Forum Discussion
Constant Average
- Anonymous5 years ago
Hi sagadgreat ,
My sample data is this.
Month
Net Sales
MonthNum
Customer
Year
Jan
100
1
A
2020
Feb
100
2
A
2020
Mar
100
3
A
2020
Apr
58
4
A
2020
May
54
5
A
2020
Jun
56
6
A
2020
Jan
50
1
B
2020
Mar
100
3
B
2020
Apr
54
4
B
2020
May
58
5
B
2020
Jun
56
6
B
2020
Jan
100
1
A
2019
Feb
50
2
A
2019
Mar
100
3
A
2019
Apr
58
4
A
2019
May
54
5
A
2019
Jun
56
6
A
2019
1.Create a measure to get the 3 month average.
3 mo ave = DIVIDE ( CALCULATE ( SUM ( 'Table'[Net Sales] ), FILTER ( ALLEXCEPT ( 'Table', 'Table'[Customer], 'Table'[Year] ), [Month] IN { "Jan", "Feb", "Mar" } ) ), 3 )2.Create a measure to get the difference between net sales and the 3 month average.
difference = [3 mo ave]-MAX('Table'[Net Sales])3.The result is as follows. Let the filter select a single select.
You can check more details from here.
Best Regards,
Stephen Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi sagadgreat ,
My sample data is this.
Month | Net Sales | MonthNum | Customer | Year |
Jan | 100 | 1 | A | 2020 |
Feb | 100 | 2 | A | 2020 |
Mar | 100 | 3 | A | 2020 |
Apr | 58 | 4 | A | 2020 |
May | 54 | 5 | A | 2020 |
Jun | 56 | 6 | A | 2020 |
Jan | 50 | 1 | B | 2020 |
Mar | 100 | 3 | B | 2020 |
Apr | 54 | 4 | B | 2020 |
May | 58 | 5 | B | 2020 |
Jun | 56 | 6 | B | 2020 |
Jan | 100 | 1 | A | 2019 |
Feb | 50 | 2 | A | 2019 |
Mar | 100 | 3 | A | 2019 |
Apr | 58 | 4 | A | 2019 |
May | 54 | 5 | A | 2019 |
Jun | 56 | 6 | A | 2019 |
1.Create a measure to get the 3 month average.
3 mo ave =
DIVIDE (
CALCULATE (
SUM ( 'Table'[Net Sales] ),
FILTER (
ALLEXCEPT ( 'Table', 'Table'[Customer], 'Table'[Year] ),
[Month] IN { "Jan", "Feb", "Mar" }
)
),
3
)
2.Create a measure to get the difference between net sales and the 3 month average.
difference = [3 mo ave]-MAX('Table'[Net Sales])
3.The result is as follows. Let the filter select a single select.
You can check more details from here.
Best Regards,
Stephen Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
- sagadgreat5 years agoHelper I
Apologies about getting back to this so late!! Excellent solution. Thanks.