Calculate if a column contains some text for each occurrence of a unique value

[tomekm]

Hello,

I'm trying to come up with a formula for the following scenario. Any help would be greatly appreciated.

For a given Primary Key in column D ("R_PK1"), IF the Primary Key is present EXACTLY 2 times in the table, AND if column B equals "BBB" in any of the 2 rows for that Primary Key, then give me "some text" else "some other text".

Below is the sample data. Thank you for your help!

Code:	Category:	Type:	R_PK1:	Output column:
abc	AAA	Shared	1111111	some text
xyz	BBB	Shared	1111111	some text
cdb	AAA	Shared	2222222	some other text
wsd	BBB	Shared	2222222	some other text
gjk	FFF	Shared	2222222	some other text

[Ashish_Mathur · Accepted solution]

Hi,

This calculated column formula works

Column = if(and(CALCULATE(COUNTROWS(Data),FILTER(Data,Data[R_PK1]=EARLIER(Data[R_PK1])))=2,CALCULATE(COUNTROWS(Data),FILTER(Data,Data[R_PK1]=EARLIER(Data[R_PK1])&&Data[Category]="BBB"))>=1),"Some Text","Some other text")

Hope this helps.

[Anonymous · Accepted solution]

Hi  @tomekm ,

Here are the steps you can follow：

1. Create calculated column.

Output column =
var _count=COUNTX(FILTER(ALL('Table'),'Table'[R_PK1:]=EARLIER('Table'[R_PK1:])),[R_PK1:])
var _category=SELECTCOLUMNS(FILTER(ALL('Table'),[R_PK1:]=EARLIER('Table'[R_PK1:])),"1",'Table'[Category:])
return
IF(
    _count=2&&"BBB" in _category,"Some Text","Some other text")

2. Result:

Best Regards,

Liu Yang

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly

[Ashish_Mathur · Accepted solution]

Hi,

This calculated column formula works

Column = if(and(CALCULATE(COUNTROWS(Data),FILTER(Data,Data[R_PK1]=EARLIER(Data[R_PK1])))=2,CALCULATE(COUNTROWS(Data),FILTER(Data,Data[R_PK1]=EARLIER(Data[R_PK1])&&Data[Category]="BBB"))>=1),"Some Text","Some other text")

Hope this helps.

[tomekm]

Thank you!

[Ashish_Mathur]

You are welcome.

[PijushRoy]

Hi @tomekm 

Please try this, (change table name and col name as per quirement)
= VAR _count2 = DISTINCTCOUNT('Table'[R_PK1])
RESULT
SWITCH(
TRUE(),
_count2 = 2 && 'Table'[Category] = "BBB", "Some Test",
"Some Other Test")

If not solved, please keep posted (paste from excel, your data is broken when I try to use)

If solve your requirement, please mark this answer as SOLUTION
If this comment helps you, please LIKE this comment/Kudos

[tomekm]

Code:	Category:	Type:	R_PK1:	Output column:
abc	AAA	Shared	1111111	some text
xyz	BBB	Shared	1111111	some text
cdb	AAA	Shared	2222222	some other text
wsd	CCC	Shared	2222222	some other text
gjk	FFF	Shared	2222222	some other text

[tomekm]

Thank you for the quick response. Unfortunately this formula didn't work for me. I'm pasting the data in tabular format here:

Code:	Category:	Type:	R_PK1:	Output column:
abc	AAA	Shared	1111111	some text
xyz	BBB	Shared	1111111	some text
cdb	AAA	Shared	2222222	some other text
wsd	CCC	Shared	2222222	some other text
gjk	FFF	Shared	2222222	some other text