03-18',18), ('2002-03-19',12); If you can assume that there is always exactly 1 day between entries then it's easy enough with a self join. If you cannot assume that, you can use a subquery with limit and order by to obtain the next record: SELECT a.ts, (SELECT b.size FROM x b WHERE b.ts > a.ts ORDER BY b.ts ASC LIMIT 1) - a.size AS difference FROM x a; ... but that'll be really slow for any significant number of entries. Another alternative if you can't assume each record is always exactly 1 day apart is to populate a temp table with the values and add a serial column that guarantees a 1 offset between values, then do a self join. I have no idea whether or not this might be faster, but thought I'd throw it out there as an alternative: CREATE TEMPORARY SEQUENCE x_seq; SELECT nextval('x_seq') AS id, ts, size INTO TEMPORARY TABLE x_temp FROM x ORDER BY ts ASC; SELECT a.ts, a.size - b.size AS diff FROM x_temp a, x_temp b WHERE a.id = b.id + 1; Note that this query doesn't give you the first record with zero difference; it returns only true differences. Here's one possible way to add your initial record: SELECT a.ts, b.size - a.size AS diff FROM x_temp a, x_temp b WHERE b.id = a.id + 1 OR (b.id = (SELECT min(id) FROM x_temp) AND a.id = b.id); -- Craig Ringer
On Sun, May 11, 2008 at 11:47 PM, Craig Ringer <craig@postnewspapers.com.au> wrote:
not really... if you have an index on the TS column.
Best regards,
--
gurjeet[.singh]@EnterpriseDB.com
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Mag Gam wrote:Without making any comments on the advisability of the structure you're
>
> ts | size| Diff
> -------------------+-----+------
> 2002-03-16 | 11 | 0
>
> 2002-03-17 | 15 | 4
> 2002-03-18 | 18 | 3
> 2002-03-19 | 12 | -6
>
>
> I need the first column to be 0, since it will be 11-11. The second
> colum is 15-11. The third column is 18-15. The fourth column is 12-18.
>
> Any thoughts about this?
trying to use, here are a few ideas.
The easy way is to use PL/PgSQL and FOR EACH .. SELECT . It's probably
going to be rather fast too as it can use a single sequential scan.
Otherwise (all examples use the following code):
CREATE TABLE x (ts timestamp, size int);
INSERT INTO x (ts, size) VALUES
('2002-03-16',11),
('2002-03-17',15),
('2002-03-18',18),
('2002-03-19',12);
If you can assume that there is always exactly 1 day between entries
then it's easy enough with a self join.
If you cannot assume that, you can use a subquery with limit and order
by to obtain the next record:
SELECT
a.ts,
(SELECT b.size FROM x b WHERE b.ts > a.ts ORDER BY b.ts ASC LIMIT 1)
- a.size AS difference
FROM x a;
... but that'll be really slow for any significant number of entries.
not really... if you have an index on the TS column.
Best regards,
--
gurjeet[.singh]@EnterpriseDB.com
singh.gurjeet@{ gmail | hotmail | indiatimes | yahoo }.com
EnterpriseDB http://www.enterprisedb.com
Mail sent from my BlackLaptop device
"Gurjeet Singh" <singh.gurjeet@gmail.com> writes: > On Sun, May 11, 2008 at 11:47 PM, Craig Ringer <craig@postnewspapers.com.au> >> If you cannot assume that, you can use a subquery with limit and order >> by to obtain the next record: >> >> SELECT >> a.ts, >> (SELECT b.size FROM x b WHERE b.ts > a.ts ORDER BY b.ts ASC LIMIT 1) >> - a.size AS difference >> FROM x a; >> >> ... but that'll be really slow for any significant number of entries. > not really... if you have an index on the TS column. The OP said this was a view, so it may well not have any easy way to provide such an index. regards, tom lane
Mag Gam wrote: > I am trying to find the difference between the size column. So the > desired output would be > > ts | size| Diff > -------------------+-----+------ > 2002-03-16 | 11 | 0 > > 2002-03-17 | 15 | 4 > 2002-03-18 | 18 | 3 > 2002-03-19 | 12 | -6 > > > I need the first column to be 0, since it will be 11-11. The second > colum is 15-11. The third column is 18-15. The fourth column is 12-18. > > Any thoughts about this? Here's one way to do this with PL/PgSQL. It's probably not the most efficient, but it does work. For this code to be safe `size' must never be NULL and `ts' must be unique across all records in the input set. CREATE OR REPLACE FUNCTION x_diff( OUT ts TIMESTAMP, OUT size INTEGER, OUT diff INTEGER) RETURNS SETOF record AS $$ DECLARE cur_x x; last_size INTEGER := null; BEGIN FOR cur_x IN SELECT * FROM x ORDER BY ts ASC LOOP ts := cur_x.ts; size := cur_x.size; IF last_sizeIS NULL THEN -- First record in set has diff `0' because the differences -- are defined againstthe previous, rather than next, -- record. diff := 0; ELSE diff := cur_x.size- last_size; END IF; last_size := cur_x.size; RETURN NEXT; END LOOP; RETURN; END; $$ LANGUAGE 'plpgsql' STRICT; If you need to constrain the range of values processed that's not too tricky - either feed the function a refcursor for a query result set to iterate over, or pass it parameters to constrain the query with a WHERE clause. The former is more flexible, the latter is easier to use. -- Craig Ringer
Thanks all. I will try some of these suggestions. <br /><br /><br /><div class="gmail_quote">On Sun, May 11, 2008 at 3:58PM, Craig Ringer <<a href="mailto:craig@postnewspapers.com.au">craig@postnewspapers.com.au</a>> wrote:<br /><blockquoteclass="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left:1ex;"><div class="Ih2E3d">Mag Gam wrote:<br /><br /> > I am trying to find the difference between the sizecolumn. So the<br /> > desired output would be<br /> ><br /> > ts | size| Diff<br /> > -------------------+-----+------<br/> > 2002-03-16 | 11 | 0<br /> ><br /> > 2002-03-17 | 15 |4<br /> > 2002-03-18 | 18 | 3<br /> > 2002-03-19 | 12 | -6<br /> ><br /> ><br /> > I needthe first column to be 0, since it will be 11-11. The second<br /> > colum is 15-11. The third column is 18-15. Thefourth column is 12-18.<br /> ><br /> > Any thoughts about this?<br /><br /></div>Here's one way to do this withPL/PgSQL. It's probably not the most<br /> efficient, but it does work. For this code to be safe `size' must never<br/> be NULL and `ts' must be unique across all records in the input set.<br /><br /> CREATE OR REPLACE FUNCTION x_diff(<br/> OUT ts TIMESTAMP,<br /> OUT size INTEGER,<br /> OUT diff INTEGER)<br /> RETURNS SETOF record AS $$<br/> DECLARE<br /> cur_x x;<br /> last_size INTEGER := null;<br /> BEGIN<br /> FOR cur_x IN SELECT * FROM x ORDERBY ts ASC LOOP<br /> ts := cur_x.ts;<br /> size := cur_x.size;<br /> IF last_size IS NULL THEN<br/> -- First record in set has diff `0' because the differences<br /> -- are defined againstthe previous, rather than next,<br /> -- record.<br /> diff := 0;<br /> ELSE<br /> diff := cur_x.size - last_size;<br /> END IF;<br /> last_size := cur_x.size;<br /> RETURNNEXT;<br /> END LOOP;<br /> RETURN;<br /> END;<br /> $$ LANGUAGE 'plpgsql' STRICT;<br /><br /> If you need toconstrain the range of values processed that's not too<br /> tricky - either feed the function a refcursor for a queryresult set to<br /> iterate over, or pass it parameters to constrain the query with a WHERE<br /> clause. The formeris more flexible, the latter is easier to use.<br /><br /> --<br /><font color="#888888">Craig Ringer<br /></font></blockquote></div><br/>