Forum Discussion
Calculating Continuation percentages
Hi,
I have a table with student registries from different years, i already have the counts by the ID column, but I want to calculate the percentage of students from each studies in each academic years that continue their studies in the next year.
Thanks in advance,
- Anonymous2 years ago
Hi ZakariAK ,
I create a table as you mentioned.
Then I create a calculated column named Percentage.
Percentage = VAR _Total = COUNT ( Students[ID] ) VAR _Continue = CALCULATE ( COUNT ( Students[ID] ), FILTER ( ALL ( Students ), 'Students'[Study Area] = EARLIER ( Students[Study Area] ) && 'Students'[Academic Year] = EARLIER ( Students[Academic Year] ) ) ) RETURN DIVIDE ( _Continue, _Total, 0 )Best Regards
Yilong Zhou
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 Replies
- bhanu_gautamSuper User
ZakariAK ,You can create a measure for continuing students
Total Students = COUNT('YourTable'[Student ID])
Continuing Students =
CALCULATE(
COUNT('YourTable'[Student ID]),
FILTER(
ALL('YourTable'),
'YourTable'[Academic Year] = EARLIER('YourTable'[Academic Year]) + 1
&& 'YourTable'[Study] = EARLIER('YourTable'[Study])
)
)
Create a new measure to calculate the percentage of students who continue their studies:
Percentage Continuing = DIVIDE([Continuing Students], [Total Students]) * 100This will display the academic year, study, total number of students, number of students continuing their studies in the next year, and the percentage of students continuing for each combination of academic year and study
- AnonymousNot applicable
Hi ZakariAK ,
I create a table as you mentioned.
Then I create a calculated column named Percentage.
Percentage = VAR _Total = COUNT ( Students[ID] ) VAR _Continue = CALCULATE ( COUNT ( Students[ID] ), FILTER ( ALL ( Students ), 'Students'[Study Area] = EARLIER ( Students[Study Area] ) && 'Students'[Academic Year] = EARLIER ( Students[Academic Year] ) ) ) RETURN DIVIDE ( _Continue, _Total, 0 )Best Regards
Yilong Zhou
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.