Starting December 3, join live sessions with database experts and the Microsoft product team to learn just how easy it is to get started
Learn moreGet certified in Microsoft Fabric—for free! For a limited time, get a free DP-600 exam voucher to use by the end of 2024. Register now
I have a table that has Order, date, Sales
Order # | Date | Sales | ||
Order #1 | 20 Aug 2019 | 100 | ||
Order #2 | 20 Aug 2019 | 200 | ||
Order #3 | 19 Aug 2019 | 100 | ||
Order #4 | 19 Aug 2019 | 100 | ||
Order #5 | 19 Aug 2019 | 200 | ||
Order #6 | 18 Aug 2019 | 200 |
First I have to find the average sales per day . So for 20 AUG 2019 average would be 150, 19 AUG 2019 = 133 and for 18 AUG average would be 200.
Then I have to find highest sales against the average
20 AUG 2019 ( 200 - 150) = 50
19 AUG 2019 (200 - 133) = 67
18 AUG 2019 ( 200- 200) = 0
So the output would be 19 AUG 2019. Many dates can have 67 so I want all those dates as output.
Solved! Go to Solution.
Hi @Ethanhunt123 ,
Create these measures
Average Sale =
AVERAGE('Table'[Sales])
MAXX Sale = MAXX(FILTER(ALL('Table'),'Table'[Date] = MAX('Table'[Date])),'Table'[Sales])
Highest Sale against Average = [MAXX Sale] - [Average Sale]
Rank Highest Sales = RANKX(ALL('Table'[Date]), [Highest Sale against Average])
Create a Visual Filter where Rank =1
Regards,
Harsh Nathani
Did I answer your question? Mark my post as a solution! Appreciate with a Kudos!! (Click the Thumbs Up Button)
you can try to create a new table to return the date value.
Table 2 =
VAR tbl=SUMMARIZE('Table','Table'[Date],"diff",max('Table'[Sales])-AVERAGE('Table'[Sales]))
VAR tbl2=ADDCOLUMNS(tbl,"rank",RANKX(tbl,[diff],,DESC))
return SELECTCOLUMNS(FILTER(tbl2,[rank]=1),"date",'Table'[Date])
Proud to be a Super User!
@Ethanhunt123 , Plot this measure with date
countx(filter(summarize(Table,table[Date], "_1", Rankx(all(Table), calculate( max(Table[Sales]) - avergage(Table[Sales])),,desn,skip)),[_1]=1),[Date])
Hi @Ethanhunt123 ,
Create these measures
Average Sale =
AVERAGE('Table'[Sales])
MAXX Sale = MAXX(FILTER(ALL('Table'),'Table'[Date] = MAX('Table'[Date])),'Table'[Sales])
Highest Sale against Average = [MAXX Sale] - [Average Sale]
Rank Highest Sales = RANKX(ALL('Table'[Date]), [Highest Sale against Average])
Create a Visual Filter where Rank =1
Regards,
Harsh Nathani
Did I answer your question? Mark my post as a solution! Appreciate with a Kudos!! (Click the Thumbs Up Button)
Starting December 3, join live sessions with database experts and the Fabric product team to learn just how easy it is to get started.
March 31 - April 2, 2025, in Las Vegas, Nevada. Use code MSCUST for a $150 discount! Early Bird pricing ends December 9th.
User | Count |
---|---|
94 | |
84 | |
83 | |
72 | |
49 |
User | Count |
---|---|
143 | |
130 | |
108 | |
64 | |
55 |