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Hi there,
I have a dataset like the table below (larger though):
ID | App | Business function | Name |
231 | BroadDisplay | Kok; Bok; Tok | Puk; Put |
245 | SmallDisplay | Stone; Bone; Home | Stuck; Cut; Duck |
Now I made subfields for 'Business function' and 'Name' like this:
SubField("Business function", ';') as Tags,
SubField("Name", ';') as Tags2,
The following problem now occurs when I want to count the Business function or Name:
ID | App | Tags (Business function) | Tags2 (Name) |
231 | BroadDisplay | Kok | Puk |
231 | BroadDisplay | Bok | Puk |
231 | BroadDisplay | Tok | Puk |
231 | BroadDisplay | Kok | Put |
231 | BroadDisplay | Bok | Put |
231 | BroadDisplay | Tok | Put |
245 | SmallDisplay | Stone | Stuck |
245 | SmallDisplay | Bone | Stuck |
245 | SmallDisplay | Home | Stuck |
245 | SmallDisplay | Stone | Cut |
245 | SmallDisplay | Bone | Cut |
245 | SmallDisplay | Home | Cut |
245 | SmallDisplay | Stone | Duck |
245 | SmallDisplay | Bone | Duck |
245 | SmallDisplay | Home | Duck |
This would mean that even with the set analysis below my business function Tok, Kok, Bok is still counted twice. And Stone, Bone, Home is counted tree times if I am correct:
count(aggr(sum(DISTINCT [Tags]), [ID]))
OR
count(aggr(sum(DISTINCT [Tags]), [Tags2]))
I somehow want to combine the DISTINCT ID and Tags2 but I don't know if that is possible. Does anyone have a clue what I am talking about and would show me the solution (A)?
Would be great help if anyone knows how!!!
Tnx
can you please send the screenshot of what you want the end result look like?
May be this:
Sum(Aggr(Count(DISTINCT Tags), Tags2, ID))
First of all tnx for the quick replies . I want the following but I just want this to happen in the set analysis:
ID | App | Tags (Business function) | Tags2 (Name) |
231 | BroadDisplay | Kok | Puk; Put |
231 | BroadDisplay | Bok | Puk; Put |
231 | BroadDisplay | Tok | Puk; Put |
245 | SmallDisplay | Stone | Stuck; Cut; Duck |
245 | SmallDisplay | Bone | Stuck; Cut; Duck |
245 | SmallDisplay | Home | Stuck; Cut; Duck |
Sunny T I will try yours in a sec, it might be the one I am looking for.