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Hello, good afternoon!
I'm making a progress board in training courses and I'm glued together so I can show the segmented progress.
Each employee must take 8 courses, those 8 courses are divided into three classifications: HSE (4 courses), Cybersecurity (2 courses) and Corporate (2 courses).
The report shows me for each row if the collaborator did x course or not, I have it as "Done" or "Pending" and also a binary column: Done = 1 and Pending = 0.
For the advancement for the HSE case, I need the % of collaborators who have already taken the 4 courses, knowing that if they did the 4 courses it would be 1 completed and if someone is missing at least 1 course, it would be 0.
I managed to make measurements that show me the following table:
Where:
Solved! Go to Solution.
Please try this measure:
HSE Advance =
VAR _vtable =
SUMMARIZE (
ALLSELECTED ( 'Consolidated Collaborators' ),
'Consolidated Collaborators'[Colaborador],
"_HSE2", [HSE2]
)
RETURN
COUNTROWS ( FILTER ( _vtable, [_HSE2] = 1 ) )
/ COUNT ( 'Consolidated Collaborators'[Email] )
The COUNTROWS ( FILTER ( _vtable, [_HSE2] = 1 ) ) returns 3(the quantity of that HSE2 = 1)
The result is as follow:
Best Regards
Zhengdong Xu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi,
Try this pattern
Measure = divide(countrows(filter(values(Data[Collaborador[),[hse]>3)),count('Consolidated Collaborators'[Email]))
Hi,
Try this pattern
Measure = divide(countrows(filter(values(Data[Collaborador[),[hse]>3)),count('Consolidated Collaborators'[Email]))
Please try this measure:
HSE Advance =
VAR _vtable =
SUMMARIZE (
ALLSELECTED ( 'Consolidated Collaborators' ),
'Consolidated Collaborators'[Colaborador],
"_HSE2", [HSE2]
)
RETURN
COUNTROWS ( FILTER ( _vtable, [_HSE2] = 1 ) )
/ COUNT ( 'Consolidated Collaborators'[Email] )
The COUNTROWS ( FILTER ( _vtable, [_HSE2] = 1 ) ) returns 3(the quantity of that HSE2 = 1)
The result is as follow:
Best Regards
Zhengdong Xu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
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