Average fo weekly distinct values

[RGL]

I need have a table with daily sales and calculate the average weekly sales for a fixed period, for L4W and for the last week.

How can I calculate the average of unique customers of each week.

For example

Week No.      DISTINTICT CUSTOMERS

Week 13                100

Week 14                  90

Week 15                120

I need the average of 100,90,120

There can be customers with cancellations or refunds, which have a negative sale amount, so I need to just count distinct customers with positive amounts.

thanks

[amitchandak]

@RGL ,

Try

averagex(Values(Table[Week No.]),[DISTINTICT CUSTOMERS])

[Anonymous · Accepted solution]

// Your customers must be stored in a dimension
// table Customer. T is your sales table, the
// fact table. There's also a Dates dimension
// that's connected to your fact table via the
// Date field. The Date table is marked as
// a Date table in the model. Fact table should
// always be hidden and the only allowed objects
// exposed in it are measures. With this star-schema
// design (which is the only correct design, by the way)
// you can do...

[Total Amount] = SUM( T[Amount] )

// This is the number of visible customers,
// not necessarily the buying customers.
[# Custs] = COUNTROWS( Customer )

// These are buying customers within those
// that are visible.
[# Buying Custs] = DISTINCTCOUNT( T[CustomerId] )
	
// Now, you want to have the number of customers
// that for the selected period in Dates have Amounts > 0.
[# Positive Cust] =
	COUNTROWS(
		FILTER(
			Customer,
			[Total Amount] > 0
		)
	)

// Now, you can create the weekly average you want.
// Bear in mind that this average reposponds to
// any filters that restrict the number of days
// in the weeks and also excludes the weeks where
// the number of positive customers is BLANK. If
// you want to include the weeks where there were
// no customers, just add 0 to [# Positive Cust]
// in the measure.
[Weekly CustNo Avg] =
	AVERAGEX(
		VALUES( Dates[Week No] ),
		[# Positive Cust]
	)

Best

D