Re: [libreoffice-users] How to return cell address/reference instead of function result with 2d operations?

On 2015-11-27 13:56, Edwar Cifuentes wrote:
> On 11/27/2015 06:11 AM, Brian Barker wrote:
> > It's Boolean expression, giving the result TRUE or FALSE. It's 
> > testing the individual cells in the range A1:C2 to determine whether 
> > they are equal to the maximum value of that range. It may seem 
> > strange to be able to write such an expression for an entire range 
> > instead of for individual cells, but that is the power of array 
> > formulae.
> >
> > Brian Barker
> I see. I entered =(A1:C2 = MAX(A1:C2)) in a separate cell and it 
> generated a 2x3 true/false table corresponding to the one I created. 
> The bottom right cell returned TRUE because it corresponds to 6, the 
> highest value in my original table.
>
> =ROW(A1:C2) generates a one column table with numbered rows.
>
> Now, the whole first argument of the ADDRESS function is 
> =(MAX((A1:C2=MAX(A1:C2))*ROW(A1:C2))), which looks like "multiplying" 
> a table by another table. What does it really mean? does it kind of 
> superimpose the two tables and return the highest value? does it 
> multiply the highest value from table 1 with the highest from table 2? 
> or what does it exactly do?

Edwar:

My apologies to all for not analyzing my solution to this issue 
properly. I use microsoft products at work, so in my continual harried 
state there, I use google to find quick solutions for people that need 
answers: Never finding time to workout the solutions myself.  In this 
case, the google groups solution was to match specific number to a 
matrix of data.  It was clearly stated in this answer that it only 
worked with no duplicated data.  I modified this solution to find the 
MAX value in the data matrix, instead of a known target value.  I played 
around with the numbers, introducing duplicates, but failed to find the 
combinations where my statement "If there are duplicate values it shows 
the last location" was not true.

After Brian's answer proved my assumption wrong, I set up a larger data 
matrix using the "RANDBETWEEN" function that would introduce duplicates 
in different locations with the matrixafter a re-calculation. I was not 
questioning Brian's answer, I just want to examine his answer more 
closely by setting up a true/false table similar to Brian's.


sorry for the confusion.

TomW


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