Forum Discussion

jpuck's avatar
jpuck
Frequent Visitor
8 years ago

Cannot use regular count and distinct count values together in chart?

Pure beginner here, only been using the desktop app for a couple of days.

 

I'm trying to make a stacked bar chart with two values (on x axis) for a few users (on y axis).  Adding two values is no problem, however I want one value to be a regular count and the other to be a distinct count.  If I do this, it actually seems to not like both types of counts being in there and instead just shows two equal bars with the regular count value.  If I make both counts either regular or distinct, then it's fine, but unfortunately my data won't work like that, one HAS to be a regular count and the other HAS to be a distinct count.

 

Is this a known/expected limitation and if so, is there a way to get around this?

3 Replies

  • Anonymous's avatar
    Anonymous
    Not applicable
    8 years ago

    Hi jpuck

     

    How are you doing this? I have taken the count and distinct count of same column and it appears correctly. Make sure you have selected the count, Distinct count in the Fields property properly.

     

    Otherwise, you can create 2 measures, 1 with count(col1) and another 1 with distinctcount(col1) and use them in the chart.

     

     

     

    Thanks

    Raj

    • jpuck's avatar
      jpuck
      Frequent Visitor
      8 years ago

      Anonymous

       

      So you didn't directly solve my problem, however your question, screenshots, and general confirmation that this actually can be done made me think differently.  I believe what's happening here is that while I'm taking two different pieces of information from my dataset, one of them has a relationship back to the other that essentially makes them the same thing based on how I'm filtering.  I have no idea if that makes any sense at all to you, but THANK YOU for responding, otherwise I'm not sure I ever would have looked at this from another angle.

       

      I still don't have it figured out but I'm convinced now it's not an impossible thing or a bug, so I'll keep working on it.  Thanks!

  • v-yuta-msft's avatar
    v-yuta-msft
    Community Support
    8 years ago

    Hi jpuck,

     

    " If I make both counts either regular or distinct, then it's fine, but unfortunately my data won't work like that, one HAS to be a regular count and the other HAS to be a distinct count."

     

    <--- What's your expected result? Could you give some example? In addtion, Could you share a mockup with some dummy data for further analysis?

     

    Regards,

    Jimmy Tao