The following bug has been logged on the website:
Bug reference: 12228
Logged by: Jonathon Lachlan-Hache
Email address: jonlachlan@gmail.com
PostgreSQL version: 9.4beta2
Operating system: MacOSX
Description:
I'm using the 9.4 RC1 and trying out the WITHIN GROUP functions. I used a
percentile_disc pattern successfully, however I think I came across a
problem with using percent_rank(). Here is my SQL:
SELECT percent_rank(datavalue) WITHIN GROUP (ORDER BY datavalue) as
pct_rank, measureid
FROM measuredata
WHERE surveyyear=2013
GROUP BY measureid;
I want to be able to run a query that programmatically displays the
'pct_rank' of datavalue. My table is organized into four columns:
organizationid
measureid
surveyyear
datavalue
The 'datavalue' column is the measure, whereas all the other columns are
attributes. When I run percent_rank() it needs to determine what percentile
is the 'datavalue', but within the scope of the same 'measureid' and
'surveyyear'.
The percent_rank() function does not appear to support this. The above query
returns the following:
ERROR: column "measuredata.datavalue" must appear in the GROUP BY clause or
be used in an aggregate function
LINE 1: SELECT percent_rank(datavalue) WITHIN GROUP (ORDER BY datava...
^
DETAIL: Direct arguments of an ordered-set aggregate must use only grouped
columns.
********** Error **********
ERROR: column "measuredata.datavalue" must appear in the GROUP BY clause or
be used in an aggregate function
SQL state: 42803
Detail: Direct arguments of an ordered-set aggregate must use only grouped
columns.
Character: 21
This doesn't make sense, because if I include 'datavalue' as a GROUP BY
clause, then my group N-size is exactly 1.
Basically, because the argument of percent_rank() must also be present in
the GROUP BY clause, there is no way to use percent_rank() to
programatically determine the percent-rank of a value within a set. And this
is likely to be the primary use-case of any kind of percent-rank function,
so it would make sense to include it in the new implementation.
Thanks so much,
Jon
