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nicolett_yuri

Help with Set Analysis - MAX

I have a table with MONTH, WEEK, DAY, PRODUCT and VALUE

I created a graph with dimension MONTH and PRODUCT, in expresson I need the VALUE of the biggest day of the month. (This value will be used to perform other account)

Problems encountered: When do the set analysis to catch the biggest day of the month, he does not respect the month dimension. Only respect when I selected.

Expression used: SUM({$<DATA = {$(=max(DATA))} >} VALOR)

I've tried to use AGGR, but had no success.

Follow the example application.

Thank you!

1 Solution

Accepted Solutions
nicolett_yuri
Author

I solved the problem without using Set Analysis:

Sum( If( DATA = Aggr( Nodistinct Max( DATA ), MES, PRODUTO ), VALOR ) )

But if someone solve otherwise post here!

View solution in original post

4 Replies
nicolett_yuri
Author

I solved the problem without using Set Analysis:

Sum( If( DATA = Aggr( Nodistinct Max( DATA ), MES, PRODUTO ), VALOR ) )

But if someone solve otherwise post here!

swuehl
MVP
MVP

It seems to me that your maximum DATA per MES and PRODUTO might be static (not depending on selections), so you could add a flag to your data model (add these lines to your script):

LEFT JOIN (Calendario) LOAD

    MES,

    PRODUTO,

    max(DATA) as DATA,

    1 as MAXDATAFLAG

Resident Calendario group by MES,PRODUTO;

and use this as expression in your chart with dimensions MES and PRODUTO:

=sum({<MAXDATAFLAG={1}>} VALOR)

danieloberbilli
Specialist II
Specialist II

I think it's not possible to use Set Analysis in the way how the data looks as there is no unique connection between DATA and MES...and I think thats the problem here. Guess you have to script it differently (e.g. sript calendar-table and data-table separately) - then set analysis should be no problem.

Cheers

Daniel

nicolett_yuri
Author

That's a good idea, but I'll not solve my problem in script because I'll need another solutions to day, week, etc.