Forum Discussion
Frequency calc / Histogram
- 8 years ago
Hi David,
Delete the column "CustID" from Summary. I found that the visual is correct now. Please check the file here: https://1drv.ms/u/s!ArTqPk2pu-BkgSqND4imKbOYtOeh
Summary = SUMMARIZE ( 'Visits', 'Visits'[Site id], "Fres", COUNT ( Visits[Site id] ) )Best Regards!
Dale
Hi dapling,
Could you please mark the proper answer as solution or share the solution if it's convenient for you? That will be a big help to the others.
Best Regards!
Dale
Sorry it's taken so long for me to get back to this...
I have shared the illustration file, with the second Owner table.
https://1drv.ms/f/s!ApwUJrTj9hAldFElMq8P1GBq-KM
Thanks a lot.
- v-jiascu-msft8 years agoMicrosoft Employee
Hi dapling,
1. Update the relationships as the picture showed.
2. Create the visuals like this:
You can check it out in this file: https://1drv.ms/u/s!ArTqPk2pu-BkgSn1uv0UV2PlYlo2.
Best Regards!
Dale
- dapling8 years agoFrequent Visitor
Hi Dale, I really appreciate your help, but I'm afraid this is still not what I'm trying to achieve.
The table needs to be capable of identifying that there is a single instance (occurence) of a site (owned by PQR) which has been visited FIVE times, none visited FOUR times, 1 visited THREE times (also PQR's site), 1 site visited TWICE (owned by MNO) and 4 sites visited ONCE (2 owned by MNO and 2 by PQR).
Freq * Site Count
[5*1] + [4 *0] + [3*1] + [2*1] + [1*4] = 14.
i.e. 14 visits in total to 7 different sites, BUT split by two different owners.
The Cust id is no longer relevant for this slice.
Confusing... Argh!!! I've saved my amended file in the same shared place as above.
Thanks again, David
- v-jiascu-msft8 years agoMicrosoft Employee
Hi David,
Delete the column "CustID" from Summary. I found that the visual is correct now. Please check the file here: https://1drv.ms/u/s!ArTqPk2pu-BkgSqND4imKbOYtOeh
Summary = SUMMARIZE ( 'Visits', 'Visits'[Site id], "Fres", COUNT ( Visits[Site id] ) )Best Regards!
Dale