Forum Discussion
davidinternship
6 years agoNew Member
Average calculation in same column but for different values
Hello everyone, I'm trying to calculate the average of some different values in the same column. And after that, using that result to divide another value in the same column, resulting a value in po...
- Anonymous6 years ago
// Of course, your notation is incorrect. // The formula should read: // // X = [2A / (B + C)] - 1 // // In mathematics IT DOES MATTER where you put // brackets. Note that such a calculation should // be performed in Power Query, not in DAX. [New Column] = // column, not a measure var __code = T[ID Code] var __a = MAXX( filter( T, T[ID Code] = __code && T[Type] = "A" ), T[Value] ) var __b = MAXX( filter( T, T[ID Code] = __code && T[Type] = "B" ), T[Value] ) var __c = MAXX( filter( T, T[ID Code] = __code && T[Type] = "C" ), T[Value] ) return DIVIDE( 2 * __a, __b + __c ) - 1 // If you want a calculated table, // you can do this: [Calc Table] = ADDCOLUMNS( VALUES( T[ID Code] ), "X", var __code = T[ID Code] var __a = MAXX( filter( T, T[ID Code] = __code && T[Type] = "A" ), T[Value] ) var __b = MAXX( filter( T, T[ID Code] = __code && T[Type] = "B" ), T[Value] ) var __c = MAXX( filter( T, T[ID Code] = __code && T[Type] = "C" ), T[Value] ) return DIVIDE( 2 * __a, __b + __c ) - 1 )Best
D
Anonymous
6 years agoNot applicable
// Of course, your notation is incorrect.
// The formula should read:
//
// X = [2A / (B + C)] - 1
//
// In mathematics IT DOES MATTER where you put
// brackets. Note that such a calculation should
// be performed in Power Query, not in DAX.
[New Column] = // column, not a measure
var __code = T[ID Code]
var __a =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "A"
),
T[Value]
)
var __b =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "B"
),
T[Value]
)
var __c =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "C"
),
T[Value]
)
return
DIVIDE(
2 * __a,
__b + __c
) - 1
// If you want a calculated table,
// you can do this:
[Calc Table] =
ADDCOLUMNS(
VALUES( T[ID Code] ),
"X",
var __code = T[ID Code]
var __a =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "A"
),
T[Value]
)
var __b =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "B"
),
T[Value]
)
var __c =
MAXX(
filter(
T,
T[ID Code] = __code
&&
T[Type] = "C"
),
T[Value]
)
return
DIVIDE(
2 * __a,
__b + __c
) - 1
)
Best
D
davidinternship
6 years agoNew Member
Thank you for the answer. It helpde me a lot.
I just needed to change little things to adapt on my need, but your answer was fundamental.
Best wishes!