Hello Joel,
Thank you for your swift reply,
I realise that the Average function does that, my problem is that i won't
know when one month will end and the other will start. ie the total columns
for april might be more or less than the number of days in april, and may
will follow straight after. My problem stays, how do I count the april
columns?
A thought I had was to count with a macro the number of columns for any one
month and derive the address of the first and last cell, subsequently feeding
it into the average function. Question is, how would I get about counting the
size of the range, and its starting and ending cell?
Janos
"Joel" wrote:
> The average function ingores empty cells. Therefroe, if you average more
> columns it doesn't change the calculation. Pick the larrgest number of
> column you will need and average all the columns even if you don't use them.
>
> =average(a5:dd5)
>
>
> "Janos" wrote:
>
> > Hello,
> >
> > I don't know if this is reallya difficult question, but its got me stumped.
> > What I am trying to achieve is to calculate a monthly average for individual
> > users. There is a table that looks like this:
> >
> > User 23/04/2007 24/04/2007
> > Mike 41% 52%
> > Micheal 39% 45%
> > Jason 58% 55%
> >
> >
> > The first column contains the user in question, and the subsequent columns
> > the data for each day. The result is that all the data for one user is in one
> > row. The only (big) problem is that new columns of data will be added on a
> > daily, or quasi daily basis. This implies that i don't know how many columns
> > would be added for one particular month (so, no predifined spacings for the
> > calculations). The averages per user need to be calculated and formatted in
> > the following format:
> >
> > User April May
> > Mike 40% 45%
> > Micheal 30% 50%
> > Jason 50% 40%
> >
> > I couldn't for the life of me think of a way to do this with a worksheet
> > function, and other than a hunch of needing nested loops or doing one loop
> > per month per user, I'm stumped.
> >
> > Any help would be greatly appreciated!
> >
> > Thanks,
> >
> > Janos
> >
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