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utkarshaggarwal's avatar
utkarshaggarwal
New Member
5 years ago
Solved

Client vs Peer Comparison

Hi,

 

I am new to Power BI and have the following use case with me:

 

  • I have 2 data sources (in Excel)
    • Source A has the data for a client for n number of criteria (Industry, Employees count, Revenue, etc.) in columns.
    • Source B has the peer data (100+ rows) for the same criteria, 1 peer per row.

I want to combine the 2 data sources in the following way:

 

  • If I create a bar chart for any criteria (eg. Average Revenue) then I should be able to compare the client with the peer in the same chart.
  • I should be able to apply filters on the peer's data in such a way that the Client's bar doesn't change and only the Peer's bar changes if I use a slicer in the dashboard.

 

Hope the use-case makes sense.  Highly appreciate any help on this.

  • utkarshaggarwal , Create a common dimension, which has the client name from A and Peer name from B

    called peer group, If need add a new column as client name in source A

     

    distinct(union(distinct(source[Client_name]), distinct(source[peer_name])))

     

    Now using this you can analyze them together

     

    if you want to select peer using a slicer, then you might need independent tables.

    refer date example

    How to use two Date/Period slicers :https://www.youtube.com/watch?v=WSeZr_-MiTg

3 Replies

  • utkarshaggarwal , Create a common dimension, which has the client name from A and Peer name from B

    called peer group, If need add a new column as client name in source A

     

    distinct(union(distinct(source[Client_name]), distinct(source[peer_name])))

     

    Now using this you can analyze them together

     

    if you want to select peer using a slicer, then you might need independent tables.

    refer date example

    How to use two Date/Period slicers :https://www.youtube.com/watch?v=WSeZr_-MiTg

  • v-xulin-mstf's avatar
    v-xulin-mstf
    Community Support

    Hi utkarshaggarwal

     

    The two tables need to have common columns and create relationships that can then achieve your expected output as:

     

    If you still have some question, please don't hesitate to let me known.‌‌

     

    Best Regards,

    Link

     

    Is that the answer you're looking for? If this post helps, then please consider Accept it as the solution. Really appreciate!

     

  • v-xulin-mstf's avatar
    v-xulin-mstf
    Community Support

    Hi utkarshaggarwal

     

    Is your issue solved?

    If the issue has been solved, please adopt the solution to help others.

    If you still have some question, please don't hesitate to let me known.‌‌

    😉

     

    Best Regards,

    Link

     

    Is that the answer you're looking for? If this post helps, then please consider Accept it as the solution. Really appreciate!