Skip to main content
cancel
Showing results for 
Search instead for 
Did you mean: 

Register now to learn Fabric in free live sessions led by the best Microsoft experts. From Apr 16 to May 9, in English and Spanish.

Reply
Jeremy19
Helper III
Helper III

Count number of identical values

Hello,

My data :

 

Player  id

A          1

A          2

B          1

C          1

C          3

 

I want to have the number of id which is identical to all the players which are selected by a filter. For example if I select A, the value is 2, if I also select B the value is 1 and if I select all three players the value is only 1 because there is only the id 1 which is common.

 

I try to count with a measure the number of common IDs between all the players but I cannot find the right formula.

 

Thanks

1 ACCEPTED SOLUTION
v-yalanwu-msft
Community Support
Community Support

Hi, @Jeremy19 ;

You could try it.

count = 
var _playnum=CALCULATE(DISTINCTCOUNT('Table'[Player]),ALLSELECTED('Table'))
var _countall=SUMX( SUMMARIZE('Table',[Player],[id],"1", IF(_playnum= CALCULATE(COUNT('Table'[id]),FILTER(ALLSELECTED('Table'),[id]=MAX('Table'[id]))),1,0)),[1])
return
DIVIDE(_countall ,_playnum)

The final output is shown below:

vyalanwumsft_0-1639535284883.pngvyalanwumsft_1-1639535290920.png

Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

View solution in original post

3 REPLIES 3
v-yalanwu-msft
Community Support
Community Support

Hi, @Jeremy19 ;

You could try it.

count = 
var _playnum=CALCULATE(DISTINCTCOUNT('Table'[Player]),ALLSELECTED('Table'))
var _countall=SUMX( SUMMARIZE('Table',[Player],[id],"1", IF(_playnum= CALCULATE(COUNT('Table'[id]),FILTER(ALLSELECTED('Table'),[id]=MAX('Table'[id]))),1,0)),[1])
return
DIVIDE(_countall ,_playnum)

The final output is shown below:

vyalanwumsft_0-1639535284883.pngvyalanwumsft_1-1639535290920.png

Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

Thingsclump
Resolver V
Resolver V

Try this. @Jeremy19 

 

 

Count of common IDs = CALCULATE(COUNTROWS(FILTER(SUMMARIZE('TABLE','TABLE'[ID],"duplicates',CALCULATE(COUNTROWS('TABLE'))),[duplicates]>1)))

 

 

thingsclump

www.thingsclump.com 

Vera_33
Resident Rockstar
Resident Rockstar

Hi @Jeremy19 

 

try it, modify the Table name and Column name accordingly

count_num = 
VAR CurPlayer =VALUES('Table1'[Player])
VAR CurCount=COUNTROWS(CurPlayer)
VAR TI=FILTER(Table1,Table1[Player] IN CurPlayer)
VAR T2=GROUPBY(TI,Table1[id],"Players",COUNTAX(CURRENTGROUP(),[Player]))
RETURN
COUNTROWS(FILTER(T2,[Players]=CurCount))

Helpful resources

Announcements
Microsoft Fabric Learn Together

Microsoft Fabric Learn Together

Covering the world! 9:00-10:30 AM Sydney, 4:00-5:30 PM CET (Paris/Berlin), 7:00-8:30 PM Mexico City

PBI_APRIL_CAROUSEL1

Power BI Monthly Update - April 2024

Check out the April 2024 Power BI update to learn about new features.

April Fabric Community Update

Fabric Community Update - April 2024

Find out what's new and trending in the Fabric Community.