Re: Help on a complex query (avg data for day of the week)

Richard (and list),

Thanks for the help! More below:

On Wed, 21 Dec 2005 09:04 pm, Richard Huxton wrote:
> Matthew Smith wrote:
> > I want to form a query that returns the average total usage for each day
> > of the week, eg:
>
> [snip]
>
> > To get this info, I am using the following query:
> >
> > select dow as day, sum(sum_data)/count(dow) as avg_usage from
> > (select extract('dow' from date_trunc('day', time)) as dow, sum(data) as
> > sum_data
> > from datalog where time >= '2005-09-11' and time <= '2005-09-25' group by
> > dow) as avg_data_per_day group by day;
> >
> > This works well, assuming that there is at least one entry in the table
> > for each day in the time period.
> >
> > The problem comes when there are days where no data is logged.
>
> 1. Calculate how many data-points each day represents
> 2. Sum the days you do have data for
> 3. Left-join #1 to #2 so you can calculate the average.
>
> I'm guessing for #1 you'd rather not have 7 UNIONs, so you might find
> generate_series() a useful function. See Ch 9.18. Set Returning Functions.

Sadly I am using 7.3, it seems that generate_series() is 8.0 and later. I am
not sure what you meant by "Calculate how many data-points each day
represents", but your mention of unions gave me an idea. It's nasty, but it
will work for what I am trying to do (sorry if it is what you meant):

I can use unions to make a list of dates and left join them into my query to
get the missing days in. This would be the same as left joining in the result
from generate_series(), but it is not as tidy.

I will make the list using something like this:

select '2005-09-11'::timestamp as date UNION select '2005-09-12'::timestamp as
date UNION select '2005-09-13'::timestamp as date UNION etc.

It's nasty, but it will hold until I can migrate to 8.1. Thanks again for the
help!

Cheers,

Matthew Smith