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Lopez0090's avatar
Lopez0090
Icon for Helper III rankHelper III
4 years ago
Solved

How to display individual values in a matrix visual

The number of "Colors" is counted for each "Time". There are two types of "Color", "Blue" and "Red", and the number of each color obtained is displayed.

It's easy enough to just count the number of colors, but I want to show the percentage of Blue out of the total number as well, as shown in the image below.

 

This is a sample, but it uses this data.

 

When I set the measure to value with the percentage calculated as a test, it is also displayed for Red and Totals.
How can I display only the percentage for Blue?

 

I think this can be reproduced in a table visual by creating a measure with 4 (count columns by color + total column + percentage column for blue only) * 2 (number of types of Times) = 8.

However, this is not very realistic as we plan to increase the number of colors and the number of Times, such as 3rd.

 

If anyone knows of a better way to do this, could you please let me know?

 

 

Best regards, Lopez

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  • Samarth_18's avatar
    Samarth_18
    4 years ago

    Hi Lopez0090 ,

     

    You can create a seperate measure for "# of Blue","# of Red","% of Blue"  like below and take direct sum of Count for total

    # of Blue = CALCULATE(sum(Table[Count]),FILTER(Table,Table[Color] = "Blue"))
    # of Red = CALCULATE(sum(Table[Count]),FILTER(Table,Table[Color] = "Red"))
    % of Blue = divide(Table[# of Blue],SUM(Table[Count]),0)

     

    Output:-

     

    Thanks,

    Samarth

1 Reply

  • Samarth_18's avatar
    Samarth_18
    Icon for Community Champion rankCommunity Champion
    4 years ago

    Hi Lopez0090 ,

     

    You can create a seperate measure for "# of Blue","# of Red","% of Blue"  like below and take direct sum of Count for total

    # of Blue = CALCULATE(sum(Table[Count]),FILTER(Table,Table[Color] = "Blue"))
    # of Red = CALCULATE(sum(Table[Count]),FILTER(Table,Table[Color] = "Red"))
    % of Blue = divide(Table[# of Blue],SUM(Table[Count]),0)

     

    Output:-

     

    Thanks,

    Samarth