Forum Discussion
Grouped by week - determine change
- 3 years ago
Hi emildeak ,
Based on your description, I have created a simple sample:
Please try:
Week Avg = var _a = WEEKDAY(MAX('Table'[Day]),2) var _b = CALCULATE(AVERAGE('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2))) return IF(_a=7,_b) Weekly Change = var _a = WEEKDAY(MAX('Table'[Day]),2) var _b = CALCULATE(SUM('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2)))// the value of current week var _c = CALCULATE(SUM('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2)-1))// the value of previous week var _d = DIVIDE(_c,_b) return IF(_a=7,_d)Output:
Since I'm not sure what your weekly change is based on, I calculated the percentage change in their total weekly value. You can change the aggregation in the measure to get the value you want, or clarify your calculation logic to me.
Best Regards,
Jianbo Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi emildeak ,
Based on your description, I have created a simple sample:
Please try:
Week Avg =
var _a = WEEKDAY(MAX('Table'[Day]),2)
var _b = CALCULATE(AVERAGE('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2)))
return IF(_a=7,_b)
Weekly Change =
var _a = WEEKDAY(MAX('Table'[Day]),2)
var _b = CALCULATE(SUM('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2)))// the value of current week
var _c = CALCULATE(SUM('Table'[Value]),FILTER(ALL('Table'),WEEKNUM([Day],2)=WEEKNUM(MAX('Table'[Day]),2)-1))// the value of previous week
var _d = DIVIDE(_c,_b)
return IF(_a=7,_d)
Output:
Since I'm not sure what your weekly change is based on, I calculated the percentage change in their total weekly value. You can change the aggregation in the measure to get the value you want, or clarify your calculation logic to me.
Best Regards,
Jianbo Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.