I know that I like to kick-start my brain by getting 'into' a list problem/challenge
in the (London) mornings,
but today I'm completely beaten. Thanks for the broad explanation, which is probably
quite meaningful, but I'm
not into breeding (could that statement be misunderstood?) and so don't understand the
terminology, but let's
Can I request some db-speak please? You already have a tbl (or more). So let's start
by asking for the table
Do you already have some SQL as a 'first attempt'? Could you also post that, together
with a note about its
short-comings/what needs to be added? That would also assist (my, feeble)
Usually in this sort of problem, the trick is to work out how to order/group the data,
and then bearing in mind
the various table-relationships, organise the join(s) and apply the requisite
statistical functions. (that's the
(if you haven't got anything started yet) Along with the definitions, how about
starting a SELECT by listing the
data you want to see, filling in the FROM clause, and then skipping to the WHERE
clause and putting in the last
or last-two criteria, eg the year/date-range to be used in the analysis - well do what
you can/makes sense to
you, so far.
That might be enough to 'inspire' a solution - or start us on the way...
> Hi All,
> I have a problem working out a suitable algorithm either in PHP or MySQL.
> Basically I have a DB that keeps track of breeding records. Each record
> has a paired data, and a split-up date.
> I need to generate some statistics to work out average numbers of pairs
> per month, averaged on a daily basis, for a given start and stop date,
> typically a year or year-to-date.
> All the algorithms I can think of are messy, where I have to loop through
> all the breeding records for every day of the year, and count how many
> pairs are breeding by seeing if the date is between the start and stop
> dates, and then average that on a monthly basis. I can't see
> that scaling very well, as there might be several hundred breeding records
> for a given year, multiplied by 365 days.
> Has anyone any hints/pointers for an efficient way to do this?
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