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I have the following table:
studnr/course/result
| 102 | A | 4,6 | |
| 103 | A | 6,6 | |
| 101 | A | 3,7 | |
| 105 | A | 9,1 | |
| 101 | A | 7 | |
| 102 | B | 4,4 | |
| 102 | B | 5,4 | |
| 104 | B | 4,6 | |
| 105 | B | 3,9 | |
| 105 | B | 6,1 | |
| 104 | B | 7,8 | |
| 101 | B | 3,1 | |
| 101 | B | 3,5 | |
| 101 | B | 7,6 | |
| 102 | B | 3,1 | |
| 102 | C | 3,3 | |
| 102 | C | 4,5 | |
| 102 | C | 6,3 | |
| 104 | C | 5,6 | |
| 105 | C | 5,6 | |
| 101 | D | 5,3 | |
| 101 | D | 4,6 | |
| 104 | D | 4,9 | |
| 105 | D | 3,1 | |
| 105 | D | 4,5 |
I need to calculate the average number of attempts needed to get a sufficient per course.
This is, quite simply, the following formula:
No. of Attempts / No. of Sufficients
I've done this by using a measure but I did not get the right answer since the No. of Attempts should only be counted if the student also has got a sufficient.
Hopefully this is enough information to go on.
Solved! Go to Solution.
@Anonymous
Is this the same question as this post?
https://community.powerbi.com/t5/Desktop/Apply-filter-on-measure/m-p/847800#M406924
If so, the solution should be the same.
Avg to sufficient =
VAR _sufficient = 5.5
VAR _Students =
FILTER (
SUMMARIZE(Results,Results[Student],Results[Course]),
CALCULATE(
MAX ( Results[Result] )
) > _sufficient
)
VAR _Classes = CALCULATE ( COUNTROWS ( Results ), _Students )
RETURN DIVIDE ( _Classes, COUNTROWS( _Students ) )
I have attached my working file for you to take a look at.
@Anonymous
Is this the same question as this post?
https://community.powerbi.com/t5/Desktop/Apply-filter-on-measure/m-p/847800#M406924
If so, the solution should be the same.
Avg to sufficient =
VAR _sufficient = 5.5
VAR _Students =
FILTER (
SUMMARIZE(Results,Results[Student],Results[Course]),
CALCULATE(
MAX ( Results[Result] )
) > _sufficient
)
VAR _Classes = CALCULATE ( COUNTROWS ( Results ), _Students )
RETURN DIVIDE ( _Classes, COUNTROWS( _Students ) )
I have attached my working file for you to take a look at.
Thank you for your answer, it is indeed correct.
Sorry for posting a duplicate question, I searched for it but could not find it.
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