Re: U_SUMMARIZE: how aggregate is calculated?
From: Dan <guntermann_at_verizon.com>
Date: Thu, 02 Dec 2004 08:14:28 GMT
Message-ID: <E9Ard.14224$lv6.6902_at_trnddc03>
Date: Thu, 02 Dec 2004 08:14:28 GMT
Message-ID: <E9Ard.14224$lv6.6902_at_trnddc03>
"Alfredo Novoa" <anovoa_at_ncs.es> wrote in message
news:04qsq09g8s60vdepi32ajt5mlvocoh5ku2_at_4ax.com...
> On Wed, 1 Dec 2004 15:19:43 -0800, "Mikito Harakiri"
> <mikharakiri_at_iahu.com> wrote:
>
>>Suppose we have one column relation which is essentially a set of
>>intervals:
>>
>>{[1,2], [3,5], [4,7]}
>>
>>and I want to project away the interval column and calculate aggregated
>>value of SUM(1). What the result there would be? 3? 2? 7?
>
> It depends on what a sum of intervals is for you.
>
> For me it would be:
>
> {1,2} U {3,4,5} U {4,5,6,7} = {1,2,3,4,5,6,7} = [1,7]
>
How about {[1,2],[3,5], [7,7]}?
- Dan