On 31/05/2013 at 22:10, Johnny Rosenberg <[email protected]> wrote:
> I obviously misunderstood the whole concept, so how is it supposed to be
> done?
Like that:
{=MAX((C1:C10000=P3)*(YEAR(D1:D10000)=YEAR(TODAY()))*E1:E10000)}
(C1:C10000=P3) will act like IF statement. It will return 10000 elements array
containing 1 (if cell matches P3) or 0 (otherwise).
The same goes to second statement, which compares years with current year.
We do not do anything to last column values.
This formula will evaluate to multiplication of arrays containing 0, 1 and
original numbers. If both conditions are true, it will not change value in E
column. If at least one condition is not true, it will effectively zero entire
row.
Then these numbers (original E values and zeroes) are fed to MAX function.
BUT array formulas on large datasets are far from being efficient. Using
database function might be better idea (basically, database functions are as
fast as array formulas or faster than them).
The basics of database functions are:
- your range of data must be structured; first row is considered header (there
should be text briefly describing content of column)
- you must repeat your header in range containing conditions. Each column must
be present at least once.
- in condition range, cells in one row represents conjunction
- in condition range, each row represents alternative
The tricky part here is, your date column contains not only year, but also
month and day. So you can't really put "2013" into criteria range and call it
a day. You must search for dates between 1.1.2013 and 31.12.2013.
If your date column contains cells with date type, then they are internally
represented by number of days since 30 December 1899. So we can get around the
issue with DATEVALUE.
Explaining what to put where would take some time, so I put spreadsheet
online. You can download it here:
<http://minio.komunikatory.pl/pliki/array-and-dmax.ods>
--
Best regards
Mirosław Zalewski
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