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Current column is [Days to Close] new custom column is Avg. Days Worked
I would the like formula to say if [days to close] is greater than or equal to 181 then 90 and if [days to close] is less than or equal to 180 then divide the days to close by 2.
the first part of my formula works but the second part isn't and I have spent too many hours trying to rework the formula.
=if [Days to Close]>=181 then 90 else if [Days to Close]<=180 then Value.Divide ([Days to Close], 2)
Solved! Go to Solution.
Create a new column using the following DAX
create a new column as
Hi @AlyssaP17 ,
In M Code
Custom Column=if [Days to Close]>=181 then 90 else if [Days to Close]<=180 then [Days to Close]/ 2
In DAX (Calculated Column)
Column =
SWITCH(
TRUE(),
Table[Days to Close] >= 181 , 90,
DIVIDE(Table[Days to CLose],2)
Regards,
Harsh Nathani
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@harshnathani Thank you, I am using M code. My formula changed and this is what I have entered. I am still getting the "Token Else expected" error.
custom column=if[Days to Close]>501 then [Days to Close]/4 else if [Days to Close]<=500 then [Days to Close]/3 else if [Days to Close]<180 then [Days to Close]/2
Do you think this would be easier in DAX?
Hi @AlyssaP17 .
Use this for M Code.
if [Days to Close]<180 then [Days to Close]/2 else if [Days to Close]<=500 then [Days to Close]/3 else [Days to Close]/4
Additionally, you can use add contional columns
To nest an if within another if we use the same if – then – else methodology, as shown below
= if [thing to to test #1] = "something else" then [do this if true] else if [thing to to test #2] = "something else" then [do this if true] else [do this if false]
https://exceloffthegrid.com/power-query-if-statements-for-conditional-logic/
Regards,
Harsh Nathani
Did I answer your question? Mark my post as a solution! Appreciate with a Kudos!! (Click the Thumbs Up Button)
Create a new column using the following DAX
let's consider your column name is days_close.
You can try this DAX in your calculated columns using Divide function,
=if([days_close]>=181, 90, if([days_close]<=180, Divide([days_close],2),0))
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