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Hi
I have produced a mathematically impossible pie chart:
Including the segments that do not have a data label, the total % = 101.72%
The data behind it is confidential, so I can't share it, but the amounts were produced by perfoming a distinct count of a values in one table with legend fields from another table.
Does anyone know any circumstances that might lead to this happening?
Thanks!
Solved! Go to Solution.
Yep, that can happen:
Name, Awesomeness
Scott, Very
Avi, Very
Tom, Not Very
Scott, Not Very
The Count := DISTINCTCOUNT(MyTable[Name])
Slice this by Awesomeness...
2 Very Awesome (Scott, Avi)
2 Not Very Awesome (Scott, Tom)
3 Total (Scott, Tom, Avi)
4/3 = 133%
Yep, that can happen:
Name, Awesomeness
Scott, Very
Avi, Very
Tom, Not Very
Scott, Not Very
The Count := DISTINCTCOUNT(MyTable[Name])
Slice this by Awesomeness...
2 Very Awesome (Scott, Avi)
2 Not Very Awesome (Scott, Tom)
3 Total (Scott, Tom, Avi)
4/3 = 133%
Thanks @Anonymous,
So, how do I get it to count "Scott" for both [Awsomeness] values, so that it totals 100%?
I have tried using COUNT instead of DISTINCTCOUNT and it returns the same reults.
I guess you would have to better explain what you want to see.
Say I had a chart of cars, and wanted to show Red, Blue and Green cars. Since some cars come in multiple colors...
Red: { Toyota, Honda }
Blue: { Toyota, Ford, Honda }
Green: { Honda }
2 (distinct) cars come in red, 3 in blue, and 1 in green. There are 3 total distinct cars.
It is correct to say 66% of cars come in red, 100% in blue, and 33% in green. The "grand total" would be 200%, but that is a meanless #... that I really wouldn't worry about.
Ok, so...
I want to show the following:
Toyota, Red
Toyota, Blue
Honda, Red
Honda, Blue
Honda, Green
Ford, Blue
Total cars: 6
Red: 2, 33%
Blue: 3, 50%
Green: 1, 17%
This, of course, means that I should use "count" instead of "distinct count". Now, when I tried this before, "count" gave the same result as "distinct count". This is because I had the "value" field filled from one table and the "legend" field from another.
Solution: When I have them both from the same table it gives me what I want.
Thanks for your help @Anonymous, you'd be a great maths teacher!
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