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HDD328
Frequent Visitor

calculated column to find difference in value per day

Hello,

I am having trouble creating a calculated column that will take the Hrs value from the current day and subtract from the previous day for all incoming data. (example data below)

HDD328_0-1733835211199.png

 

I've tried multiple different ways, but can't seem to get the expected result. Any help is appreciated!

1 ACCEPTED SOLUTION
lkalawski
Super User
Super User

Hi @HDD328 ,

 

I suggest doing this type of calculation using a measure, because it is more optimal.

I have assumed Sum Hrs as Current Date, but this can be modified depending on the context of the task:

Cur Day Hours = 
Sum('Table'[Hrs])

 

Just calculate the Hrs value for the previous day:

Prv Day Hours = 
VAR __Date = MAX('Table'[DATE])
VAR __Unit = SELECTEDVALUE('Table'[Unit])
VAR __PrevDay = __Date - 1
VAR PreviousHrs = 
    CALCULATE(
        MAX('Table'[Hrs]),
        FILTER(
            ALL('Table'),
            'Table'[Unit] = SELECTEDVALUE('Table'[Unit]) &&
            'Table'[DATE] = MAX('Table'[DATE]) - 1
        )
    )
RETURN PreviousHrs

and then calculate the difference between today and yesterday:

Diff = 
VAR __Result = [Cur Day Hours] - [Prv Day Hours]

RETURN IF(ISBLANK([Prv Day Hours]), BLANK(), __Result)

 

The result:

lkalawski_0-1733838632794.png

PBI_SuperUser_Rank@1x.pngResident Rockstar | Former Super User
If I helped, please accept the solution and give kudos! 
Connect with me
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View solution in original post

2 REPLIES 2
BITomS
Solution Supplier
Solution Supplier

Hi @HDD328 ,

 

I agree with @lkalawski that where possible, try to use measures. But for reference, this is how you could create a calculated column:

 

CalCol =
'Table'[Hrs]
- CALCULATE (
SUM ( 'Table'[Hrs] ),
ALL ( 'Table' ),
'Table'[UNIT] = EARLIER ( 'Table'[UNIT] ),
'Table'[DATE]
= EARLIER ( 'Table'[DATE].[Date] ) - 1
)

 

Hope this helps!

lkalawski
Super User
Super User

Hi @HDD328 ,

 

I suggest doing this type of calculation using a measure, because it is more optimal.

I have assumed Sum Hrs as Current Date, but this can be modified depending on the context of the task:

Cur Day Hours = 
Sum('Table'[Hrs])

 

Just calculate the Hrs value for the previous day:

Prv Day Hours = 
VAR __Date = MAX('Table'[DATE])
VAR __Unit = SELECTEDVALUE('Table'[Unit])
VAR __PrevDay = __Date - 1
VAR PreviousHrs = 
    CALCULATE(
        MAX('Table'[Hrs]),
        FILTER(
            ALL('Table'),
            'Table'[Unit] = SELECTEDVALUE('Table'[Unit]) &&
            'Table'[DATE] = MAX('Table'[DATE]) - 1
        )
    )
RETURN PreviousHrs

and then calculate the difference between today and yesterday:

Diff = 
VAR __Result = [Cur Day Hours] - [Prv Day Hours]

RETURN IF(ISBLANK([Prv Day Hours]), BLANK(), __Result)

 

The result:

lkalawski_0-1733838632794.png

PBI_SuperUser_Rank@1x.pngResident Rockstar | Former Super User
If I helped, please accept the solution and give kudos! 
Connect with me
Linkedin

 

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