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Anonymous
Not applicable

Does averaging from the visualization field area apply differently than creating a measure?

I have a card in my visualization 

bamamatting_0-1668610003108.png

bamamatting_1-1668610045729.png

Is average calculating correctly or do I have to use a count of which values are 100% and which are 0% and then divide to find the average?

1 ACCEPTED SOLUTION
v-xiaosun-msft
Community Support
Community Support

Hi @Anonymous ,

 

According to your description, here is my suggestion.

If you are using a card and want to calculate the value average of the field, you can use the system's built-in average function which calculates correctly.

vxiaosunmsft_0-1668653923594.png

If you want to calculate the count average of 100% and 0%, you can use a count of which values are 100% and which are 0% and then divide to find the average.

count of 100% = CALCULATE(COUNT('Table (2)'[Column1]),FILTER('Table (2)','Table (2)'[Column1]= 1))
count100%/total = 'Table (2)'[count of 100%]/COUNT('Table (2)'[Column1])

vxiaosunmsft_1-1668654620706.png

 

Best Regards,
Community Support Team _ xiaosun

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

1 REPLY 1
v-xiaosun-msft
Community Support
Community Support

Hi @Anonymous ,

 

According to your description, here is my suggestion.

If you are using a card and want to calculate the value average of the field, you can use the system's built-in average function which calculates correctly.

vxiaosunmsft_0-1668653923594.png

If you want to calculate the count average of 100% and 0%, you can use a count of which values are 100% and which are 0% and then divide to find the average.

count of 100% = CALCULATE(COUNT('Table (2)'[Column1]),FILTER('Table (2)','Table (2)'[Column1]= 1))
count100%/total = 'Table (2)'[count of 100%]/COUNT('Table (2)'[Column1])

vxiaosunmsft_1-1668654620706.png

 

Best Regards,
Community Support Team _ xiaosun

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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