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Reply
KoenStam
Regular Visitor

Flexible indexed values

(A) I have a relatively simple table with 2 column, one for the date and one containing daily market prices (of aluminium).

(B) I made a graph showing the daily prices over the time period covered by my data.

(C) Subsequently, I made a slicer for the date, making it possible to showing a more narrow time period. Besides, I made 2 measures, one showing the first date of the selected time period and one for the last date of my selected time period:

 

Firstdate = FIRSTDATE('Primary Aluminium'[Date])

Lastdate = LASTDATE('Primary Aluminium'[Date])

 

(D) I made 2 measures showing the related first and last price of the selected time period, using a visual level filter for the date.

 

Question> Now, what I want is to show on the graph is not the actual price per day, but the indexed price. Meaning that the starting price is 1 and the prices of the following days are shown relative to the price of the first day (e.g. instead of 2000 and 2200, it will show 1 and 1.1 for the first two days). The difficulty is that when I change the selected time period, the starting date (and therefore starting value changes), therefore the prices of the following days have to be divided by a different starting value. How to do this?

 

image.png

 

image.png

 

 

 

 

 

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

@KoenStam,

 

You may add a measure as shown below.

Measure =
DIVIDE (
    SUM ( 'Primary Aluminium'[Close] ),
    CALCULATE (
        SUM ( 'Primary Aluminium'[Close] ),
        FIRSTDATE ( ALLSELECTED ( 'Primary Aluminium'[Date] ) )
    )
)
Community Support Team _ Sam Zha
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

2 REPLIES 2
v-chuncz-msft
Community Support
Community Support

@KoenStam,

 

You may add a measure as shown below.

Measure =
DIVIDE (
    SUM ( 'Primary Aluminium'[Close] ),
    CALCULATE (
        SUM ( 'Primary Aluminium'[Close] ),
        FIRSTDATE ( ALLSELECTED ( 'Primary Aluminium'[Date] ) )
    )
)
Community Support Team _ Sam Zha
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

Many thanks, your reply solved my question!

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