topic Re: COUNT DISTINCT users ID, but only select their last result in DAX Commands and Tips

COUNT DISTINCT users ID, but only select their last result

NumeroENAP — Wed, 12 Aug 2020 13:22:27 GMT

Hi,

I have someting of the kind :

UserID	Name	DT_EVAL	Conformity
Markov67	Andrei Markov	2020-08-10	conform
Markov67	Andrei Markov	2020-08-08	not conform
Plekanec14	Thomas Plekanec	2020-08-10	not conform

I want my measure to return me : 2

Reason : I only want the last evaluation of every user.

Thanks

Re: COUNT DISTINCT users ID, but only select their last result

Greg_Deckler — Mon, 10 Aug 2020 18:40:29 GMT

NumeroENAP - Perhaps use MAX to get the greatest date in the table and then FILTER down to that date, use DISTINCT and COUNTROWS, so like:

Measure = COUNTROWS(DISTINCT(FILTER('Table',[Date] = MAX('Table'[Date])))

Re: COUNT DISTINCT users ID, but only select their last result

NumeroENAP — Mon, 10 Aug 2020 20:18:12 GMT

Ok, it's nice, and it's defenitely the closest I've been from the solution, but, it's not exactly what I need.

By exemple, if I need to group it by conformity :

Conformity	Distinct User
Conform	1
Not conform	2

But, with what you gave me, even if it gives me a table with what I need (raw), but I can't group it the way I want.

Am I clear enough?

Thanks!

Re: COUNT DISTINCT users ID, but only select their last result

Greg_Deckler — Tue, 11 Aug 2020 04:41:47 GMT

NumeroENAP - Wait, now I am confused, where did conform/not comform come from? Perhaps we should back-up. Please first check if your issue is a common issue listed here: https://community.powerbi.com/t5/Community-Blog/Before-You-Post-Read-This/ba-p/1116882

Also, please see this post regarding How to Get Your Question Answered Quickly: https://community.powerbi.com/t5/Community-Blog/How-to-Get-Your-Question-Answered-Quickly/ba-p/38490

The most important parts are:
1. Sample data as text, use the table tool in the editing bar
2. Expected output from sample data
3. Explanation in words of how to get from 1. to 2.

Re: COUNT DISTINCT users ID, but only select their last result

NumeroENAP — Wed, 12 Aug 2020 13:24:21 GMT

I need to have a measure that tells me exactly :

How many employees have a «conform» or «not conform» evaluation at their last evaluation ;
I also need that measure to be flexible, so I can use it in others visualisation, with other columns of the same table.

UserID	Name	DT_EVAL	Conformity
Markov67	Andrei Markov	2020-08-10	Conform
Markov67	Andrei Markov	2020-08-08	Not conform
Plekanec14	Thomas Plekanec	2020-08-10	Not conform

Thanks!

Re: COUNT DISTINCT users ID, but only select their last result

Anonymous — Wed, 12 Aug 2020 15:45:05 GMT

// You have to have a Calendar table // that's DISCONNECTED from the T table. // Now, this is a measure that for a // selected period of time from Calendar // tells you the number of 'Conform' // emps at their last evaluation BEFORE OR ON // the last day of the selected period. // I assume an employee is fully identified // by UserID and that each employee can only // have 1 entry on any specific day. This // measure does fully respond to slicers, so // be careful to correctly interpret the results. [# Conform Emps] = var __lastVisibleDate = MAX( Calendar[Date] ) var __empsWithConformAsLastEval = FILTER( VALUES( T[UserID] ), var __latestRecordForUser = CALCULATETABLE( TOPN(1, filter( T, T[DT_EVAL] <= __lastVisibleDate ), T[DT_EVAL], DESC ) ) var __conformityIsConform = NOT ISEMPTY( FILTER( __latestRecordForUser, T[Conformity] = "conform" ) ) return __conformityIsConform ) return COUNTROWS( __empsWithConformAsLastEval )

Re: COUNT DISTINCT users ID, but only select their last result

NumeroENAP — Wed, 12 Aug 2020 23:27:14 GMT

In the end, I went with a non-performing option, but it works great.

First of all, I created a calculated column :

LAST_EVAL_DT = VAR LastID = RES[ID]

RETURN

MAXX(

FILTER( ALL('RES') , 'RES'[ID] = LastID) ,

'RES'[DT_EVAL])

After that, I created a column that I could filter easily with a COUNTROWS :

QC = IF (RES[DT_EVAL] = RES[LAST_EVAL_DT] , 1 , BLANK() )

Thank you all for your efforts.