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cmatcovich
New Contributor

accmulation in multidimensional pivot table

Hi,

I need to build a pivot table like the following one. I tried with RangeSum(Above...)) but it doesn't work.

Can somebody help me? Thanks

FiscalYearMonthWeekDateRevenuesYear Accumulated RevenuesMonth Accumulated Revenues
2012Jun101/06/2012101010
02/06/2012112121
03/06/2012123333
204/06/2012134646
05/06/2012146060
06/06/2012157575
07/06/2012169191
08/06/201217108108
09/06/201218126126
10/06/201219145145
311/06/201220165165
12/06/201221186186
13/06/201222208208
14/06/201223231231
15/06/201224255255
16/06/201225280280
17/06/201226306306
418/06/201227333333
19/06/201228361361
20/06/201229390390
21/06/201230420420
22/06/201231451451
23/06/201232483483
24/06/201233516516
525/06/201234550550
26/06/201235585585
27/06/201236621621
28/06/201237658658
29/06/201238696696
30/06/201239735735
Jul501/07/20124077540
602/07/20124181681
03/07/201242858123
04/07/201243901166
05/07/201244945210
06/07/201245990255
07/07/2012461036301
08/07/2012471083348
709/07/2012481131396
10/07/2012491180445
11/07/2012501230495
12/07/2012511281546
13/07/2012521333598
14/07/2012531386651
15/07/2012541440705
816/07/2012551495760
17/07/2012561551816
18/07/2012571608873
19/07/2012581666931
20/07/2012591725990
21/07/20126017851050
22/07/20126118461111
923/07/20126219081173
24/07/20126319711236
25/07/20126420351300
26/07/20126521001365
27/07/20126621661431
28/07/20126722331498
29/07/20126823011566
1030/07/20126923701635
31/07/20127024401705
1 Solution

Accepted Solutions
whiteline
Honored Contributor II

Re: accmulation in multidimensional pivot table

use this expression for accumulation:

=aggr(rangesum(above(sum(Revenue), 0, rowno())), _Date)

2 Replies
whiteline
Honored Contributor II

Re: accmulation in multidimensional pivot table

use this expression for accumulation:

=aggr(rangesum(above(sum(Revenue), 0, rowno())), _Date)

cmatcovich
New Contributor

Re: accmulation in multidimensional pivot table

It works!!!!

I need to study deeply how "aggr" function works.

Thanks a lot!

Corrado

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